Physics 111 ectue 1 Static Equilibium SJ 8th Ed.: Chap 1.1 1.3 Oveview - Equilibium Defined Conditions fo Equilibium Cente of Gavity Defined inding it When do mass cente and CG not coincide? Static Equilibium Examples 1.1 The Rigid Object in Equilibium: The Conditions fo Equilibium 1. Moe on the Cente of Gavity 1.3 Examples of Rigid Objects in Static Equilibium 1.4 Elastic Popeties of Solids (Moduli)
The Equilibium Conditions No new physics hee apply Newton s second law Special case of nd aw with acceleations equal to zeo So net foces and toques ae zeo as well i,ext net dp dt ma p constant I { net,ext.eq. 0 } THEN a 0 fist condition met ist condition fo equilibium: i, ext 0 Identify foces, wite equations setting thei sum 0 τ d dt i,ext τ net Ι α constant I { τ net,ext.eq. 0 } THEN α 0 second condition met Second condition fo equilibium τ i, ext 0 Identify toques, wite equations setting thei sum 0
MECHANICA EQUIIBRIUM No change in tanslational o otational state ISOATED SYSTEM constant momentum, zeo net foce, constant angula momentum, zeo net toque zeo linea acceleation of paticle o cm of igid body, zeo angula acceleation of igid body DYNAMIC VERSUS STATIC EQUIIBRIUM: Dynamic: Σ 0 and Στ 0...BUT... p and/o ae not 0 Static: In addition, p 0 and 0 STABIITY O STATIC EQUIIBRIUM Imagine displacing system a small amount: does it etun? stable unstable neutal EXAMPES: book on table static neutal puck sliding on ice dynamic -- ceiling fan on dynamic -- ceiling fan off static neutal ca on staight oad dynamic -- ca on cuve not equilibium -- ladde leaning against wall static unstable (fiction) stable (foot in goove)
Definition: The cente of gavity of a body is the point about which all of the toques due to gavitational foces alone add up to zeo balance a ule see-saw a box about to tip 1 m m 1 m < m 1 Gavitational toques cancel about the CG A single suppot foce at the CG ceates zeo added toque (moment am 0)... and can cancel the weight foces (linea equilibium) CG often coincides with mass cente (CM)... but may not if gavity vaies acoss the body The actual weight can be eplaced by the weight of a single paticle at the CG A single non-gavitational nomal foce applied at the CG can poduce tanslational equilibium
Cente of gavity fo a sledge hamme 1.1. The cente of gavity fo a sledge hamme lies on the centeline of the handle, close to the head, at the X mak. Suppose you saw acoss the handle though the cente of gavity, cutting the ax in two pieces. You then weigh both pieces. Which of the following will you find? x A) The handle piece is heavie than the head piece B) The head piece is heavie than the handle piece C) The two pieces ae equally heavy D) The compaative weights depend on moe infomation Definition: The cente of gavity of a body is the point about which all of the toques due to gavitational foces sum to zeo
Cente of gavity: Summay The sum of gavitational foces acting on all the vaious mass elements is equivalent to a single gavitational foce (the total weight) acting though a single point called the cente of gavity (CG) The net toque due to gavitational foces on an object of mass M equals the foce Mg acting at the cente of gavity of the object. If g is unifom ove the object, then the cente of gavity of the object coincides with its cente of mass. If the object has constant density and is symmetical, the cente of gavity coincides with its geometic cente.
