ERRATA BELT CONVEYORS FOR BULK MATERIALS, 7th edition. As of February 1, 2015

Size: px
Start display at page:

Download "ERRATA BELT CONVEYORS FOR BULK MATERIALS, 7th edition. As of February 1, 2015"

Transcription

1 Conveyor Equipment Manufacturers Association 567 Strand Ct., Suite, Naples, Florida Tel: (39) Fax: (39) Web Site: ERRATA BELT CONVEYORS FOR BULK MATERIALS, 7th edition As of February 1, 015 New Errata Items are listed below in 'Red'. The pages following the listed Errata will reflect the 'error' in red on the left hand side and then the corrected page will be on the right hand side. CHAPTER 4 Page 69 Move close parentheses in equation 4.15 Change of equation reference in A meaning (Equation 4.1 should read Equation 4.5 ) Page 7 Change of figure reference in figure 4.1 (Figure 4.4 should read Figure 4.11 or 4.13 ) Page 74 Change of equation reference in equation 4.5 (Equation 4.6 should read Equation 4.15 ) Change in elements of equation 4.6 (b c should read w s in denominator) Change in elements of equation 4.8 as follow: Page 75 Change of figure reference in figure 4.9 (Figure 4.5 should read Figure 4.3 or Figure 4.1 with A=A s ) After value of A s, add w s = Change in elements of d s formula as follow: Page 76 CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED Lacks a Square function (It should read ) in equation 4.3 Page 78 Change of equation reference in figure 4.37 (Figure 4.8 should read Equation 4.11 and 4.13 ) Remove one of sub w s (should read ) in the formula of A f. THE VOICE OF THE CONVEYOR INDUSTRY OF THE AMERICAS

2 Page 80 - Change table reference in figure 4.38 (Table 4.44 should read Table 4.43 ) CHAPTER 5 Page Change in figure (It should read K3A 500 (rpm)/n (rpm)) CHAPTER 6 Page Change in elements of formula in figure 6.15 as follow: Page Change Css = x m ss to x μ ss in nomenclature table. Page Change Ris to Rris in equation 6.5 Page Add ft(m) at the end of Sin meaning (It should read n ft (m)) Page Change in elements of formulas in figure 6.40 as follow: Page Change in the KbiR-L meaning (Equation 6.60 should read Equation 6.57 ) Page With: Dm Equation 6.70 should read dm from Equation 4.17 for Dm using A from Equation Pages for As, = bulk density, Si = idler spacing CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED - Corrections in table 6.47 to Type Rubber values from constant a1 to the end. Page Type 1 table should be labeled Type 3 and Type 3 table should be labeled Type 1 in table THE VOICE OF THE CONVEYOR INDUSTRY OF THE AMERICAS

3 Page In figure 6.50, Typos (9.4 o C) should read (-9.4 o C) in operating temperature. Page Pj should to be added in KbiR-S formula - T0 = 9.4 C should read T = C - Move the 3 lines starting at From table 6.47 up to ahead of s = - Replace (15 F - 3 F)/1.8 with -9.4 C; 9.4 C in the numerator should read -9.4 C - The value of s should read in xf formula Page Pj should to be added in KbiR-S formula - Move From F calculation:... down to directly above "Since the value wiw - Move the finish xp calculation below of From Table C should read C in xp formula ( Places) - Eliminate from xp calculations Page Remove Pj in Tbi formula - Add formula for KbiR-S including Pj below Calculate KbiR-S - Delete entire line "BW = 48 in..." - Move the line "Xld =..." to below "Use Rbi = 1.0 line - Add x Pj; x and change the result to = in KbiR-S formula and move it up to under For fabric belts: csd =... - Remove x Pj in the Note and 0.5 should read Page Change in wrrir and wrl meanings (lbf/im should read lbf/in ). CHAPTER 8 Page 38 - In Equation 8.33, in superscript should be added in D formula as follow: CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED CHAPTER 1 Page In Figure 1.73, ɸ in the drawing should be ɸi THE VOICE OF THE CONVEYOR INDUSTRY OF THE AMERICAS

4 Page 55 - In Figure 1.95, Replace: 'Vs=Tangential Velocity, fps, of the cross sectional area center of gravity of the load shape' with: Vs=Velocity of the load cross section used for plotting the trajectory Page In Figure 1.109, Greek letters are incorrect, there is a font error for phi describing the angle of incline of the conveyor. It is showing the 'capital phi' not in lower case as it should be. (ϕ should be φ ( places) and γ should be ϒ (1 place)). Page In the paragraph between Equation and Equation 1.11; the reference to Table 4.4, should read Table In the Figure 1.110, this should read: Vs = Velocity of the load cross section used for plotting the trajectory: 1. Belt velocity, V, is used as the velocity of the material at its center of mass if the discharge point is at the tangency of the belt to discharge pulley (Vs = V). Velocity of the material at its center of mass, Vcg, is used as the velocity of the material for all other conditions of discharge after the point of belt to discharge pulley tangency. (Vs= Vcg) APPENDIX B Page Change definition of Vs to be consistent with changes in Chapter 1 Trajectories. Vcg also added CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED THE VOICE OF THE CONVEYOR INDUSTRY OF THE AMERICAS

5 CAPACITIES, BELT WIDTHS AND SPEEDS 4 CEMA Standard Cross Sectional Area, A s s, for standard CEMA three equal roll troughing idlers based on the average CEMA center roll length circular surcharge surface and the CEMA standard belt edge. wmc calculated from b w and with b we set to the standard dimensions: A s = BW r s sch sin ( s) cos( s ) + b c b wmc sin ( ) sin( ) cos + b wmc ( ) Equation 4.15 A s, CEMA standard cross sectional area w = b c + b wmc cos( ) Equation 4.16 with standard cross sectional area A s A A s = Standard material cross sectional area based on design criteria [ft (m )] (Equation 4.1) = CEMA Standard Cross Sectional Area, bulk material cross sectional area based on three equal roll CEMA troughing idler, the surcharge angle with circular top surface, and standard edge distance [ft (m )] BW = Belt width, [in (mm)] b c b d = Dimensionless ratio of the effective upper surface of the belt above the center roll to the belt width, BW = Dimensionless ratio of maximum depth of material above the belt at the center roll to the belt width, BW b we = Dimensionless ratio of the standard edge distance to the belt width, BW b wmc = Dimensionless ratio of the length of material on the wing roll to the belt width, BW d m w = Dimensionless ratio of depth of the material above the belt at the center roll to the belt width, BW = Dimensionless ratio of the widest part of the load to the belt width, BW = Idler trough angle, (degrees when used with a trig function, otherwise radians) s = Material surcharge angle, (degrees when used with a trig function, otherwise radians) r sch = Dimensionless ratio of the radius tangent to the surcharge angle at the belt edge to the belt width, BW CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED d m = b wmc sin( ) + b c sin s ( ) move close parenthesis to the end cos + ( ) b wmc sin( s ) ( 1cos( s ) Equation 4.17 d m standard cross sectional area, A s 69

6 CAPACITIES, BELT WIDTHS AND SPEEDS 4 CEMA Standard Cross Sectional Area, A s s, for standard CEMA three equal roll troughing idlers based on the average CEMA center roll length circular surcharge surface and the CEMA standard belt edge. wmc calculated from b w and with b we set to the standard dimensions: A s = BW r s sin ( s) cos( s ) + b c b wmc sin ( ) sin( ) cos + b wmc ( ) Equation 4.15 A s, CEMA standard cross sectional area w = b c + b wmc cos( ) Equation 4.16 with standard cross sectional area A s A = Standard material cross sectional area based on design criteria [ft (m )] (Equation 4.5) A s = CEMA Standard Cross Sectional Area, bulk material cross sectional area based on three equal roll CEMA troughing idler, the surcharge angle with circular top surface, and standard edge distance [ft (m )] BW = Belt width, [in (mm)] b c b d = Dimensionless ratio of the effective upper surface of the belt above the center roll to the belt width, BW = Dimensionless ratio of maximum depth of material above the belt at the center roll to the belt width, BW b we = Dimensionless ratio of the standard edge distance to the belt width, BW b wmc = Dimensionless ratio of the length of material on the wing roll to the belt width, BW d m w = Dimensionless ratio of depth of the material above the belt at the center roll to the belt width, BW = Dimensionless ratio of the widest part of the load to the belt width, BW = Idler trough angle, (degrees when used with a trig function, otherwise radians) s = Material surcharge angle, (degrees when used with a trig function, otherwise radians) r sch = Dimensionless ratio of the radius tangent to the surcharge angle at the belt edge to the belt width, BW CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED d m = b wmc sin( ) + b c sin s ( ) cos + ( ) b wmc sin( s ) ( 1cos( s ) Equation 4.17 d m standard cross sectional area, A s 69

7 4 CAPACITIES, BELT WIDTHS AND SPEEDS Example: Non-standard Edge Distance Assume: CEMA standard three equal roll troughing idler, Q = 1800 tph, V = 500 fpm, m = 60 lbf/ft 3 Given: BW = 48.0 inches, = 35 degrees, s = 0 degrees From Figure 4.4: b c = and b w = A = a' b' c' Q = V m 1800 t lbf 000 h t 500 ft = min lbf min h ft 3 = cos() sin( s ) s sin( s ) cos( s ) ( ) 3,600,000 lbf h =.0 ft lbf 1,800,000 h ft ( ) + cos( ) sin( ) = ( ) ) (0.340) = = cos = b c sin( ) + b c ( ) sin( s ) ( sin s ( s) cos( s ) = ( 0.340) ( ) = = = b c A BW sin( s ) sin s ( s) cos( s ) ( ) in = - ft ( 48.0) = = b wmc = -b' + Figure 4.11 or 4.13 ( b' ) - 4 a' c' = ( ) (0.1166) a' = = b we = b w - b wmc = = B we = b we BW = in =.6 in (65 mm) CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED Figure 4.1 Example of calculating non standard belt edge distance from known idler, belt width and cross sectional area, A 7

