Ch12Word - Chapter 12 CHAPTER 12 Static Equilibrium...

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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  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  Page 1 Use Word 6.0c or later to view Macintosh picture. Use Word 6.0c or later to view Macintosh picture. Use Word 6.0c or later to view Macintosh picture.
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Chapter  12 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 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 .
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