Example: ind the balance point (CG) fo a level see-saw x m 1 m s balance point CG BD N s - s m 1 g m g Balance means: ind the CG and put a suppot foce N ( total weight) thee: inea acceleation is zeo ( static tanslational equilibium ) Angula acceleation is also zeo ( otational equilibium ) To locate CG coodinate s Pick likely CG location. Set sum of toques aound it 0, including only gavitational foces Suppot foce N applied at CG exets zeo toque, povides linea equilibium 1) τ cg 0 1 + m gs m g( s) N x 0 s m ( m 1 + m ) locates cg Ex: et m 1 m then s 1/3 ) y 0 (m 1 + m )a 0 N m 1 g m g N (m 1 + m ) g total weight If an object is in equilibium, then the net toque 0 fo any axis chosen (moe below). o an axis not at the CG, toque due to N is not zeo but still cancels Copyight the gavitational R. Janow Sping toques 01
ocating CG s of igid bodies pevious suspension point T -Pick body up (extenal foce T) and let it swing feely, come to equilibium position, and stop (fiction). - Body comes to est with CG below o above suspension point, o/w toque would not zeo and body would not be in equilibium. - Mak a vetical line at suspension point - Repeat pocess fo a diffeent suspension point - The CG is whee the lines intesect. If gavitation field is unifom, CG & CM coincide but may not be at geometical symmety points unless density is unifom. o unifom sphee, cube, disc, ectangle,.. CG at cente o cone. cylinde, ba, etc CG along axis of symmety o composite objects, beak into shapes with symmety STABIITY WEIGHT ACTS AT CG W N Block tips if CG is to the left of all suppot points in base. Net toque then can not 0
o Equilibium Poblems: Does it matte which otation axis you choose fo calculating toques? No. Choose any convenient axis. I net 0 THEN τ net is the same aound evey paallel axis..so... I τ net 0 fo one axis THEN it is zeo fo evey othe axis O Poof: a v O v i Compae otation axis at O with one shifted to O fo a system of point paticles { m i } v ' i v v O i p i angula m i i i ' + a shift axis v ' v v ' v [ i + a] p i i p i + a momentum O p by a to O' v i about oigin axis O angula momentum elative to axis O angula momentum elative to axis O o' linea momentum P v of mass cente cm Take the time deivative of each tem, use second law (otational and linea) dt d dt v v If i 0 then τ net a dp dt d O O' cm τ v net, o o' ext + τ net, v is the same about any axes O o O' Note also: if P cm 0, the angula momentum is the same aound all paallel axes net, Rightmost tem vanishes if net Σ i 0
Rules fo solving equilibium poblems ist Condition: Second Condition: net 0 τ net 0 Choose any convenient axis fo toque calculation. ist condition implies that esult will be the same (see poof) fo any otation axis Calculate toques due to weight of igid bodies by placing thei weights at thei cente of gavity locations (see poof) f s µ s Use static fiction foce whee needed: N o plane statics (flat wold) All foces lie in x-y plane. z always equals zeo 0 i,x, net,y i, y net,x 0 All toques lie along +/- z axis. τ x, τ y 0 τ net,z τ i, z 0
When can a system be in equilibium? 1.. The figue shows five sketches of a unifom od with two o moe foces acting pependicula to it. The magnitudes of the foces can be adjusted to any value needed, except zeo. o which of the sketches can the od be in tanslational equilibium? 1 A) 1,,3,4,5 B) 4 C), 3, 4, 5 D) 3, 4, 5 E) 4, 5 3 1.3. o which of the situations can the od be in oveall static equilibium? 4 Net toque is the same fo any convenient axis when net foce 0 5
Method fo Solving Equilibium Poblems ESSENTIAY THE SAME AS OR SOVING SECOND AW PROBEMS Daw sketch, label it, decide what is in o out of the system. Daw fee body diagams. Include foces ON the body being analyzed. Show point of application to indicate toques. Choose axes and the +/- sense fo otations. Replace foces by thei x, y, z components Choose a otation axis/point to use in calculating toques. o All choices yield the same net toque, so long as the ist equilibium condition applies but... o Some choices simplify the solution. ook fo ways to give zeo moment am (zeo toque) to ielevant 0 0 o toublesome foces. Apply x, y, τ 0 ( dimensions) z Substitute the actual foces and toques into the equations. Count the unknowns; make sue thee ae N equations when thee ae N unknowns. Constaint equations ae often needed. Solve the set of simultaneous equations algebaically as fa as is easonable befoe substituting numbes. Ty to intepet esulting equations intuitively. Check that numeical answes make sense, have easonable magnitudes, physical units, etc.