8 4 CAPACITIES, BELT WIDTHS AND SPEEDS Example: Non-standard Edge Distance Assume: CEMA standard three equal roll troughing idler, Q = 1800 tph, V = 500 fpm, m = 60 lbf/ft 3 Given: BW = 48.0 inches, = 35 degrees, s = 0 degrees From Figure 4.11 or 4.13: b c = and b w = A = a' b' c' Q = V m 1800 t lbf 000 3,600,000 lbf h t 500 ft = h =.0 ft min lbf lbf ,800,000 min h ft 3 h ft ( ) + cos( ) sin( ) = cos() sin( s ) sin s ( s) cos( s ) ( ) = ( ) ) (0.340) = = cos = b c sin( ) + b c ( ) sin( s ) ( s sin( s ) cos( s ) = ( 0.340) ( ) = = b c A = BW sin( s ) s sin( s ) cos( s ) ( ) in = - ft ( 48.0) = = b wmc = -b' + ( b' ) - 4 a' c' = ( ) (0.1166) a' = = CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED b we = b w - b wmc = = B we = b we BW = in =.6 in (65 mm) Figure 4.1 Example of calculating non standard belt edge distance from known idler, belt width and cross sectional area, A 7

9 4 CAPACITIES, BELT WIDTHS AND SPEEDS s, is greater than the standard center roll effective width, b c 4.15 If w s > b c recalculate A s (Equation 4.6) using: b wmc = b s b c cos ( ) Equation 4.5 d ms = A BW b s s sin s ( ) - cot ( s) ( ) sin - ( ) 4 b c - b s cos( ) Equation 4.6 d ms Ws d s = b b s c b s Ws D ms = d ms BW Equation 4.7 D ms wws d w tan( )+b ms + b s 1 sin( s ) 1 tan( s ) Equation 4.8 d s CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED 74

10 4 CAPACITIES, BELT WIDTHS AND SPEEDS s, is greater than the standard center roll effective width, b c If w s > b c recalculate A s (Equation 4.15) using: b wmc = w s b c cos ( ) Equation 4.5 d ms = A BW w s s sin s ( ) - cot ( s) ( ) sin - ( ) 4 b c - w s cos( ) Equation 4.6 d ms d s = w b s c w s D ms = d ms BW Equation 4.7 D ms tan( )+ d ms + w s 1 sin( s ) 1 tan( s ) Equation 4.8 d s CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED 74

11 CAPACITIES, BELT WIDTHS AND SPEEDS Figure 4.3 or Figure 4.1 with A =A s 4 Example: Height of Bulk Material between Skirtboards, D s and D ms Given: BW = 48.0 inches, =35 degrees, s = 0 degrees From Figure 4.5 b c = b s = A s = 1.8 ft W s = Calculate the heigth of material rubbing on the skirtboards: A s BW b s s sin( s ) - cot ( sin s) - ( ) b ( c- b s ) 4 cos( ) d ms = b s in ft - 1 (48.0) (0.340) - (.7475) = ( 0.376) -( ) ) [ (.7)( ] = = ( ) = D ms = d ms BW = =.5 in (6.5 mm) W s d W s Calculate the maximum depth of material between the skirboards: d s = b s b c tan( )+b ms + b s 1 sin( s ) 1 tan( s ) = x = = D s = d s BW = x 48.0 = 10. in (3 mm) Figure 4.9 make experience based choices to modify the recommended measurements. The width of the skirtboards as, samplers or dust collection, or anticipated mistracking. Multiple loading points on a belt require either skirting the multiple load points as one continuous skirtboard, or making successive skirtboards wider in the direction of belt travel. The lump size discussion in this chapter governs the belt width and therefore skirtboard should be generous enough to handle the lump sizes, and to allow for the material volume in the turbulent loading area having a loose bulk density. The height and length of the skirtboard is often aid in controlling dust exiting the skirtboarded area. CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED 75

12 CAPACITIES, BELT WIDTHS AND SPEEDS 4 Example: Height of Bulk Material between Skirtboards, D s and D ms Given: BW = 48.0 inches, =35 degrees, s = 0 degrees From Figure 4.3 (or Figure 4.1 with A = A s ) b c = w s = A s = 1.8 ft Calculate the height of material rubbing on the skirtboards: A s BW b s s sin( s ) - cot ( sin s) - ( ) b ( c- b s ) 4 cos( ) d ms = b s in ft - 1 (48.0) (0.340) - (.7475) = ( 0.376) -( ) ) = 0.115[ (.7)( ] = ( ) = D ms = d ms BW = =.5 in (6.5 mm) Calculate the maximum depth of material between the skirtboards: d s = w s b c tan( )+b + w s 1 sin( s ) 1 d ms tan( s ) = x = = D s = d s BW = x 48.0 = 10. in (3 mm) Figure 4.9 make experience based choices to modify the recommended measurements. The width of the skirtboards as, samplers or dust collection, or anticipated mistracking. Multiple loading points on a belt require either skirting the multiple load points as one continuous skirtboard, or making successive skirtboards wider in the direction of belt travel. The lump size discussion in this chapter governs the belt width and therefore skirtboard should be generous enough to handle the lump sizes, and to allow for the material volume in the turbulent loading area having a loose bulk density. The height and length of the skirtboard is often aid in controlling dust exiting the skirtboarded area. CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED 75

13 4 CAPACITIES, BELT WIDTHS AND SPEEDS 100% Full, Edge To Edge, Cross Sectional Area, A f porting the conveyor system should be designed for the dead loads plus the live material load as if the belt f, d f and r schf are introduced for calculations for a we wmc w and b w is used to calculate A f and d f. The s with b we set to zero. d f Figure 4.30 A f, Cross sectional area dimensionless nomenclature for 100% full, edge to edge, Equation 4.31 r schf A f = BW Equation 4.3 A f b c w f A f s r schf CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED b w 1-cos r schf = ( ( ) b c + cos ( ) sin( s ) r schf s - sin ( s) cos( s ) + b c b w xsin( ) +b sin( )cos w ( ) ß d f = b w sin( )+ b c +b w cos ( ) 1 sin( s ) - 1 tan( s ) Equation 4.33 d f 76

14 4 CAPACITIES, BELT WIDTHS AND SPEEDS 100% Full, Edge To Edge, Cross Sectional Area, A f porting the conveyor system should be designed for the dead loads plus the live material load as if the belt f, d f and r schf are introduced for calculations for a we wmc w and b w is used to calculate A f and d f. The s with b we set to zero. d f Figure 4.30 A f, Cross sectional area dimensionless nomenclature for 100% full, edge to edge, Equation 4.31 r schf A f = BW Equation 4.3 A f b c w f A f s r schf CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED b w 1-cos r schf = ( ( ) b c + cos ( ) sin( s ) r schf s - sin ( s) cos( s ) + b c b w xsin( ) +b w ß sin( )cos ( ) d f = b w sin( )+ b c +b w cos ( ) 1 sin( s ) - 1 tan( s ) Equation 4.33 d f 76

15 4 CAPACITIES, BELT WIDTHS AND SPEEDS Example: 100% Full, Edge to Edge, Belt Cross Sectional Area, A f Given: BW = 48.0 inches, =35 degrees, s = 0 degrees From Figure 4.8 b c = b w = cos r schf = ( ( )) b c + cos( ) sin( s ) ( = ) = = A f = BW r schf s - sin ( s) cos( s ) + b c b w sin( ) +b sin( ) cos ww x ( ) 4608 = in + [ ] ft = 3.0 [ ]= =.6 ft (0.4 m ) d f = b w sin( )+ b c +b w cos( ) 1 sin( s ) - 1 tan( s ) = = = D f = d f BW = = 1.4 in (314 mm) Equation 4-11 and 4-13 Figure 4.37 CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED 78

16 4 CAPACITIES, BELT WIDTHS AND SPEEDS Example: 100% Full, Edge to Edge, Belt Cross Sectional Area, A f Given: BW = 48.0 inches, =35 degrees, s = 0 degrees From Equations 4.11 and 4.13 b c = b w = cos r schf = ( ( )) b c + cos( ) sin( s ) ( = ) = = A f = BW r schf s - sin ( s) cos( s ) + b c b w sin( ) +b sin( ) cos w x ( ) 4608 = in + [ ] ft = 3.0 [ ]= =.6 ft (0.4 m ) d f = b w sin( )+ b c +b w cos( ) 1 sin( s ) - 1 tan( s ) = = = D f = d f BW = = 1.4 in (314 mm) Figure 4.37 CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED 78

17 4 CAPACITIES, BELT WIDTHS AND SPEEDS General Applications: Capacity Derating one conveyor to a common cause of transfer point spillage and plugging is the time it takes for the discharged load to settle down and reach the receiving belt speed and direction. It is common practice, in and bulk material degradation. Coal Fired Power Generating Plant: Capacity Derating plants and handling other bulk materials subject to degradation and the hazards associated with spillage, leakage and dust generation. It is common practice not to load conveyors handling these bulk materials to their capacity in order to reduce degradation, accommodate surge loads and to reduce spillage and Example: Capacity Derating Required capacity: Q = 400 tph Bulk Material Properties: m = 90 lbf ft s= 0deg 3 ft Initial design choices: BW = 48 in = 35 deg V = 600 Angle of incline, = 0 deg min Calculate conveyed cross sectional area, A (Ref. Equation 4.5) 400 t Q A = = h 1 h lbf ,000 lbf 60 min t V m 600 ft = min = 1.48 ft lbf lbf 90 54,000 min ft 3 min ft 4.43 Derate loading of cross section to 85%, DF = 1.18 Minimum A s = A DF = 1.48 ft 1.18 = 1.75 ft (0.16 m ) From Table 4.44 at = 35 and s = 0 deg: 48-inch belt, A s = ft (0.168 m ) The initial design choice (BW = 48 in and V = 600 fpm) appears appropriate from a capacity standpoint Figure 4.38 Capacity derating example CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED 80

18 4 CAPACITIES, BELT WIDTHS AND SPEEDS General Applications: Capacity Derating one conveyor to a common cause of transfer point spillage and plugging is the time it takes for the discharged load to settle down and reach the receiving belt speed and direction. It is common practice, in and bulk material degradation. Coal Fired Power Generating Plant: Capacity Derating plants and handling other bulk materials subject to degradation and the hazards associated with spillage, leakage and dust generation. It is common practice not to load conveyors handling these bulk materials to their capacity in order to reduce degradation, accommodate surge loads and to reduce spillage and Example: Capacity Derating Required capacity: Q = 400 tph Bulk Material Properties: m = 90 lbf ft s= 0deg 3 ft Initial design choices: BW = 48 in = 35 deg V = 600 Angle of incline, = 0 deg min Calculate conveyed cross sectional area, A (Ref. Equation 4.5) 400 t Q A = = h 1 h lbf ,000 lbf 60 min t V m 600 ft = min = 1.48 ft lbf lbf 90 54,000 min ft 3 min ft Derate loading of cross section to 85%, DF = 1.18 Minimum A s = A DF = 1.48 ft 1.18 = 1.75 ft (0.16 m ) From Table 4.4 at = 35 and s = 0 deg: 48-inch belt, A s = ft (0.168 m ) The initial design choice (BW = 48 in and V = 600 fpm) appears appropriate from a capacity standpoint Figure 4.38 Capacity derating example CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED 80

