CHAPTER 8b Static Equilibrium Units

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1 CHAPTER 8b Static Equilibrium Units The Cnditins fr Equilibrium Slving Statics Prblems Stability and Balance Elasticity; Stress and Strain The Cnditins fr Equilibrium An bject with frces acting n it, but that is nt mving, is said t be in equilibrium. The first cnditin fr equilibrium is that the frces alng each crdinate axis add t zer. Example 1: Calculate the tensins FA and F B in the tw crds that are cnnected t the vertical crd supprting the 00-kg chandelier. Fy FAsin 60 (00 kg)( g) 0 (00 kg) g (00 kg)( g) FA (31 kg)( g) 60N sin Fx FB FA cs 60 0 F F cs 60 (31 kg)( g)(0.500) (115 kg) g 1130N B The magnitudes f FA and FB determine the strength f crd r wire that must be used. In this case, the wire must be able t hld mre than 30 kg. A The Cnditins fr Equilibrium The secnd cnditin f equilibrium is that there be n trque arund any axis; the chice f axis is arbitrary. 1

2 Slving Statics Prblems The previus technique may nt fully slve all statics prblems, but it is a gd starting pint. Example : A unifrm 1500-kg beam, 0.0m lng, supprts a 15,000-kg printing pres 5.0 m frm the right supprt clumn. Calculate the frce n each f the vertical supprt clumns. (10.0 m)(1500 kg) g (15.0 m)(15,000 kg) g (0.0 m) F B 0 F (1,000 kg) g 118,000 N B F F (1500 kg) g (15,000 kg) g F 0 y A B F (4500 kg) g 44,100N A If a frce in yur slutin cmes ut negative (as F A will here), it just means that it s in the ppsite directin frm the ne yu chse. This is trivial t fix, s dn t wrry abut getting all the signs f the frces right befre yu start slving. If there is a cable r crd in the prblem, it can supprt frces nly alng its length. Frces perpendicular t that wuld cause it t bend. Example 3: A unifrm beam,.0 m lng with mass m = 5.0 kg, is munted by a hinge n a wall. The beam is held in a hrizntal psitin by a cable that makes an angle The beam supprts a sign f mass M = 8.0 kg suspended frm its end. Determine the cmpnents f the frce FH that the hinge exerts n the beam, and the tensin FT in the supprting cable. The sum f the frces in the vertical (y) directin is F 0 The sum f the frces in the hrizntal (x) directins is F 0 y F F mg Mg 0 F Hy Hx x Ty FTx 0

3 Chse trques that tends t rtate the beam cunterclckwise as psitive. The weight mg f the beam acts as its center. Slve fr F 0 ( F )(.0 m) mg(1.10 m) 0 Hy Hy 1.10m.0m FHy mg (0.500)(5.0 kg)(9.80 m/ s ) 13N Since tensin FT in the cable acts alng the cable 30.0, tan F / F Ty Tx r F F tan F (tan 30.0 ) 0.577F Ty Tx Tx Tx F m M g F kg m s N N Ty ( ) Hy (53.0 )(9.80 / ) F F / N Tx TY F F 687N Hx Tx F F F 793N T Tx Ty Example 4: A unifrm hrizntal beam 5.00 m lng and weighing 300N is attached t a wall by a pin cnnectin that allws the beam t rtate. Its far end is supprted by a cable that makes an angle f with the hrizntal. If a persn weighing 600N stands 1.50 m frm the wall, find the tensin in the cable. i R B M T 0 0 W ( L / ) W (1.50 m) TLsin(53 ) 0 i Beam Man Substitute L = 5.00 m and the weights, slving fr T; (300 N)(.50 m) (600 N)(1.50 m) ( T sin 53.0 ) 0 T 413N F R T cs x x F R W W T sin y y Beam Man Sub the value f T and the weights; R 49 N R 570N x y 3