Example: Mechanical Advantage of a pulley system The pulley system is used to aise a weight slowly at constant speed ( a 0 ). Use the equilibium conditions and fee body diagams to find the tension in each cable and the lifting foce. The pulleys ae massless and fictionless. Solution equies fist condition only At each pulley, the sum of upwad the sum of downwad foces T 3 T T T 3 T 4 T 3 T B T A T A T A T 1 T 1 W 9800 N T T T 4900 N 1 T 3 T T 3 T T T 4 3 4 450 4900 N N T B W 9800 N Mechanical advantage (weight lifted/foce exeted) W T 3 9800 450 4
Example: Angle of the chai lift cable θ The chai lift is at the middle of the cable span as shown. It s weight causes the cable to deflect by an angle θ fom the hoizontal on both sides of the chai. The pulleys ae massless and fictionless. The skie weighs 78 kg. W 00 N ind the angle θ when the hanging weight is 00 N as shown. m 78 kg ee body diagam θ T T θ T is unifom and T 00 N y 0 T sin( θ ) 78 x 9. 8 sin( θ ) 0. 1737 T x 00 θ 10. 0 o
Example: Distibuting weight between font and ea wheels of a ca PP10603-08: A ca whose mass 1360 kg has 3.05 m between its font and ea axles. Its cente of gavity is located 1.78 m behind the font axle. The ca is stationay on level gound. ind the magnitude of the foce fom the gound on each font and ea wheel (assuming equal foces on both sides of the ca). Solution: apply fist and second equilibium conditions. Calculate toques using axis though ea wheel contact point with the gound. 3.05 m R 1360 kg x 9.8 13,38 N. R and ae foces on each wheel y 0 R + x 0 τ o R x 0 ( - d) + x d 1.78 m 0 Toques about ea wheels ( - d) - R 13,38 x (3.05 - x 3.05 13,38-5550 1.78) 775 N. R 3889 N.
Example: A mete stick balances hoizontally on a knife-edge at the 50.0 cm mak. When two 5.0 g coins ae attached ove the 1.0 cm mak the stick now balances at the 45.5 cm mak. What is the mass of the mete stick? Poblem PP10603-11: 0 cm 100 cm x 1 x x 3 N Mg m x 5 gm.01 kg x 1 0.1 m x.455 m X 3 0.5 m M mete stick mass APPY EQUIIBRIUM CONDITIONS: y 0 N Mg N + Mg Choose axis fo toques at 0 cm (left end of stick). Any othe axis yields same net toque Mete stick alone acts like a mass M concentated at its own CG at x 3 0.5 m τ 0 o 0 0 M - x 1 + Nx Mgx 3 - x 1 + x + Mgx Mgx 3 m[ x - x 1 ] + M[ x x 3 ] m[ x - x 1 ]. 455. 1 0. 01 7. 44 [ x x ]. 500. 455 x 3 M 74. 4 gm 10
ind the foces on the pivot 1.4. The sketches show fou ovehead views of unifom disks that can slide o otate on a fictionless floo. Thee foces act on each disk, eithe at the im o at the cente. The foce vectos otate along with the disks, so the foce aangements emain the same. Which disks ae in equilibium? 1 3 3 4 A) 1,, 3, 4 B) 1, 3, 4 C) 3 D) 1, 4 E) 3, 4 0 τ net 0 net
O Example 1: Massless beam suppoting a weight The.4 m. long weightless beam shown in the figue is suppoted on the ight by a cable that makes an angle of 50 o with the hoizontal beam. A 3 kg mass hangs fom the beam 1.5 m fom the pivot point on the left. a) Calculate the toque caused by the hanging mass. b) Detemine the cable tension that poduces equilibium θ 50 o, x x m 3 kg 1.5 m,.4 Choose: otation axis at O Daw BD of beam m y x o x θ T Pat a) τ Pat b) mass x 3 x 9. 8 x 1. 5 470 τ 0 T y o - 0 x fo + component T equilibiu m y of T to beam Tsin( θ ) N.m T x sin( θ ) 56 N.