19 BELT CONVEYOR IDLERS 5 Step No. 3: K Effect Of Load On Predicted Bearing L 10 Life R ) is less than the CEMA load rating of the class of idler selected the earing L life will increase. K Factor Figure 5.30 K L Step No. 4: K 3A Effect Of Belt Speed On Predicted Bearing L 10 Life CEMA L life ratings are ased on r. lower seeds increase life and faster seeds decrease life. igre. shows this relationshi. K 3A Factor CIL (Calulated Idler Load) ILR (Idler Load Rating) K = CIL ILR 1 K = CIL ILR 500 (rpm) L 10 n (rpm) Belt Speed Roll Speed = Roll Circumference (rpm) Figure 5.31 K 3A ct o t o rct ar L 10 o CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED 3.3 K 3A (Roller Bearings) 3.0 (Ball Bearings) 109

20 BELT CONVEYOR IDLERS 5 Step No. 3: K Effect Of Load On Predicted Bearing L 10 Life R ) is less than the CEMA load rating of the class of idler selected the earing L life will increase. K Factor Figure 5.30 K L Step No. 4: K 3A Effect Of Belt Speed On Predicted Bearing L 10 Life CEMA L life ratings are ased on r. lower seeds increase life and faster seeds decrease life. igre. shows this relationshi. K 3A Factor CIL (Calulated Idler Load) ILR (Idler Load Rating) K = CIL ILR 1 K = CIL ILR K 3A 500 (rpm) n (rpm) Belt Speed Roll Speed = Roll Circumference (rpm) Figure 5.31 K 3A ct o t o rct ar L 10 o CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED 3.3 (Roller Bearings) 3.0 (Ball Bearings) 109

21 6 BELT TENSION AND POWER ENGINEERING T nenergy Gravity paths due to the potential energy change in the bulk material and belt for a height change H n. The tension is sensitive to the direction of travel so that with uphill movement the tension increases and a downhill or negative slope angle causes reduction in this component of tension along the conveyance direction as gravity pulls the conveyor down the slope. Gravity or potential energy is considered to have a continuous effect on tension along the length of any side belt cancel each other out from the perspective of total conveyor T e but need to be included in circuit calculations to identify the local tension at any point. T Hn = H n ( W b + W m ) Equation 6.14, Flight tension change due to elevate n ), the belt and the load Where: T Hn = Tension change in fligh "n" due to lift H n = Elevation change in flight "n" W b = Weight of the belt per unit length from manufacturer W m = Weight of the bulk material on the belt per unit length 5.9 8, ,958 T H5 = H 5 ( W b + W m ) = 44.0 ft (6.3 lbf lbf Add ) = 7,68.8 lbf (3,304 kgf) ft ft Figure 6.15 Example calculation of tension needed to elevate material in flight 5 n as the net change in elevation Bulk Material Acceleration Work or kinetic energy must be provided to the bulk material to accelerate it to match the speed of the belt. The accelerating force is provided by the belt through an increase in tension at the loading point(s) in the direction of belt movement. Using the amount of Kinetic Energy added to the bulk material allows the calculation of belt tension effects without concern for the acceleration rate or the dynamics involved with impact, although these can be important issues for belt and chute wear and material degradation. CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED 148

22 6 BELT TENSION AND POWER ENGINEERING T nenergy Gravity paths due to the potential energy change in the bulk material and belt for a height change H n. The tension is sensitive to the direction of travel so that with uphill movement the tension increases and a downhill or negative slope angle causes reduction in this component of tension along the conveyance direction as gravity pulls the conveyor down the slope. Gravity or potential energy is considered to have a continuous effect on tension along the length of any side belt cancel each other out from the perspective of total conveyor T e but need to be included in circuit calculations to identify the local tension at any point. T Hn = H n ( W b + W m ) Equation 6.14, Flight tension change due to elevate n ), the belt and the load Where: T Hn = Tension change in flight "n" due to lift H n = Elevation change in flight "n" W b = Weight of the belt per unit length from manufacturer W m = Weight of the bulk material on the belt per unit length T H5 = H 5 ( W b + W m ) = 5.9 ft (6.3 lbf lbf ) = 8,738.4 lbf (3,958 kgf) ft ft Figure 6.15 Example calculation of tension needed to elevate material in flight 5 n as the net change in elevation Bulk Material Acceleration Work or kinetic energy must be provided to the bulk material to accelerate it to match the speed of the belt. The accelerating force is provided by the belt through an increase in tension at the loading point(s) in the direction of belt movement. Using the amount of Kinetic Energy added to the bulk material allows the calculation of belt tension effects without concern for the acceleration rate or the dynamics involved with impact, although these can be important issues for belt and chute wear and material degradation. CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED 148

23 BELT TENSION AND POWER ENGINEERING 6 Skirtboard Seal Friction A skirt seal which rides on the belt is commonly used to contain dust and small particles. The calculation predicts the resistance as the product of a friction factor and the unit normal force between the moving belt and the seal without inuence from the material loading. The values are provided below apply to a generic rubber edge seal as shown in igure. for ight n that is sealed along its full length on both sides. Where: F ss F ss L n Figure 6.0 Skirtboard seal drag on conveyor belt Equation 6.1 T ssn, Calculation of skirtboard seal drag T ssn = Tension change due to belt sliding on skirtboard sealed flight, "n" C ss = m ss F ss R rss Frictional resistance to the belt movement μ ss = Sliding friction coefficient between belt and seal rubber (dimensionless) F ss = Effective normal force between belt and seal = Modifying Factor (dimensionless) R rss μ T ssn = C ss L n R rss L 1 = 15.0 ft μ ss = 1.0 F ss = 3.0 lbf / ft R rss = 1.0 C ss = lbf x 1.0 = 6.0 lbf/ft ft T ss1 = C ss L 1 R rss = 6.0 lbf 15.0 ft 1.0 = 90.0 lbf (40.9 kgf) ft CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED Figure 6. T ss1, Skirtboard seal example calculation arious specialty sealing products are available to perform this function with varying performance life and drag. Typical values for design are μ ss. and ss = 3.0 lbfft. kgf/m) of skirt seal for conventional slab rubber skirt board seals shown in igure.0. ss is calculated by multiplying by a factor of because it is assumed that both sides of the belt have a skirt seal. Therefore an estimate of.0 lbf/ft.0 151

24 BELT TENSION AND POWER ENGINEERING 6 Skirtboard Seal Friction A skirt seal which rides on the belt is commonly used to contain dust and small particles. The calculation predicts the resistance as the product of a friction factor and the unit normal force between the moving belt and the seal without inuence from the material loading. The values are provided below apply to a generic rubber edge seal as shown in igure. for ight n that is sealed along its full length on both sides. Where: F ss F ss L n Figure 6.0 Skirtboard seal drag on conveyor belt Equation 6.1 T ssn, Calculation of skirtboard seal drag T ssn = Tension change due to belt sliding on skirtboard sealed flight, "n" C ss = µ ss F ss R rss Frictional resistance to the belt movement μ ss = Sliding friction coefficient between belt and seal rubber (dimensionless) F ss = Effective normal force between belt and seal = Modifying Factor (dimensionless) R rss T ssn = C ss L n R rss L 1 = 15.0 ft μ ss = 1.0 F ss = 3.0 lbf / ft R rss = 1.0 C ss = lbf x 1.0 = 6.0 lbf/ft ft T ss1 = C ss L 1 R rss = 6.0 lbf 15.0 ft 1.0 = 90.0 lbf (40.9 kgf) ft CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED Figure 6. T ss1, Skirtboard seal example calculation arious specialty sealing products are available to perform this function with varying performance life and drag. Typical values for design are μ ss. and ss = 3.0 lbfft. kgf/m) of skirt seal for conventional slab rubber skirt board seals shown in igure.0. ss is calculated by multiplying by a factor of because it is assumed that both sides of the belt have a skirt seal. Therefore an estimate of.0 lbf/ft.0 151

25 BELT TENSION AND POWER ENGINEERING (0.68) Drag for One Roll lbf-in (N-m) 5.0 (0.57) 4.0 (0.45) 3.0 (0.34).0 (0.3) 1.0 (0.11) 0 0 K iv = Slope Figure 6.4 Drag from a single idler roll rpm T ir = K iv R riv ( N i rpm)+ K is R is Equation 6.5 ir, Drag from a single idler roll T isn = K it T ir n r Equation 6.6 T isn, Idler seal friction CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED S in L n ris D r 153

26 BELT TENSION AND POWER ENGINEERING (0.68) Drag for One Roll lbf-in (N-m) 5.0 (0.57) 4.0 (0.45) 3.0 (0.34).0 (0.3) 1.0 (0.11) 0 0 K iv = Slope Figure 6.4 Drag from a single idler roll rpm T ir = K iv R riv ( N i rpm)+ K is R r is Equation 6.5 ir, Drag from a single idler roll T isn = K it T ir n r Equation 6.6 T isn, Idler seal friction CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED S in L n D r 153

27 6 BELT TENSION AND POWER ENGINEERING Where: T ir = Change in tension for single roll from idler seal resistance lbf (N) T isn = Change in tension in flight "n" from idler seal resistance lbf (N) D r = Idler roll diameter in (mm) K iv = Slope of torsional speed curve per roll lbf-in N-m rpm Table 6.9 rpm K is = Seal torsional resistance per roll at 500 rpm lbf-in (N-m) Table 6.9 K it = Temperature correction factor (dimensionless) Figure 6.7 K itb = Curve fit constants for temperature correction [R 1 (K 1 )] L n n r N i R ris R riv S in K IT Note: R = Rankine, K = Kelvin [T R = F (T K = C K)] = Length of flight "n" ft (m) = Number of rolls per idler set = Actual rpm of idler based on diameter and belt speed rpm = Modifying Factor for seal torsional resistance (dimensionless) = Modifying Factor for torsional speed effect (dimensionless) = Carrying or return idler spacing in flight "n" ft(m) CEMA C&D CEMA E CEMA Historical (-40) (-8) (-18) (-4) (4) (16) (7) (38) (49) Temperature F ( C) Figure 6.7 K it Teerre rrei r re r ier r idler seal drag and it is accounted for by the multiplying factor, K it ember products hae been independently tested and the euations published reect seal drag change with temperature. This comparison of results with the CEMA Historical correction is graphically shown in igure.. CEMA published K it values should only be used with published K is and K iv values. Testing shows designs can vary widely and using design specic K is or K iv values with the K it calculated by euations in igure. can misrepresent true performance. CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED 154

