ANSWER TO CHAPTER 5 HOMEWORK

1.  While the box is sliding down, friction will be up the plane, opposing the motion. From the force diagram for the box, we have SF = ma:

                    x-component: mgsinq - Ffr = ma;

                    y-component: FN - mgcosq = 0.

          From the x-equation, we have

                    Ffr = mgsinq -ma = m(gsinq -a)

                        = (1.50kg)[(9.80m/s2)sin30° -(0.30m/s2)]

                        =  69N.

          Because the friction is kinetic, we have

                   Ffr = mkFN = mkmgcosq:

                  69N = mk(15.0kg)(9.80m/s2)cos30°, which gives  mk = 0.54.

  
 2.  (a) we write SF = ma from the force diagram for the snow while it is stationary on the roof:

                      x-component:  mgsinq - Ffr = 0;

                     y-component:   FN - mgcosq = 0.

               When we combine the equations, we get

                     Ffr = mgsinq < mksFN = msmgcosq.

                Thus we have 

                     ms > tan q = tan30° = 0.58.

           (b) We write SF = ma from the force diagram for snow while it is sliding on the roof:

                      x-component: mgsinq - mkFN = ma;

                      y-component: FN - mgcosq = 0.

                 Thus a = g(sinq - mkcosq)

                            = (9.80m/s2)[sin30° - (0.2)cos30°] =3.20m/s2.

                 For the motion on the roof, we have

                         v12  =  vo2 + 2a(x -xo) = 0 + 2(3.20m/s2)(5.0m),

                  which gives

                         v1 =  5.7m/s.

               (c)  The motion when the snow leaves the roof is projectile motion, with an initial velocity of v1 = 5.7m/s at 30° below the horizontal. If we use the new coordinate system shown, we have

                         vx = v1cosq = (5.7m/s)cos30° = 4.9m/s;

                         vy2 = (-v1sinq)2 + 2gh = [-(5.7m/s)sin30°]2 + 2(-9.80m/s2)(-10.0m),

                     Which gives vy = -14.3m/s. The speed of the snow is

                          v = (vx2 + vy2)1/2 = [(4.9m/s)2 + (-14.3m/s)2]1/2 =  15m/s.

 

                          

  
3. (a) For the hanging box we can write SF = ma:

                          y-component:  m2g - FT = 0.

             For the box on the table we can write SF = ma:

                          x-component: FT - Ffr = 0;

                          y-component:  FN - m1g = 0.

             When we combine the equations, we have

                          Ffr = FT = m2g < msFN = msm1g.

              Thus we have

                          m1 > m2/ms = (2.0kg)/(0.40) =  5.0kg.

         (b) The acceleration is zero, so the only change is that the friction is kinetic:

                          m1 > m2/mk = (2.0kg)/(0.30) =  6.7kg.

 

 
 4.  The static friction force provides the centripetal acceleration. We write SF = ma from the force diagram for the coin:      

                          x-component:  Ffr  = mv2/R;

                          y-component:  FN -mg = 0.

           The highest speed without sliding requires Ffr,max = msFN.

           The maximum speed before sliding is 

                          vmax = 2pR/Tmin =  2pRfmax

                                  = 2p(0.120m)(50/min)(60s/min) = 0.628m/s

             Thus we have

                          msmg = mvmax2/R

                          ms(9.80m/s2) = (0.628m/s)2/(0.120m), which gives ms =  0.34.

 

  
 5. At the top of the hill, the normal force is upward and the weight is downward, which we select as the positive direction.

         (a) We write SF =ma from the force diagram for the car:

                           mcarg - FNcar = mv2/R;

                          (1000kg)(9.80m/s2) - FNcar = (1000kg)(20m/s)2/(100m),

               which gives      FNcar = 5.8 x103N.

          (b) When we apply a similar analysis to the driver, we have

                           (70kg)(9.80m/s2) - FNdriver = (70kg)(20m/s)2/(100m),                 

              which gives       FNdriver  = 4.1 x 102N.

          (c) For the normal force to be equal to zero, we have

                            (1000kg)(9.80m/s2) - 0 = (1000kg)v2/(100m),

                which gives     v = 31m/s   (110km/h or 70mi/h).                                

  
*6. The velocity is constant, so the acceleration is zero.

 (a) From the force diagram for the bicycle, we can write SF =ma:

                            x-component: mgsinq - FD = 0, or

                           mgsinq  = cv2;

                          (80.0kg)(9.80m/s2)sin7.0° = c[(9.5km/h)(/3.6ks/h)]2, which gives

                          c = 14kg/m.

(b) We have an additional force in SF =ma:

                          x-component: F + mgsinq - FD = 0, so

                           F = cv2 - mgsinq

                              = (13.7kg/m)[(25km/h)/(3.6ks/h)]2 - (80kg)(9.80m/s2)sin7.0° =  5.7 x 102N.

  

*7. If the triangular block is pushed so that the small block tends to move up the incline, the static friction force on the small block will be down the incline, as shown. 

We choose a coordinate system with the x-axis in the direction of the acceleration a of the triangular block. 

from the force diagram for the small block we have

                            x-component: FNsinq  + Ffrcosq  = max;

                            y-component: FNcosq - mg - Ffrsinq = may.

The top block will not slide until Ffr > mFN. As long as this is not true, ax = a, and ay = 0. Thus we find the limiting acceleration by using these conditions:

                           FNcosq - mg - mFNsinq = 0, or FN = mg/(cosq  - msinq  );

                           FNsinq  + mFNcosq = mamax, or         

                           amax,= FN(sinq  + mcosq )/m = g(sinq + mcosq)/(cosq - msinq).

From the force diagram for the triangular block to not slide, and thus the minimum force to make the small block slide, is

                           Fmin = FNsinq  + Ffrcosq + Mamax = (m + M)amax 

                                   = (m + M)g(sinq + mcosq)/(cosq - mssinq).

 

 

HOME  SEND QUESTIONS