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Unformatted text preview: Chapter 12 CHAPTER 12 Static Equilibrium; Elasticity and Fracture 1. From the force diagram for the sapling we can write ? F x = F 1 F 2 sin 20 F 3 cos = 0; 380 N (255 N) sin 20 F 3 cos = 0, or F 3 cos = 293 N. ? F y = F 2 cos 20 F 3 sin = 0; F 3 sin = (255 N) cos 20 = 240 N. Thus we have F 3 = [(293 N) 2 + (240 N) 2 ] 1/2 = 379 N . tan = (240 N)/(293 N) = 0.818, = 39.3. So = 180 = 141 . 2. From the force diagram for the junction we can write ? F x = F 2 F 1 sin 45 = 0. This shows that F 1 > F 2 , so we take F 1 to be the maximum. ? F y = F 1 cos 45 Mg = 0; Mg = (1150 N) sin 45 = 813 N . 3. We choose the coordinate system shown, with positive torques clockwise. For the torque from the persons weight about the point B we have B = MgL = (56 kg)(9.80 m/s 2 )(3.0 m) = 1.6 10 3 m N . 4. We choose the coordinate system shown, with positive torques clockwise. For the torque from the persons weight about the point A we have A = Mgx ; 1000 m N = (56 kg)(9.80 m/s 2 ) x , which gives x = 1.82 m . 5. We choose the coordinate system shown, with positive torques clockwise. We write ? = I about the point A from the force diagram for the leg: ? A = MgD F T L = 0; (15.0 kg)(9.80 m/s 2 )(0.350 m) F T (0.805 m), which gives F T = 63.9 N. Page 1 F 1 F 2 20 x y F 3 F 1 F 2 45 x y M g U s e W o r d 6 .0 c o r l a te r to v i e w M a c i n to s h p i c tu r e . U se W o rd 6.0c or l ater to vi ew M aci ntos h pic ture. U s e W o r d 6 . 0 c o r l a t e r t o v i e w M a c i n t o s h p i c t u r e . Chapter 12 Because there is no acceleration of the hanging mass, we have F T = mg , or m = F T / g = (63.9 N)/(9.80 m/s 2 ) = 6.52 kg . 6. We choose the coordinate system shown, with positive torques clockwise. We write ? = I about the support point A from the force diagram for the board and people: ? A = m 1 g ( L d ) + m 2 gd = 0; (23.0 kg)(10.0 m d ) + (67.0 kg) d = 0, which gives d = 2.56 m from the adult ....
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 Winter '07
 Whitten
 Force, Static Equilibrium

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