28 6 BELT TENSION AND POWER ENGINEERING Where: T ir = Change in tension for single roll from idler seal resistance lbf (N) T isn = Change in tension in flight "n" from idler seal resistance lbf (N) D r = Idler roll diameter in (mm) K iv = Slope of torsional speed curve per roll lbf-in N-m rpm Table 6.9 rpm K is = Seal torsional resistance per roll at 500 rpm lbf-in (N-m) Table 6.9 K it = Temperature correction factor (dimensionless) Figure 6.7 K itb = Curve fit constants for temperature correction [R 1 (K 1 )] L n n r N i R ris R riv S in K IT Note: R = Rankine, K = Kelvin [T R = F (T K = C K)] = Length of flight "n" ft (m) = Number of rolls per idler set = Actual rpm of idler based on diameter and belt speed rpm = Modifying Factor for seal torsional resistance (dimensionless) = Modifying Factor for torsional speed effect (dimensionless) = Carrying or return idler spacing in flight "n" ft(m) CEMA C&D CEMA E CEMA Historical (-40) (-8) (-18) (-4) (4) (16) (7) (38) (49) Temperature F ( C) Figure 6.7 K it Teerre rrei r re r ier r idler seal drag and it is accounted for by the multiplying factor, K it ember products hae been independently tested and the euations published reect seal drag change with temperature. This comparison of results with the CEMA Historical correction is graphically shown in igure.. CEMA published K it values should only be used with published K is and K iv values. Testing shows designs can vary widely and using design specic K is or K iv values with the K it calculated by euations in igure. can misrepresent true performance. CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED 154

29 6 BELT TENSION AND POWER ENGINEERING There are no generally available tabulated indentation values for various belt constructions for either the small or large sample methods. CEMA does not endorse any particular method so long as it accurately predicts the indentation resistance on a single idler for different temperatures and loads and can be used to determine T bin for specic belt constructions being considered in the conveyor design. ndentation loss is usually a maor factor on long overland conveyors. The magnitude of the indentation loss for short horiontal or inclined conveyors is usually not a signicant loss component in the total tension requirement. While the indentation qualities of the belt covers are an important consideration for energy consumption, it is important to balance other requirements for the cover design such as abrasion resistance or ame retardancy hen considering rubber compounds including a lo rolling resistance (LRR) conveyor belt cover. Equation 6.37 T bin, Tension increase from viscoelastic indentation reaction between roller and belt Figure 6.38 ivalent load distribtion from tree roll idler cross sectional area Two methods for K bir are provided for the small sample (K bir-s ) and large sample (K bir-l ) methods. To arrive at T bin it is necessary to adjust for the uneven loading (igure.) on the rollers to arrive at an average pressure between the belt and roller. The equation for c wd is derived from the geometry of the cross sectional area based on the load area, A and represents a correction to the average line load, w iw. Equation 6.39 c wd, Load distribution factor T bin = K bir c wd (W b + W m ) L n R rbi D m s A Area Distribution BW Equivalent Load Distribution c wd = X ld BW s CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED ß w iw Division sign X ld = m S i X ldref X ldref = 5. lbf in 36,000 N mm Equation 6.40 X ld, Loading pressure adjustment factor 160

30 6 BELT TENSION AND POWER ENGINEERING There are no generally available tabulated indentation values for various belt constructions for either the small or large sample methods. CEMA does not endorse any particular method so long as it accurately predicts the indentation resistance on a single idler for different temperatures and loads and can be used to determine T bin for specic belt constructions being considered in the conveyor design. ndentation loss is usually a maor factor on long overland conveyors. The magnitude of the indentation loss for short horiontal or inclined conveyors is usually not a signicant loss component in the total tension requirement. While the indentation qualities of the belt covers are an important consideration for energy consumption, it is important to balance other requirements for the cover design such as abrasion resistance or ame retardancy hen considering rubber compounds including a lo rolling resistance (LRR) conveyor belt cover. Equation 6.37 T bin, Tension increase from viscoelastic indentation reaction between roller and belt Figure 6.38 ivalent load distribtion from tree roll idler cross sectional area Two methods for K bir are provided for the small sample (K bir-s ) and large sample (K bir-l ) methods. To arrive at T bin it is necessary to adjust for the uneven loading (igure.) on the rollers to arrive at an average pressure between the belt and roller. The equation for c wd is derived from the geometry of the cross sectional area based on the load area, A and represents a correction to the average line load, w iw. Equation 6.39 c wd, Load distribution factor T bin = K bir c wd (W b + W m ) L n R rbi D m s A Area Distribution BW Equivalent Load Distribution c wd = X ld BW s CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED ß w iw Equation 6.40 X ld, Loading pressure adjustment factor 160

31 BELT TENSION AND POWER ENGINEERING 6 Where: T bin = Tension increase from viscoelastic deformation of belt cover rubber K birs = Viscoelastic characteristic of belt cover rubber from the small sample method Equation 6.4 K bir-l = Viscoelastic characteristic of belt cover rubber from the large sample method Equation 6.60 c wd = Load distribution factor (dimensionless) R rbi = Modifying factor (dimensionless) S i = Idler spacing [ft (m)] X ld = Loading pressure adjustment factor (dimensionless) = Troughing angle (deg) s = Surcharge angle (deg) = Bulk density of conveyed material m Small Sample Indentation Loss Method Rubber indentation energy loss aries ith the idler rolls indentation into the belt oer thiness and with the nominal deformation work as affected by the idler roll radius and the normal load. Just as important is the degree to which the rubber reacts elastically to return the energy of deformation to the system. This is affected by the rubber composition, the amount of deformation or strain, the rubber temperature and, to a lesser degree, the belt speed. The rubber composition is a design variable through the viscoelastic concepts of Storage Modulus and Loss Modulus with their ratio, known as tan delta, as an index of the rubbers loss characteristic. These properties are best obtained from harmonic testing and vary with frequency, temperature and strain, paralleling the idler indentation of interest. The loss may be then considered to be the area within the steady cycles of stress and strain along the transient path of the indentation. The contribution of the rubber material is incorporated into the indentation prediction with K bir-s. n effect, it sets the width of the ellipse in igure.. or a particular rubber, the applicable value varies with the temperature, belt speed and loading. This requires a series of calculations with application details and a set of numerical values for the particular rubber. Constants for several example rubber cover compounds are provided for evaluation and consideration. Belt manufacturer should be contacted for selection, application and specication of compounds for specic applications. Stress Compression Steady Cycle Transient Cycle Indentation Loss CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED Tension 6.57 Negative 0 Strain Positive Figure

32 BELT TENSION AND POWER ENGINEERING 6 Where: T bin = Tension increase from viscoelastic deformation of belt cover rubber K birs = Viscoelastic characteristic of belt cover rubber from the small sample method Equation 6.4 K bir-l = Viscoelastic characteristic of belt cover rubber from the large sample method Equation 6.57 c wd = Load distribution factor (dimensionless) R rbi = Modifying factor (dimensionless) S i = Idler spacing [ft (m)] X ld = Loading pressure adjustment factor (dimensionless) = Troughing angle (deg) s = Surcharge angle (deg) = Bulk density of conveyed material m Small Sample Indentation Loss Method Rubber indentation energy loss aries ith the idler rolls indentation into the belt oer thiness and with the nominal deformation work as affected by the idler roll radius and the normal load. Just as important is the degree to which the rubber reacts elastically to return the energy of deformation to the system. This is affected by the rubber composition, the amount of deformation or strain, the rubber temperature and, to a lesser degree, the belt speed. The rubber composition is a design variable through the viscoelastic concepts of Storage Modulus and Loss Modulus with their ratio, known as tan delta, as an index of the rubbers loss characteristic. These properties are best obtained from harmonic testing and vary with frequency, temperature and strain, paralleling the idler indentation of interest. The loss may be then considered to be the area within the steady cycles of stress and strain along the transient path of the indentation. The contribution of the rubber material is incorporated into the indentation prediction with K bir-s. n effect, it sets the width of the ellipse in igure.. or a particular rubber, the applicable value varies with the temperature, belt speed and loading. This requires a series of calculations with application details and a set of numerical values for the particular rubber. Constants for several example rubber cover compounds are provided for evaluation and consideration. Belt manufacturer should be contacted for selection, application and specication of compounds for specic applications. Stress Compression Steady Cycle Transient Cycle Indentation Loss CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED Tension Negative 0 Strain Positive Figure

33 BELT TENSION AND POWER ENGINEERING 6 The contribution of the rubber material is incorporated into the indentation prediction with K bir-s - the temperature, belt speed and loading. This requires a series of calculations with application details and a set of constant values for the particular rubber. Constants for several example rubber cover compounds are provided for evaluation and consideration. Belt manufacturer should be contacted for selection, ap- Where: Where: F = b 1 + [b (x F )] + [b 3 (x F )] + [b 4 (x 3 F )] (dimensionless) b 5 + [b 6 (x F )] + x F Equation 6.44 F, Normalized indentation factor x F = - C 1 ( T T 0 ) C + ( T T 0 ) + log ( v u)- s (dimensionless) s = a 1 + [a (x s )] + [a 3 (x s )] + [a 4 (x 3 s )] (dimensionless) T = Operating temperature ( o C) v u = Belt speed m s Note: v umust be in m s units in x F equation 1/3 w x S = iw (dimensionless) w max d m from Equation 4.17 for D m using A from Equation 4.5 for A s With: D m Equation 6.70, m = bulk density, S i = idler spacing w max = 85.5 lbf in Note : 50,000 N m For constants a i, b i, C 1, C and T 0 see Table 6.47 w iw = D m m + W b S BW i Equation 6.45 w iw CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED BW = Belt width in (mm) D m = Maximum depth of material on three roll idler in (mm) m = Bulk density lbf kgf ft 3 S i = Idler spacing ft (m) m 3 W b = Belt weight per unit length lbf ft N m 163

