ANSWER TO CHAPTER 12 HOMEWORK

1.    We choose the coordinate system shown, with positive torques clockwise.

       We write St =Ia about the point A from the force diagram for the two beams:

               StA = mg1/4L + Mg1/2L - FN2L = 0,which gives

               FN2 = 1/4mg + 1/2Mg

                      = 1/4[1/2(1100kg)](9.80m/s2) + 1/2(110kg)(9.80m/s2)

                      = 6.74 x 103N.

         We write SFy = may from the force diagram for the two beams:

               FN1 + FN2 -Mg - mg = 0, which gives

               FN1 = Mg + mg - FN2

                      = (1100kg)(9.80m/s2) + 1/2(1100kg)(9.80m/s2) - 6.74 x 103N 

                      = 9.43 x 103N.

  

 

2.  We choose the coordinate system shown, with positive torques clockwise.

         We write St = Ia about the point A from the force diagram for the beam:

                  StA = -(FTsina)L + Mg1/2L

                  -FT sin40° + (30kg)(9.80m/s2))1/2 = 0,

          which gives   FT = 2.3 x 102N.

          Note that we find the torque produced by the tension by finding the torques produced by the components.

          We write SF = ma from the force diagram for the beam:

                     SFx = FWx - FTcosa = 0;

                          FWx - (2.29 x 102N)cos40° = 0, which gives FWx = 175N.

                     SFy = FWy + FTsina - Mg =0;

                          FWy + (2.29 x 102N)sin40° - (30kg)(9.80m/s2) = 0, which gives FWy = 147N.

            For the magnitude of FW we have

                       FW = (FWx2 + FWy2)1/2 = [(175N)2 + (147N)2]1/2 = 2.3 x 102N.

            We find the direction from 

                       tanq = FWy/FWx = (147N)/(175N) = 0.84, which gives q = 40°

            Thus the force at the wall is   FW = 2.3 x 102N, 40° above the horizontal.

 

 

3.  The pressure needed is determined by the bulk mudulus:

                          DP = -BDV/Vo = -(90 x 109N/m2)(-0.10 x 10-2) = 9.0 x 107N/m2.

          This is (9.0 x 107N/m2)/(1.0 x 105N/m2.atm) = 9.0 x 102atm.

4.  We find the shear force applied along one edge of tahe square plaate from

                          G = Stress / Strain = (F/A)/(DL/Lo);

                          80 x 109N/m2 =  [F/(0.50m)(0.011m)]/(0.050), which gives F = 2.2 x 107N.

5.  (a)  We want the maximum stress to be (1/7.0) of the tensile strength:

                          Stressmax = F/Amin =(Tensile strength)/7.0;

                          (320kg)(9.80m/s2)/Amin = (500 x 106N/m2)/7.0, which gives Amin = 4.4 x 10-5m2.    

         (b)   We find the change in length from

                          Strain = DL/Lo = Stress/E, or

                          DL = (Stress)Lo/E = [(500 x 106N/m2)/7.0](7.5m)/(200 x 109N/m2) = 2.7 x 10-3m = 2.7mm.     

*6. We choose the coordinate system shown, with positive torques clockwise.

We write SFx = Max from the force diagram for the lamp:

                         FP - Ffr = 0, which gives FP = Ffr.

We write SFy = May from the force diagram fo the lamp:

                         FN - Mg =0, which gives FN = Mg.

If the lamp slides without acceleration, we have  FP = Ffr = mFN = mMg.

We write St = Ia about the center of the base from the force diagram for the lamp:

                       St = FPy -FNd = mMgy - Mgd =0, or y = d/m.

The maximum height without tipping will be when d is maximum, which is L:

                      ymax = L/m = (0.10m)/(0.20) = 0.50m = 50cm.

 
*7. We find the acceleration, and then the force, required to stop:

                      v2 = v02 + 2as;

                      0 = (60m/s)2 + 2a(1.0m); which gives a = -1.8 x 103m/s2.

The required force is 

                     F = ma = (75kg)(-1.8 x 103m/s2) = 1.35 x 105N.

This will produce a stress of 

                    Stress = F/A = (1.35 x 105N)/(0.30m2) = 4.5 x 105N/m2.

Because this is less than the ultimate strength of 5 x 105N/m2, it is possible to escape serious injury.

 

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