ANSWER TO CHAPTER 13 HOMEWORK

1.  From the masses we have

                     mwater = 98.44g - 35.00g = 63.44g;

                     mfluid = 88.78g- 35.00g = 53.78g.

          Because the water and the fluid occupy the same volume, we have

                     SGfluid = rfluid/rwater = mfluid/mwater = (53.78g)/(63.44g) = 0.8477.

2.  Because the force from the pressure on the cylinder supports the automobile, we have

                     PA = mg;

                    (17.0aatm)(1.013 x 105N/m2.atm)1/4p(24.5 x 10-2m)2 = m(9.80m/s2), which gives m = 8.28 x 103kg.

          Note that we use gauge pressure because there is atmospheric pressure on the outside of the cylinder.

3.  When we apply St = Ia  to the lever about the pivot point B, we have

                   F2L - F1L = 0,   or  F1 = 2F.

         Because the pressure in the hydraulic fluid is constant, we have

                   F2/A2 = F1/A1,  or

                   F2 = F1(A2/A1) = 2F(D2/D1)2.

         From the forces on the large cylinder we have

                   Fsample = F2.

          Thus the pressure on the sample is

                  Psample = Fsample/Asample

                       = 2F(D2/D1)2/Asample

                       = 2(300N)[(10.0cm)/(2.0cm)]2

                                    /(4.0 x 10-4m2)

                       = 3.8 x 107N/m2.

  

4.  (a)  The buoyant force on the completely submerged diver is

                      Fbuoy = rwatergV   

                          = (1.025 x 103kg/m3)(9.80m/s2)(65.0 x 10-3m3) = 653N.

          (b)  With the positive direction upward, the net force is

                      Fnet = Fbuoy - mg = 653N - (63.0kg)(9.80m/s2) = +36N.

                The net force is up, so the diver will   float.

5.  From the equation of continuity we have

                      Flow rate = A1v1 = A2v2,  or v2 = (A1/A2)v1 = (D1/D2)2v1.

          From Bernoulli's equation for the horizontal pipe we have

                      P1 + 1/2rv12 + rgy1 = P2 + 1/2rv22 + rgy2;

                      P1 + 1/2rv12 + 0 = P2 + 1/2r(D1/D2)4v12 + 0, which gives

                      v12 = 2(P1 - P2)/(r[(D1/D2)4 - 1];

                            = 2[(32 x103Pa) - (24 x 103Pa)]/(1.00 x 103kg/m3)[(6.0cm/4.0cm)4 - 1],

          which gives v1 = 1.98m/s.

           Thus the volume rate of flow is

                      DV/Dt = A1v1 = 1/4pD12v1 = 1/4p(6.0 x 10-2m)2(1.98m/s) = 5.6 x 10-3m3/s.

*6. The minimum mass of lead will suspend the wood under water.

Because the net force is zero, we have

                    Fnet = 0 = Flead + Fwood - mleadg - mmwoodg, or

                    rwatergVlead + rwatergVwood = mleadg + mwoodg;

                   rwaterg(mlead/rlead) +  rwaterg(mwood/rwood) = mleadg + mwoodg.

We can rearrange this:

                  mlead[1 - (1/SGlead)] = mwood[(1/SGwood) - 1];

     &nnbsp;            mlead[1 - (1/1.13)] = (3.15kg)[(1/0.50) - 1], which gives

                 mlead = 3.46kg.

 
*7 (a) From the equation of continuity we have

                  Flow rate = A1v1 = A2v2,  or v1 = (A2/A1)v2.

From Bernoulli's equation we have

                  P1 + 1/2rv12 + rgy1 = P2 + 1/2rv22 + rgy2; 

                 Patm + 1/2r(A2/A1)2v22 + 0 = Patm + 1/2rv22 + rgh, which gives

                 v2 = {2gh/[(A2/A1)2 - 1]}1/2.

We know that the height of the liquid decreases, so                      

                 dh/dt = -v2 =  -[2ghA12/(A22 - A12 )]1/2.

(b) We integrate to find the height as a function of time:

                ò h0h dh/Ö h = ò 0tÖ  [2gA12/(A22 - A12 )]d<t;

               1/2(Ö h  - Ö h0) = - Ö  [2gA12/(A22 - A12 )] t, or

               Ö h  = Ö h0 - Ö  [gA12/2(A22 - A12 )] t.

(c) The two area are

               A1 = 1/4pD12 = 1/4p(0.50 x 10-2m)2 = 1.96 x 10-5m2.

               A2 = V/h0 = (1.0 x 10-3m3)/(9.4 x 10-2m) = 1.06 x 10-2m2.

We find the time to empty from

               h1/2 = h01/2 - [gA12/2(A22 - A12 )]1/2 t;

               0 = (9.4 x 10-2m)1/2 - {(9.80m/s2)(1.96 x 10-5m2)2/2[(1.06 x 10-2m2) - (1.96 x 10-5m2)2]}1/2t,

which gives t = 75s.

 

HOME  SEND QUESTIONS