34 BELT TENSION AND POWER ENGINEERING 6 The contribution of the rubber material is incorporated into the indentation prediction with K bir-s - the temperature, belt speed and loading. This requires a series of calculations with application details and a set of constant values for the particular rubber. Constants for several example rubber cover compounds are provided for evaluation and consideration. Belt manufacturer should be contacted for selection, ap- Where: Where: F = b 1 + [b (x F )] + [b 3 (x F )] + [b 4 (x 3 F )] (dimensionless) b 5 + [b 6 (x F )] + x F Equation 6.44 F, Normalized indentation factor x F = - C 1 ( T T 0 ) C + ( T T 0 ) + log ( v u)- s (dimensionless) s = a 1 + [a (x s )] + [a 3 (x s )] + [a 4 (x 3 s )] (dimensionless) T = Operating temperature ( o C) v u = Belt speed m s Note: v umust be in m s units in x F equation x S = w iw w max 1/3 (dimensionless) With: d m from Equation 4.17 for D m using A from Equation 4.5 for A s, m = bulk density, S i = idler spacing w max = 85.5 lbf in Note : 50,000 N m For constants a i, b i, C 1, C and T 0 see Table 6.47 w iw = D m m + W b S BW i Equation 6.45 w iw CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED BW = Belt width in (mm) D m = Maximum depth of material on three roll idler in (mm) m = Bulk density lbf kgf ft 3 m 3 S i = Idler spacing ft (m) W b = Belt weight per unit length lbf ft N m 163

35 6 BELT TENSION AND POWER ENGINEERING Where: P = c + [c (x )] + [c (x 1 P 3 P)] + [c 4 (x 3 P )] (dimensionless) c 5 + [c 6 (x P )] + x P x P = - C 1 ( T T 0 ) + log( v C + ( T T 0 ) u ) (dimensionless) Notes : For constants c i see Table 6.48 For constants C 1, C and T 0 see Table 6.47 Equation 6.46 P, Rubber strain level adjustment T = Operating temperature ( o C) v u = Belt speed m s Note: v u must be in m s units in x F equation tual loading strain which is commonly in the range of nonlinear stress strain behavior. classes of conveyor belt rubber covers. These values are incorrect Constant Default Rubber Type 1 Rubber Type Rubber Type 3 Rubber T 0 ( C) E 0 (N/m ) C C a a a a b b b b b b CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED Table 6.47 bir-s F factor 164

36 6 BELT TENSION AND POWER ENGINEERING Where: P = c + [c (x )] + [c (x 1 P 3 P)] + [c 4 (x 3 P )] (dimensionless) c 5 + [c 6 (x P )] + x P x P = - C 1 ( T T 0 ) + log( v C + ( T T 0 ) u ) (dimensionless) Notes : For constants c i see Table 6.48 For constants C 1, C and T 0 see Table 6.47 Equation 6.46 P, Rubber strain level adjustment T = Operating temperature ( o C) v u = Belt speed m s Note: v u must be in m s units in x F equation tual loading strain which is commonly in the range of nonlinear stress strain behavior. classes of conveyor belt rubber covers. Constant Default Rubber Type 1 Rubber Type Rubber Type 3 Rubber T 0 ( C) E 0 (N/m ) C C a a a a b b b b b b CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED Table 6.47 bir-s F factor 164

37 BELT TENSION AND POWER ENGINEERING 6 c i Intervals: Data for Type 1 and Type 3 was transposed Each c i constant is linearly interpolated for w iw, relative to the seven levels of w ref and extrapolated past the last row. Cover Compound w ref (N/m) c 1 c c 3 c 4 c 5 c 6 (dimensionless) Default Type 1 Type Type CEMA 7th ed. Belt Book ERRATA-FEBRUARY 1, 015-COPYRIGHTED Table 6.48 bir-s P factor at several w ref values compounds for common covers for more conservative designs that operate at lower temperatures. Type 1 constants are for a low rolling resistance rubber cover compound considered for applications where 165

Conveyor Equipment Manufacturers Association

Conveyor Equipment Manufacturers Association Conveyor Equipment Manufacturers Association 567 Strand Ct., Suite, Naples, Florida 34110 Tel: (39) - 514-3441 Fax: (39) - 514-3470 Web Site: http://www.cemanet.org Belt Book, 7th Edition Errata items

More information

BELT TENSIONS & HP RESULTS

BELT TENSIONS & HP RESULTS 13 12 120' 3 SECTIONS @ 100' ARC 300' 7 SECTIONS @ 100' LOADING #1 1200 TPH 120' 204' ~ 1432.4' RADIUS 1696.1' CONVEYOR INPUT DATA BELT TENSIONS & HP RESULTS BELT WIDTH, inches = CAPACITY, tph = DENSITY,

More information

Chain Conveyors Practical Calculations

Chain Conveyors Practical Calculations PDHonline Course M360 (1PDH) Chain Conveyors Practical Calculations (Metric and US Systems) Jurandir Primo, PE 2009 PDH Online PDH Center 5272 Meadow Estates Drive Fairfax, VA 22030-6658 Phone & Fax: 703-988-0088

More information

BELT TENSION, N & POWER RESULTS CONVEYOR INPUT DATA BELT CONVEYOR NO._ GENERAL ARRANGEMENT METRIC EXAMPLE HEAD DRIVE PRO-BELT MEXPHEAD 1

BELT TENSION, N & POWER RESULTS CONVEYOR INPUT DATA BELT CONVEYOR NO._ GENERAL ARRANGEMENT METRIC EXAMPLE HEAD DRIVE PRO-BELT MEXPHEAD 1 13 16 36.576 m 3 SECTIONS @ 30.480 m ARC 131.0 ~ RADIUS 91.440 m 4 36.576 m 24.384 m 62.179 m ~ 436.6 m RADIUS 517.-580 m CONVEYOR INPUT DATA BELT WIDTH. mm = 1600 C.AP ACITY, Mtph = 2200 DENSITY. kg/m"3

More information

Dynamics of High-modulus Belts with Distributed Mass and Friction

Dynamics of High-modulus Belts with Distributed Mass and Friction Dynamics of High-modulus Belts with Distributed Mass and Friction A. Harrison 1,2 1 Faculty of Engineering, University of Newcastle, NSW 2308, Australia 2 Conveyor Science Pty. Ltd., Lisarow, NSW 2250,

More information

Local Stresses in Belt Turnovers in Conveyor Belt

Local Stresses in Belt Turnovers in Conveyor Belt Local Stresses in Belt Turnovers in Conveyor Belt By: Conveyor Dynamics, Inc. personnel Conveyor Dynamics, Inc. 1111 W. Holly Street, Suite A Bellingham, WA 985 USA Abstract Belt turnovers are commonly

More information

Errata for the First Printing of Exploring Creation With Physics, 2 nd Edition

Errata for the First Printing of Exploring Creation With Physics, 2 nd Edition Errata for the First Printing of Exploring Creation With Physics, 2 nd Edition With the help of students and teachers, we have found a few typos in the first printing of the second edition. STUDENT TEXT

More information

BULK SOLID LOADING PROFILES AND LOAD STABILITY DURING BELT CONVEYING

BULK SOLID LOADING PROFILES AND LOAD STABILITY DURING BELT CONVEYING BULK SOLID LOADING PROFILES AND LOAD STABILITY DURING BELT CONVEYING A.W. ROBERTS EMERITUS PROFESSOR AND DIRECTOR, CENTER FOR BULK SOLIDS AND PARTICULATE TECHNOLOGIES, THE UNIVERSITY OF NEWCASTLE, AUSTRALIA.

More information

A Comparison of Calculated and Measured Indentation Losses in Rubber Belt Covers

A Comparison of Calculated and Measured Indentation Losses in Rubber Belt Covers Whitepaper Conveyor Belting A Comparison of Calculated and Measured Indentation Losses in Rubber Belt Covers Energy loss due to indentation rolling resistance is a major factor in belt conveyor design.

More information

Overview. Dry Friction Wedges Flatbelts Screws Bearings Rolling Resistance

Overview. Dry Friction Wedges Flatbelts Screws Bearings Rolling Resistance Friction Chapter 8 Overview Dry Friction Wedges Flatbelts Screws Bearings Rolling Resistance Dry Friction Friction is defined as a force of resistance acting on a body which prevents slipping of the body

More information

Exam 2 Phys Fall 2002 Version A. Name ID Section

Exam 2 Phys Fall 2002 Version A. Name ID Section Closed book exam - Calculators are allowed. Only the official formula sheet downloaded from the course web page can be used. You are allowed to write notes on the back of the formula sheet. Use the scantron

More information

MECTROL CORPORATION 9 NORTHWESTERN DRIVE, SALEM, NH PHONE FAX TIMING BELT THEORY

MECTROL CORPORATION 9 NORTHWESTERN DRIVE, SALEM, NH PHONE FAX TIMING BELT THEORY MECTRO CORPORATION 9 NORTHWESTERN DRIVE, SAEM, NH 03079 PHONE 603-890-55 FAX 603-890-66 TIMING BET THEORY Copyright 997, 999, 00 Mectrol Corporation. All rights reserved. April 00 Timing Belt Theory Introduction

More information

Polyflex JB and Micro-V Belt Drive Selection Procedures

Polyflex JB and Micro-V Belt Drive Selection Procedures Polyflex JB and Micro-V Belt Drive Selection Procedures How to Design Polyflex JB and Micro-V Belt Drives Note: The upcoming drive selection and engineering sections provide information for Polyflex JB

More information

Announcements. Principle of Work and Energy - Sections Engr222 Spring 2004 Chapter Test Wednesday

Announcements. Principle of Work and Energy - Sections Engr222 Spring 2004 Chapter Test Wednesday Announcements Test Wednesday Closed book 3 page sheet sheet (on web) Calculator Chap 12.6-10, 13.1-6 Principle of Work and Energy - Sections 14.1-3 Today s Objectives: Students will be able to: a) Calculate

More information

AP* Circular & Gravitation Free Response Questions

AP* Circular & Gravitation Free Response Questions 1992 Q1 AP* Circular & Gravitation Free Response Questions A 0.10-kilogram solid rubber ball is attached to the end of a 0.80-meter length of light thread. The ball is swung in a vertical circle, as shown

More information

Physics 141 Rotational Motion 2 Page 1. Rotational Motion 2

Physics 141 Rotational Motion 2 Page 1. Rotational Motion 2 Physics 141 Rotational Motion 2 Page 1 Rotational Motion 2 Right handers, go over there, left handers over here. The rest of you, come with me.! Yogi Berra Torque Motion of a rigid body, like motion of

More information

HSC PHYSICS ONLINE B F BA. repulsion between two negatively charged objects. attraction between a negative charge and a positive charge

HSC PHYSICS ONLINE B F BA. repulsion between two negatively charged objects. attraction between a negative charge and a positive charge HSC PHYSICS ONLINE DYNAMICS TYPES O ORCES Electrostatic force (force mediated by a field - long range: action at a distance) the attractive or repulsion between two stationary charged objects. AB A B BA

More information

Principles Of Engineering. Part A

Principles Of Engineering. Part A Principles Of Engineering Final Examination Part A Fall 2007 Student Name: Date: Class Period: Total Points: /40 Converted Score: /50 Page 1 of 11 Directions: Circle the letter of the response that best

More information

7.6 Journal Bearings

7.6 Journal Bearings 7.6 Journal Bearings 7.6 Journal Bearings Procedures and Strategies, page 1 of 2 Procedures and Strategies for Solving Problems Involving Frictional Forces on Journal Bearings For problems involving a

More information

Torque/Rotational Energy Mock Exam. Instructions: (105 points) Answer the following questions. SHOW ALL OF YOUR WORK.

