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Unformatted text preview: Chapter 12 Page 1 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 α = 0; 380 N – (255 N) sin 20° – F 3 cos α = 0, or F 3 cos α = 293 N. ? F y = F 2 cos 20° – F 3 sin α = 0; F 3 sin α = (255 N) cos 20° = 240 N. Thus we have F 3 = [(293 N) 2 + (240 N) 2 ] 1/2 = 379 N . tan α = (240 N)/(293 N) = 0.818, α = 39.3°. So θ = 180° – α = 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 person’s weight about the point B we have τ 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 person’s weight about the point A we have τ 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 ? τ = I α about the point A from the force diagram for the leg: ? τ 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. 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 F 1 F 2 20° θ α x y F 3 F 1 F 2 45° x y M g d L M g A Chapter 12 Page 2 kg . 6. We choose the coordinate system shown, with positive torques clockwise. We write ? τ = I α about the support point A from the force diagram for the board and people: ? τ 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 . 7. We choose the coordinate system shown, with positive torques clockwise. We write ? τ = I α about the support point A from the force diagram for the board and people: ? τ A = – m 1 g ( L – d ) – Mg ( ! L – d ) + m 2 gd = 0; – (23.0 kg)(10 m – d ) – (12.0 kg)(5.0 m – d ) + (67.0 kg) d = 0, which gives d = 2.84 m from the adult . 8. ( a ) We choose the coordinate system shown, with positive torques clockwise. For the torques about the point B we have ? τ B = F 1 d + MgD = 0; F 1 (1.0 m) + (56 kg)(9.80 m/s 2 )(3.0 m) = 0, which gives F 1 = – 1.6 × 10 3 N (down) . For the torques about the point A we have ? τ A = – F 2 d + Mg ( D + d )= 0; F 2 (1.0 m) = (56 kg)(9.80 m/s 2 )(3.0 m + 1.0 m), which gives F 2 = 2.2 × 10 3 N (up) . ( b ) For the torques about the point B we have ? τ B = F 1 d + MgD + mg [ ! ( D + d ) – d ] = 0; F 1 (1.0 m) = – (56 kg)(9.80 m/s 2 )(3.0 m) – (35 kg)(9.80 m/s 2 )(1.0 m), which gives F 1 = – 2.0 × 10 3 N (down) ....
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This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.
 Spring '09
 Fielding
 Physics, Force, Static Equilibrium

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