4 Example 5: A 5.0-m-lng ladder leans against a wall at a pint 4.0 m abve a cement flr. The ladder is unifrm and has mass m = 1.0 kg. Assuming the wall is frictinless (but the flr is nt), determine the frces exerted n the ladder by the flr and by the wall. The wall, since it is frictinless, can exert a frce nly perpendicular t the wall, and is labeled frce FW. The cement flr exerts a frce FC which has bth hrizntal and vertical frce cmpnents: FCX is frictinal and FCY is the nrmal frce. Gravity exerts a frce mg (1 kg)(9.80 m/ s ) 118N n the ladder at its midpint. The y-cmpnent f the frce equatin is Fy FCy mg 0 S we have: F mg 118N The x-cmpnent f the frce equatin is The ladder tuches the flr a distance Cy Fx FCx FW 0 x (5.0 m) (4.0 m) 3.0m frm the wall. The lever arm fr mg is half this, r 1.5m, and the lever arm fr FW is 4.0m. We get (4.0 m) F (1.5 m) mg 0 F W W (1.5 m)(1.0 kg)(9.80 m/ s ) 4.0m 44N Fr the x-cmpnent f the frce equatin F F 44N Cx W FC (44 N) (118 N) 16N 1 tan (118 N/ 44 N) 70 Example 6: A unifrm ladder 10.0 m lng and weighing 50.0 N rests against a smth vertical wall. If the ladder is just n the verge f slipping when it makes a 50.0 angle with the grund, find the cefficient f static frictin between the ladder and grund. F F P 0 F P x fr fr F F 50.0N 0 F 50.0N y n n 0 i fr Fn grav P 00 (50.0 N)(5.00 m)sin 40.0 P(10.0 m)sin P 1.0N 1.0 N F F n s (50.0 N) fr smax s s 1.0N s N 4

5 Elasticity; Stress and Strain Hke s law: the change in length is prprtinal t the applied frce. This prprtinality hlds until the frce reaches the prprtinal limit. Beynd that, the bject will still return t its riginal shape up t the elastic limit. Beynd the elastic limit, the material is permanently defrmed, and it breaks The change in length f a stretched bject depends nt nly n the applied frce, but als n its length and crss-sectinal area, and the material frm which it is made. The material factr is called Yung s mdulus, and it has been measured fr many materials. The Yung s mdulus is then the stress divided by the strain. Example 7: A 1.60-m-lng steel pian wire has a diameter f 0.0 cm. Hw great is the tensin in the wire if it stretches 0.5 cm when tightened? 6 A r (3.14)( m) 3.14x10 m L 11 F E 0.005m 6 (.0x10 N / m ) ( ) 980 L x m N 1.60m Yung s mdulus fr steel is: 9 E 00x10 N / m 5

6 Example 8: The steel pian wire frm example prblem 7, was 1.60 m lng with a diameter f 0.0 cm. apprximately what tensin frce wuld break it? 3 r 0.10cm 1.0x10 m F 6 500x10 N / m A 6 3 F (500x10 N / m )( )(1.0 x10 m) 1600N Additinal Calculatin: Example 9: A 5.00-m lng diving bard f negligible mass is supprted by tw pillars. One pillar is at the left end f the diving bard, the ther is 1.50-m away. Find the frces exerted by the pillars when a 90.0-kg diver stands at the far end f the bard. F F F mg 0 y 1y y F 1 1y F y 3 y (0) 0 ( d) mg( L) F (0) F ( d) mg(l) 0 1y y F mg L d kg m s m m 940N ( / ) (90.0 )(9.81 / )(5.00 /1.50 ) F1y mg Fy (90.0 kg)(9.81 m / s ) 940N 060N 6

7 CHAPTER 9 STATIC EQUILIBRIUM CONCEPTS 1. Stress is applied frce per crss-sectinal area.. Strain is the rati f the change in length t the riginal length. 3. A trque is an influence which tends t change the rtatinal mtin f an bject. 4. The lever arm is defined as the perpendicular distance frm the axis f rtatin t the line f actin f the frce. 5. An bject at equilibrium has n net influences t cause it t mve, either in translatin (linear mtin) r rtatin. 6. The weight that will cause a wire f diameter d t stretch a given distance, fr a fixed length f wire, is prprtinal t d The slpe f the straight line shwn the graph is called the Yung s mdulus. 9. A rcket mves thrugh uter space with a cnstant velcity 9.0 m/s. The net frce acting n it is zer. 7

8 10. A persn weighing 800 N stands with ne ft n each f tw bathrm scales. If ne scale reads 500 N, the ther will read 300 N. 11. A machine is a device used t change the magnitude r directin f a frce. 1. A machine may nt be used t create additinal energy. PHYSICSINMOTION 8

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