Torque/Rotational Energy Mock Exam. Instructions: (105 points) Answer the following questions. SHOW ALL OF YOUR WORK. AP Physics C Spring, 2017 Torque/Rotational Energy Mock Exam Name: Answer Key Mr. Leonard Instructions: (105 points) Answer the following questions. SHOW ALL OF YOUR WORK. (22 pts ) 1. Two masses are attached

More information

SOLUTION (17.3) Known: A simply supported steel shaft is connected to an electric motor with a flexible coupling.

SOLUTION (17.3) Known: A simply supported steel shaft is connected to an electric motor with a flexible coupling. SOLUTION (17.3) Known: A simply supported steel shaft is connected to an electric motor with a flexible coupling. Find: Determine the value of the critical speed of rotation for the shaft. Schematic and

More information

Fig Use Fig. 3.1 to state the physical properties of this metal. In your answer, you should use appropriate technical terms, spelled correctly.

Fig Use Fig. 3.1 to state the physical properties of this metal. In your answer, you should use appropriate technical terms, spelled correctly. 1 (a) Fig. 3.1 shows the stress against strain graph for a metal X up to its breaking point. stress X 0 0 Fig. 3.1 strain Use Fig. 3.1 to state the physical properties of this metal. In your answer, you

More information

GCE Advanced Level 2014

GCE Advanced Level 2014 GCE Advanced Level 2014 Combined Mathematics I Model Paper 04 PART A (Answer all questions) Time 3 hrs 1. A cyclist rides along a straight path with uniform velocity u and passes a motor car, which is

More information

Chapter 4. 4 Forces and Newton s Laws of Motion. Forces and Newton s Laws of Motion

Chapter 4. 4 Forces and Newton s Laws of Motion. Forces and Newton s Laws of Motion Chapter 4 Forces and Newton s Laws of Motion PowerPoint Lectures for College Physics: A Strategic Approach, Second Edition 4 Forces and Newton s Laws of Motion Slide 4-2 Slide 4-3 1 Slide 4-4 Weight is

More information

Chapter 6 Energy and Oscillations

Chapter 6 Energy and Oscillations Chapter 6 Energy and Oscillations Conservation of Energy In this chapter we will discuss one of the most important and fundamental principles in the universe. Energy is conserved. This means that in any

More information

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics. Physics 8.01 Fall Problem Set 2: Applications of Newton s Second Law Solutions

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics. Physics 8.01 Fall Problem Set 2: Applications of Newton s Second Law Solutions MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01 Fall 2012 Problem 1 Problem Set 2: Applications of Newton s Second Law Solutions (a) The static friction force f s can have a magnitude

More information

F 2 = 26 N.What third force will cause the object to be in equilibrium (acceleration equals zero)?

F 2 = 26 N.What third force will cause the object to be in equilibrium (acceleration equals zero)? FLEX Physical Science AP Physics C Mechanics - Midterm 1) If you set the cruise control of your car to a certain speed and take a turn, the speed of the car will remain the same. Is the car accelerating?

More information

66 Chapter 6: FORCE AND MOTION II

66 Chapter 6: FORCE AND MOTION II Chapter 6: FORCE AND MOTION II 1 A brick slides on a horizontal surface Which of the following will increase the magnitude of the frictional force on it? A Putting a second brick on top B Decreasing the

More information

Physics for Scientists and Engineers 4th Edition, 2017

Physics for Scientists and Engineers 4th Edition, 2017 A Correlation of Physics for Scientists and Engineers 4th Edition, 2017 To the AP Physics C: Mechanics Course Descriptions AP is a trademark registered and/or owned by the College Board, which was not

More information

Inclined plane with protractor and pulley, roller, weight box, spring balance, spirit level, pan and thread.

Inclined plane with protractor and pulley, roller, weight box, spring balance, spirit level, pan and thread. To find the downward force, along an inclined plane, acting on a roller due to gravity and study its relationship with the angle of inclination by plotting graph between force and sin θ. Inclined plane

More information

AP PHYSICS 1. Energy 2016 EDITION

AP PHYSICS 1. Energy 2016 EDITION AP PHYSICS 1 Energy 2016 EDITION Copyright 2016 National Math + Initiative, Dallas, Texas. All rights reserved. Visit us online at www.nms.org. 1 Pre-Assessment Questions Consider a system which could

More information

Dynamics 4600:203 Homework 09 Due: April 04, 2008 Name:

Dynamics 4600:203 Homework 09 Due: April 04, 2008 Name: Dynamics 4600:03 Homework 09 Due: April 04, 008 Name: Please denote your answers clearly, i.e., box in, star, etc., and write neatly. There are no points for small, messy, unreadable work... please use

More information

Motion. Definition a change of position

Motion. Definition a change of position Potential energy Definition stored energy an object has because of its position Characteristics the higher up an object is, the greater its potential energy Example book sitting on the desk Kinetic energy

More information

Circular Orbits. Slide Pearson Education, Inc.

Circular Orbits. Slide Pearson Education, Inc. Circular Orbits The figure shows a perfectly smooth, spherical, airless planet with one tower of height h. A projectile is launched parallel to the ground with speed v 0. If v 0 is very small, as in trajectory

More information

Section 6.1 Angles and Radian Measure Review If you measured the distance around a circle in terms of its radius, what is the unit of measure?

Section 6.1 Angles and Radian Measure Review If you measured the distance around a circle in terms of its radius, what is the unit of measure? Section 6.1 Angles and Radian Measure Review If you measured the distance around a circle in terms of its radius, what is the unit of measure? In relationship to a circle, if I go half way around the edge

More information

Physics. Assignment-1(UNITS AND MEASUREMENT)

Physics. Assignment-1(UNITS AND MEASUREMENT) Assignment-1(UNITS AND MEASUREMENT) 1. Define physical quantity and write steps for measurement. 2. What are fundamental units and derived units? 3. List the seven basic and two supplementary physical

More information

Aristotle s Ideas of Motion. Conceptual Physics 11 th Edition. Galileo s Concept of Inertia. Aristotle s Ideas of Motion. Galileo s Concept of Inertia

Aristotle s Ideas of Motion. Conceptual Physics 11 th Edition. Galileo s Concept of Inertia. Aristotle s Ideas of Motion. Galileo s Concept of Inertia Aristotle s Ideas of Motion Conceptual Physics 11 th Edition Chapter 2: NEWTON S FIRST LAW OF MOTION INERTIA Natural motion (continued) Straight up or straight down for all things on Earth. Beyond Earth,

More information

UNIVERSITY OF MANITOBA

UNIVERSITY OF MANITOBA PAGE NO.: 1 of 6 + Formula Sheet Equal marks for all questions. No marks are subtracted for wrong answers. Record all answers on the computer score sheet provided. USE PENCIL ONLY! Black pen will look

More information

Level 3 Cambridge Technical in Engineering

Level 3 Cambridge Technical in Engineering Oxford Cambridge and RSA Level 3 Cambridge Technical in Engineering 05822/05823/05824/05825 Unit 3: Principles of mechanical engineering Sample Assessment Material Date - Morning/Afternoon Time allowed:

More information

Objectives. Power in Translational Systems 298 CHAPTER 6 POWER

Objectives. Power in Translational Systems 298 CHAPTER 6 POWER Objectives Explain the relationship between power and work. Explain the relationship between power, force, and speed for an object in translational motion. Calculate a device s efficiency in terms of the

More information

E 490 FE Exam Prep. Engineering Mechanics

E 490 FE Exam Prep. Engineering Mechanics E 490 FE Exam Prep Engineering Mechanics 2008 E 490 Course Topics Statics Newton s Laws of Motion Resultant Force Systems Moment of Forces and Couples Equilibrium Pulley Systems Trusses Centroid of an

More information

AP* Physics B: Newton s Laws YOU MAY USE YOUR CALCULATOR

AP* Physics B: Newton s Laws YOU MAY USE YOUR CALCULATOR AP* Physics B: Newton s Laws Name: Period: YOU MAY USE YOUR CALCULATOR CLEARLY SHOW THE METHOD YOU USED AND STEPS INVOLVED IN ARRIVING AT YOUR ANSWERS. It is to your advantage to do this, because you may

More information

Test, Lesson 7 Waves - Answer Key Page 1

Test, Lesson 7 Waves - Answer Key Page 1 Test, Lesson 7 Waves - Answer Key Page 1 1. Match the proper units with the following: W. wavelength 1. nm F. frequency 2. /sec V. velocity 3. m 4. ms -1 5. Hz 6. m/sec (A) W: 1, 3 F: 2, 4, 5 V: 6 (B)

More information

PHYS 124 Section A1 Mid-Term Examination Spring 2006 SOLUTIONS

PHYS 124 Section A1 Mid-Term Examination Spring 2006 SOLUTIONS PHYS 14 Section A1 Mid-Term Examination Spring 006 SOLUTIONS Name Student ID Number Instructor Marc de Montigny Date Monday, May 15, 006 Duration 60 minutes Instructions Items allowed: pen or pencil, calculator