Example 1, continued: ind the foces on the pivot o the peceding poblem, find the x component of the foce at the pivot point. What equations can you use? y x o x x 0 θ T T x Tcos( θ ) x - x 165 N o θ 50, x 1.5 m, T x sin( θ ) 56 N..4 m ist Condition: Second Condition: net 0 τ net 0 o the peceding poblem, find the y component of the foce at the pivot point. What equations can you use? y 0 y - + Tsin( θ ) y - Tsin( θ ) 117 N
Example : As in pevious poblem but beam has mass now O o x x y x / M B g m 3 kg BD of beam θ T The.4 m. long beam shown in the figue is suppoted on the ight by a cable that makes an angle of 50 o with the hoizontal beam. A 3 kg mass hangs fom the beam 1.5 m fom the pivot point on the left. M B mass of beam 30 kg acts as point mass at CG of beam. Choose: otation axis at O a) Detemine the cable tension needed to poduce equilibium b) ind the foces at point O θ 50 o Pat a), x τ 0 T y T 1.5 o - m, 0 x component x fo M + sin( B of M.4 m, equilibiu m g T B θ ) g + to T CG at / y beam 448 N. Tsin( θ ) Pat b) x 0 x T cos( θ ) x T cos( θ ) 88 N. y 0 y + T sin( θ ) M B g B y + M g T sin( θ ) 64 N.
Example 3: A unifom beam, of length and mass m 1.8 kg, is at est with its ends on two scales (see figue). A unifom block, with mass M.7 kg, is at est on the beam, with its cente a distance / 4 fom the beam's left end. What do the scales ead? Solve as equilibium system Axis at o is convenient Beam weight acts as if it is at CG of beam (x /) o BD
Example 3: Solution Scale eadings ae nomal foces and R 1: : 3:,block Mg g, beam g x 0 no impact R y 0 + Mg Choose axis fo toques at left of beam τ Solve 0 x 0 - Mg - + R 4 R Mg + Mg 4.7 1.8 R + + R 15 N. 4 x 9.8 4 x 9.8 15.44 om : Mg 3 1 Mg + Mg + 4 4 8.68 9 N. Block and beam epesented as paticles at thei CMs (CGs)
Example 4: A safe whose mass is M 430 kg is hanging by a ope fom a boom with dimensions a 1.9 m and b.5 m. The boom consists of a hinged beam and a hoizontal cable that connects the beam to a wall. The unifom beam has a mass m of 85 kg; the mass of the cable and ope ae negligible. (a) What is the tension T c in the cable; i. e., what is the magnitude of the foce T c on the beam fom the hoizontal cable? (b) What is the magnitude of the foce at the hinge? v T c T h
Example 4 Solution The beam is the system to analyze The beam s weight acts at it s CG x 0 h T c h T y 0 v Mg Mg T Choose hinge point O as axis fo toques - Eliminates v and h in toque equation τ o 0 T c a T b m 85 Tc [ + M ] g [ + T c 6093 b a N 430 b ] c 9.8 T c x h.5 1.7 v 6093 v + Mg (85 + v 5047 N 430) 9.8 [ 1/ h + ] v 791 N N T c T h b a cot( θ )
Example 5: Does the wheel oll ove the cub? Poblem: In the figue, foce is applied hoizontally at the axle of the wheel. The wheel's adius is and its mass is m. ind the minimum foce needed to aise the wheel ove a cub of height h. Why such lage wheels? - h s P x P y N Baie when h > Calculate toques aound contact point with cub