More information

Engineering Mechanics Prof. U. S. Dixit Department of Mechanical Engineering Indian Institute of Technology, Guwahati Introduction to vibration

Engineering Mechanics Prof. U. S. Dixit Department of Mechanical Engineering Indian Institute of Technology, Guwahati Introduction to vibration Engineering Mechanics Prof. U. S. Dixit Department of Mechanical Engineering Indian Institute of Technology, Guwahati Introduction to vibration Module 15 Lecture 38 Vibration of Rigid Bodies Part-1 Today,

More information

Physics 20 Practice Problems for Exam 1 Fall 2014

Physics 20 Practice Problems for Exam 1 Fall 2014 Physics 20 Practice Problems for Exam 1 Fall 2014 Multiple Choice Short Questions (1 pt ea.) Circle the best answer. 1. An apple falls from a tree and hits the ground 5 meters below. It hits the ground

More information

Circular motion minutes. 62 marks. theonlinephysicstutor.com. facebook.com/theonlinephysicstutor Page 1 of 22. Name: Class: Date: Time: Marks:

Circular motion minutes. 62 marks. theonlinephysicstutor.com. facebook.com/theonlinephysicstutor Page 1 of 22. Name: Class: Date: Time: Marks: Circular motion 2 Name: Class: Date: Time: 67 minutes Marks: 62 marks Comments: Page 1 of 22 1 A lead ball of mass 0.25 kg is swung round on the end of a string so that the ball moves in a horizontal circle

More information

DSC HW 3: Assigned 6/25/11, Due 7/2/12 Page 1

DSC HW 3: Assigned 6/25/11, Due 7/2/12 Page 1 DSC HW 3: Assigned 6/25/11, Due 7/2/12 Page 1 Problem 1 (Motor-Fan): A motor and fan are to be connected as shown in Figure 1. The torque-speed characteristics of the motor and fan are plotted on the same

More information

Final Exam April 26, 2016

Final Exam April 26, 2016 PHYS 050 Spring 016 Name: Final Exam April 6, 016 INSTRUCTIONS: a) No books or notes are permitted. b) You may use a calculator. c) You must solve all problems beginning with the equations on the Information

More information

WORK.ENERGY.POWER. Challenging MCQ questions by The Physics Cafe. Compiled and selected by The Physics Cafe

WORK.ENERGY.POWER. Challenging MCQ questions by The Physics Cafe. Compiled and selected by The Physics Cafe WORK.ENERGY.POWER hallenging MQ questions by The Physics afe ompiled and selected by The Physics afe 1 A mass on a smooth horizontal table is attached by two light springs to two fixed supports. The mass

More information

AP Physics First Nine Weeks Review

AP Physics First Nine Weeks Review AP Physics First Nine Weeks Review 1. If F1 is the magnitude of the force exerted by the Earth on a satellite in orbit about the Earth and F2 is the magnitude of the force exerted by the satellite on the

More information

Precalculus 1st Semester Review 2013

Precalculus 1st Semester Review 2013 Precalculus 1st Semester Review 2013 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Which relation is a function of x? 2. Given, find. 3. Find the zero(s)

More information

University of Guelph. Department of Physics

University of Guelph. Department of Physics Surname Given Names Student Number University of Guelph Department of Physics PHYS*1020DE Introductory Physics Instructor: R.L. Brooks Midterm Examination 26 February 2003 90 Minutes INSTRUCTIONS: This

More information

Pre-Comp Review Questions- 8 th Grade

Pre-Comp Review Questions- 8 th Grade Pre-Comp Review Questions- 8 th Grade Section 1- Units 1. Fill in the missing SI and English Units Measurement SI Unit SI Symbol English Unit English Symbol Time second s. Temperature K Fahrenheit Length

More information

Applied Fluid Mechanics

Applied Fluid Mechanics Applied Fluid Mechanics 1. The Nature of Fluid and the Study of Fluid Mechanics 2. Viscosity of Fluid 3. Pressure Measurement 4. Forces Due to Static Fluid 5. Buoyancy and Stability 6. Flow of Fluid and

More information

The Rheology Handbook

The Rheology Handbook Thomas G. Mezger The Rheology Handbook For users of rotational and oscillatory rheometers 2nd revised edition 10 Contents Contents 1 Introduction 16 1.1 Rheology, rheometry and viscoelasticity 16 1.2 Deformation

More information

Pressure Head: Pressure head is the height of a column of water that would exert a unit pressure equal to the pressure of the water.

Pressure Head: Pressure head is the height of a column of water that would exert a unit pressure equal to the pressure of the water. Design Manual Chapter - Stormwater D - Storm Sewer Design D- Storm Sewer Sizing A. Introduction The purpose of this section is to outline the basic hydraulic principles in order to determine the storm

More information

Chapter 4. Answer Key. Physics Lab Sample Data. Mini Lab Worksheet. Tug-of-War Challenge. b. Since the rocket takes off from the ground, d i

Chapter 4. Answer Key. Physics Lab Sample Data. Mini Lab Worksheet. Tug-of-War Challenge. b. Since the rocket takes off from the ground, d i Chapter 3 continued b. Since the rocket takes off from the ground, d i 0.0 m, and at its highest point, v f 0.0 m/s. v f v i a t f (d f d i ) 0 v i a t f d f v i d f a t f (450 m/s) ( 9.80 m/s )(4.6 s)

More information

Mechanisms Simple Machines. Lever, Wheel and Axle, & Pulley

Mechanisms Simple Machines. Lever, Wheel and Axle, & Pulley Mechanisms Simple Machines Lever, Wheel and Axle, & Pulley Simple Machines Mechanisms that manipulate magnitude of force and distance. The Six Simple Machines Lever Wheel and Axle Pulley The Six Simple

More information

PUMP SYSTEM ANALYSIS AND SIZING. BY JACQUES CHAURETTE p. eng.

PUMP SYSTEM ANALYSIS AND SIZING. BY JACQUES CHAURETTE p. eng. PUMP SYSTEM ANALYSIS AND SIZING BY JACQUES CHAURETTE p. eng. 5 th Edition February 2003 Published by Fluide Design Inc. www.fluidedesign.com Copyright 1994 I TABLE OF CONTENTS Introduction Symbols Chapter

More information

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Diagram 1 A) B - A. B) A - B. C) A + B. D) A B.

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Diagram 1 A) B - A. B) A - B. C) A + B. D) A B. Exam Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) In the diagram shown, the unknown vector is 1) Diagram 1 A) B - A. B) A - B. C) A + B.

More information

Conveyor Chain Exclusive for Specific Conveyance. Conveyor Chain with Attachments for Conveying Bulk Materials

Conveyor Chain Exclusive for Specific Conveyance. Conveyor Chain with Attachments for Conveying Bulk Materials Exclusive for Specific Conveyance The previous section describes that by combining with various attachments, the DK can be used for almost all general applications. This section describes the developed

More information

AP PHYSICS 1 UNIT 4 / FINAL 1 PRACTICE TEST

AP PHYSICS 1 UNIT 4 / FINAL 1 PRACTICE TEST AP PHYSICS 1 UNIT 4 / FINAL 1 PRACTICE TEST NAME FREE RESPONSE PROBLEMS Put all answers on this test. Show your work for partial credit. Circle or box your answers. Include the correct units and the correct

More information

Practice Exam 2. Multiple Choice Identify the choice that best completes the statement or answers the question.

Practice Exam 2. Multiple Choice Identify the choice that best completes the statement or answers the question. Practice Exam 2 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. A roller-coaster car has a mass of 500.0 kg when fully loaded with passengers. At the bottom

More information

Potential Energy and Conservation of Energy Chap. 7 & 8

Potential Energy and Conservation of Energy Chap. 7 & 8 Level : AP Physics Potential Energy and Conservation of Energy Chap. 7 & 8 Potential Energy of a System see p.191 in the textbook - Potential energy is the energy associated with the arrangement of a system

More information

Lecture 18. In other words, if you double the stress, you double the resulting strain.

Lecture 18. In other words, if you double the stress, you double the resulting strain. Lecture 18 Stress and Strain and Springs Simple Harmonic Motion Cutnell+Johnson: 10.1-10.4,10.7-10.8 Stress and Strain and Springs So far we ve dealt with rigid objects. A rigid object doesn t change shape

More information

is acting on a body of mass m = 3.0 kg and changes its velocity from an initial

is acting on a body of mass m = 3.0 kg and changes its velocity from an initial PHYS 101 second major Exam Term 102 (Zero Version) Q1. A 15.0-kg block is pulled over a rough, horizontal surface by a constant force of 70.0 N acting at an angle of 20.0 above the horizontal. The block

More information

PVP2006-ICPVT

PVP2006-ICPVT Proceedings of PVP26-ICPVT-11 26 ASME Pressure Vessels and Piping Division Conference July 23-27, 26, Vancouver, BC, Canada PVP26-ICPVT-11-93731 REINVESTIGATION OF A HEAT EXCHANGE FLANGE LEAK Mike Porter

More information

l Every object in a state of uniform motion tends to remain in that state of motion unless an

l Every object in a state of uniform motion tends to remain in that state of motion unless an Motion and Machine Unit Notes DO NOT LOSE! Name: Energy Ability to do work To cause something to change move or directions Energy cannot be created or destroyed, but transferred from one form to another.

More information

CHAPTER 6 TEST REVIEW -- MARKSCHEME

CHAPTER 6 TEST REVIEW -- MARKSCHEME Force (N) AP PHYSICS Name: Period: Date: 50 Multiple Choice 45 Single Response 5 Multi-Response Free Response 3 Short Free Response 2 Long Free Response DEVIL PHYSICS BADDEST CLASS ON CAMPUS AP EXAM CHAPTER

More information

PRACTICE TEST for Midterm Exam

PRACTICE TEST for Midterm Exam South Pasadena AP Physics PRACTICE TEST for Midterm Exam FORMULAS Name Period Date / / d = vt d = v o t + ½ at 2 d = v o + v 2 t v = v o + at v 2 = v 2 o + 2ad v = v x 2 + v y 2 = tan 1 v y v v x = v cos

More information

Exam 1 Solutions. PHY 2048 Spring 2014 Acosta, Rinzler. Note that there are several variations of some problems, indicated by choices in parentheses.

Exam 1 Solutions. PHY 2048 Spring 2014 Acosta, Rinzler. Note that there are several variations of some problems, indicated by choices in parentheses. Exam 1 Solutions Note that there are several variations of some problems, indicated by choices in parentheses. Problem 1 Let vector a! = 4î + 3 ĵ and vector b! = î + 2 ĵ (or b! = î + 4 ĵ ). What is the

More information

NEWTON S LAWS OF MOTION, EQUATIONS OF MOTION, & EQUATIONS OF MOTION FOR A SYSTEM OF PARTICLES

NEWTON S LAWS OF MOTION, EQUATIONS OF MOTION, & EQUATIONS OF MOTION FOR A SYSTEM OF PARTICLES NEWTON S LAWS OF MOTION, EQUATIONS OF MOTION, & EQUATIONS OF MOTION FOR A SYSTEM OF PARTICLES Objectives: Students will be able to: 1. Write the equation of motion for an accelerating body. 2. Draw the

More information

Unit 1 Lesson 1.1 Mechanisms. Simple Machines. The Six Simple Machines. The Six Simple Machines. Project Lead The Way, Inc.

Unit 1 Lesson 1.1 Mechanisms. Simple Machines. The Six Simple Machines. The Six Simple Machines. Project Lead The Way, Inc. Mechanisms Simple Machines Lever, Wheel and Axle, and Pulley 2012 Simple Machines Mechanisms that manipulate magnitude of force and distance. The Six Simple Machines Lever Wheel and Axle Pulley The Six

More information

INTERNATIONAL INDIAN SCHOOL DAMMAM

INTERNATIONAL INDIAN SCHOOL DAMMAM INTERNATIONAL INDIAN SCHOOL DAMMAM SECOND TERMINAL EXAMINATION 2011 Class - XI Subject - Physics (Theory) Time: 3 hours Max. Marks 70 General Instructions 1. All questions are compulsory. Symbols have

More information

TABLE OF CONTENTS SECTION TITLE PAGE 2 PRINCIPLES OF SEISMIC ISOLATION OF BRIDGES 3

TABLE OF CONTENTS SECTION TITLE PAGE 2 PRINCIPLES OF SEISMIC ISOLATION OF BRIDGES 3 TABLE OF CONTENTS SECTION TITLE PAGE 1 INTRODUCTION 1 2 PRINCIPLES OF SEISMIC ISOLATION OF BRIDGES 3 3 ANALYSIS METHODS OF SEISMICALLY ISOLATED BRIDGES 5 3.1 Introduction 5 3.2 Loadings for the Analysis

More information

Forces. Unit 2. Why are forces important? In this Unit, you will learn: Key words. Previously PHYSICS 219

Forces. Unit 2. Why are forces important? In this Unit, you will learn: Key words. Previously PHYSICS 219 Previously Remember From Page 218 Forces are pushes and pulls that can move or squash objects. An object s speed is the distance it travels every second; if its speed increases, it is accelerating. Unit

More information

3. Kinetics of Particles

3. Kinetics of Particles 3. Kinetics of Particles 3.1 Force, Mass and Acceleration 3.3 Impulse and Momentum 3.4 Impact 1 3.1 Force, Mass and Acceleration We draw two important conclusions from the results of the experiments. First,

More information

Forces & Motion Balanced & Unbalanced Forces, Newton s First Law

Forces & Motion Balanced & Unbalanced Forces, Newton s First Law Forces & Motion 11.1 Balanced & Unbalanced Forces, Newton s First Law Forces Change Motion Force: push or pull Any time you change the motion of an object you use a force. 3 Major Types of forces: 1. Contact

More information

AP Physics C: Mechanics

AP Physics C: Mechanics 2017 AP Physics C: Mechanics Scoring Guidelines 2017 The College Board. College Board, Advanced Placement Program, AP, AP Central, and the acorn logo are registered trademarks of the College Board. Visit

More information

Physics 207 Lecture 9. Lecture 9

Physics 207 Lecture 9. Lecture 9 Lecture 9 Today: Review session Assignment: For Thursday, Read Chapter 8, first four sections Exam Wed., Feb. 18 th from 7:15-8:45 PM Chapters 1-7 One 8½ X 11 note sheet and a calculator (for trig.) Place:

More information

Lecture 4. Newton s 3rd law and Friction

Lecture 4. Newton s 3rd law and Friction Lecture 4 Newton s 3rd law and Friction Newtons First Law or Law of Inertia If no net external force is applied to an object, its velocity will remain constant ("inert"). OR A body cannot change its state

More information

i. Indicate on the figure the point P at which the maximum speed of the car is attained. ii. Calculate the value vmax of this maximum speed.

i. Indicate on the figure the point P at which the maximum speed of the car is attained. ii. Calculate the value vmax of this maximum speed. 1. A 0.20 kg object moves along a straight line. The net force acting on the object varies with the object's displacement as shown in the graph above. The object starts from rest at displacement x = 0

More information

Dynamics of Machinery

Dynamics of Machinery Dynamics of Machinery Two Mark Questions & Answers Varun B Page 1 Force Analysis 1. Define inertia force. Inertia force is an imaginary force, which when acts upon a rigid body, brings it to an equilibrium

More information

Lecture 10 Mechanical Energy Conservation; Power

Lecture 10 Mechanical Energy Conservation; Power Potential energy Basic energy Lecture 10 Mechanical Energy Conservation; Power ACT: Zero net work The system of pulleys shown below is used to lift a bag of mass M at constant speed a distance h from the

More information

Dynamics Review Outline

Dynamics Review Outline Dynamics Review Outline 2.1.1-C Newton s Laws of Motion 2.1 Contact Forces First Law (Inertia) objects tend to remain in their current state of motion (at rest of moving at a constant velocity) until acted

More information

Torsion of shafts with circular symmetry

Torsion of shafts with circular symmetry orsion of shafts with circular symmetry Introduction Consider a uniform bar which is subject to a torque, eg through the action of two forces F separated by distance d, hence Fd orsion is the resultant

More information

Chapter 1: Trigonometric Functions 1. Find (a) the complement and (b) the supplement of 61. Show all work and / or support your answer.

Chapter 1: Trigonometric Functions 1. Find (a) the complement and (b) the supplement of 61. Show all work and / or support your answer. Trig Exam Review F07 O Brien Trigonometry Exam Review: Chapters,, To adequately prepare for the exam, try to work these review problems using only the trigonometry knowledge which you have internalized

More information

N - W = 0. + F = m a ; N = W. Fs = 0.7W r. Ans. r = 9.32 m

N - W = 0. + F = m a ; N = W. Fs = 0.7W r. Ans. r = 9.32 m 91962_05_R1_p0479-0512 6/5/09 3:53 PM Page 479 R1 1. The ball is thrown horizontally with a speed of 8 m>s. Find the equation of the path, y = f(x), and then determine the ball s velocity and the normal

More information

Physics 121, Final Exam Do not turn the pages of the exam until you are instructed to do so.

Physics 121, Final Exam Do not turn the pages of the exam until you are instructed to do so. , Final Exam Do not turn the pages of the exam until you are instructed to do so. You are responsible for reading the following rules carefully before beginning. Exam rules: You may use only a writing

More information

Pre Comp Review Questions 7 th Grade

Pre Comp Review Questions 7 th Grade Pre Comp Review Questions 7 th Grade Section 1 Units 1. Fill in the missing SI and English Units Measurement SI Unit SI Symbol English Unit English Symbol Time second s second s. Temperature Kelvin K Fahrenheit

More information

Physics I (Navitas) EXAM #2 Spring 2015

Physics I (Navitas) EXAM #2 Spring 2015 95.141 Physics I (Navitas) EXAM #2 Spring 2015 Name, Last Name First Name Student Identification Number: Write your name at the top of each page in the space provided. Answer all questions, beginning each

More information

Isaac Newton ( )

Isaac Newton ( ) Isaac Newton (1642-1727) In the beginning of 1665 I found the rule for reducing any degree of binomial to a series. The same year in May I found the method of tangents and in November the method of fluxions

More information

FORCES AND MOTION UNIT TEST. Multiple Choice: Draw a Circle Completely around the ONE BEST answer.

FORCES AND MOTION UNIT TEST. Multiple Choice: Draw a Circle Completely around the ONE BEST answer. Name: Date: Period: FORCES AND MOTION UNIT TEST Multiple Choice: Draw a Circle Completely around the ONE BEST answer. 1. A force acting on an object does no work if a. a machine is used to move the object.

More information

PHYSICS 231 INTRODUCTORY PHYSICS I

PHYSICS 231 INTRODUCTORY PHYSICS I PHYSICS 231 INTRODUCTORY PHYSICS I Lecture 6 Last Lecture: Gravity Normal forces Strings, ropes and Pulleys Today: Friction Work and Kinetic Energy Potential Energy Conservation of Energy Frictional Forces

More information

Exam 2--PHYS 101--F17

Exam 2--PHYS 101--F17 Name: Exam 2--PHYS 0--F7 Multiple Choice Identify the choice that best completes the statement or answers the question.. A ball is thrown in the air at an angle of 30 to the ground, with an initial speed

More information

(a) On the dots below that represent the students, draw and label free-body diagrams showing the forces on Student A and on Student B.

(a) On the dots below that represent the students, draw and label free-body diagrams showing the forces on Student A and on Student B. 2003 B1. (15 points) A rope of negligible mass passes over a pulley of negligible mass attached to the ceiling, as shown above. One end of the rope is held by Student A of mass 70 kg, who is at rest on

More information

Topic 2: Mechanics 2.3 Work, energy, and power

Topic 2: Mechanics 2.3 Work, energy, and power Essential idea: The fundamental concept of energy lays the basis upon which much of science is built. Nature of science: Theories: Many phenomena can be fundamentally understood through application of

More information

Foundations with D f equal to 3 to 4 times the width may be defined as shallow foundations. TWO MAIN CHARACTERISTICS ULTIMATE BEARING CAPACITY

Foundations with D f equal to 3 to 4 times the width may be defined as shallow foundations. TWO MAIN CHARACTERISTICS ULTIMATE BEARING CAPACITY oundation Analysis oundations with D f eual to 3 to 4 times the width may be defined as shallow foundations. TWO MAI CHARACTERISTICS o Safe against overall shear failure o Cannot undergo excessive displacement,

More information

Physics 201 Lecture 16

Physics 201 Lecture 16 Physics 01 Lecture 16 Agenda: l Review for exam Lecture 16 Newton s Laws Three blocks are connected on the table as shown. The table has a coefficient of kinetic friction of 0.350, the masses are m 1 =

More information