Example 5: Can the wheel oll ove the cub? Poblem: In the figue, foce is applied hoizontally at the axle of the wheel. The wheel's adius is and its mass is m. ind the minimum foce needed to aise the wheel ove a cub of height h. - h s BD of the Wheel Take toques aound hee Geomety: s ( h) h h as long as > Equations fo equilibium: y N + P y 0 x P x 0 τ s Ns ( h ) 0 h N P x P y o > h, gows as h. becomes infinite if h. o < h vehicle is stopped. As wheel is about to lift off and oll, N 0 P y P x ( - h) s h h h h fo > cannot pull it ove cub When does? Ans: fo h ( - h) h
Example 6: Climbing a ladde A ladde whose length 1 m and mass m 45 kg leans against a fictionless) wall. Its top is at height h 9.3 m above the pavement that the lowe end ests on. The pavement is not fictionless. The ladde's own mass cente is / 3 fom the lowe end. A fiefighte of mass M 7 kg climbs the ladde until he cente of mass is s / fom the lowe end. S ind the magnitudes of the foces on the ladde fom the wall and the pavement. ind how fa up the ladde the fiefighte can climb befoe it slips. The static fiction coefficient between ladde and pavement is 0.. a cos(θ) h sin(θ) θ h sin -1 ( -1 o θ ) sin ( 9.3/1) 50.8 cot( θ ).815 tan( θ ) 1.7
Example 5 solution Climbing a adde S S x iefighte is a distance S fom the base s x s cos( θ py nomal ) foce at base µ px static fiction foce s Mg fiefighte s ladde' weight s weight, ladde CG py at /3 w foce of wall on ladde No fiction Equilibium conditions x 0 w px y 0 py Mg Mg + (45 py + Choose axis fo toques at base of ladde O τ o 0 cos( 3 θ ) 7) + w hoizontal w px 9.8 py Mgs sin( cos( θ 1146 N ) θ )
adde Example Solution, continued Solve toque equation fo w cos( θ ) w + Mgs 3 sin( θ ) m s g W + M 3 tan( θ Note: 1. As s gows, so does W PX (fiction) needed to stay in equilibium.. As θ 0, fictional foce needed becomes huge, ladde slips Example: o S / (halfway up): Set impending motion condition: Solve toque equation fo s: m M g w + 408 N o px 3 tan( 50. 8 ) How fa up the ladde can the fiefighte climb befoe it slips, assuming µ s 0. Mgs cos( θ ) w sin( θ ) cos( θ s µ s (M + Mg m)g tan( θ ) m M. (1 + 45/7) 1 x 1.7 s. 9.3 m px µ s py µ s w + 3 3 45 7 1 3 ) ) [ M m ] g Getting halfway up equies µ s >> 0.
Example 7: Rock Climbe In the figue a ock climbe with mass m 55 kg ests duing a chimney climb, pessing only with he shouldes and feet against the walls of a fissue of width w 1.0 m. He cente of mass is a hoizontal distance d 0.0 m fom the wall against which he shouldes ae pessed. The coefficient of static fiction between he shoes and the wall is µ 1 1.1, and between he shouldes and the wall it is µ 0.70. To est, the climbe wants to minimize he hoizontal push on the walls. The minimum occus when he feet and he shouldes ae both on the vege of sliding. µ 1 µ (a) What is the minimum hoizontal push on the walls? (impending motion) (b) What should the vetical distance h be between the shouldes and feet, in ode fo he to emain stationay?
Solution of Example 7: Rock Climbe ind the minimum hoizontal push on the walls. The climbe is the system to analyze x 0 N N' N y 0 f 1 + f N' o impending slippage: f 1 µ 1 N f µ N ( µ 1 + µ ) N N ( µ + µ 1 ) 55 1.1 x 9.8 + 0.7 300 What should the vetical distance h be between shouldes and feet? Choose axis fo toques at point O N τ o 0 f x 0 + N'x 0 + d + hn f 1 w h µ Nw 1 N d µ 1 (w d) µ This choice of h poduces impending motion d 0.74m O w 1.0 m