PHY 206 Spring 2003 HW #13 Solutions
11.
A hydraulic life is used to raise a car of mass 1500 kg on a shaft of radius 8 cm. The piston
radius is 1 cm. How much force must be exerted to lift the car?
The pressures at the shaft and the piston are equal, so
F1 F2

. The area is proportional to the
A1 A2
radius squared, so the force needed to lift a weight of 1500 kg × 9.81 m/s2 is 229 N.
14.
**A 1500-kg car rests on four tires, each of which is inflated to a gauge pressure of 200 kPa. What
is the area of contact of each tire with the road, if the four tires support the weight equally?
The relevant pressure is given Pgauge = 200 kPa = 2.0105 N/m2. This is the pressure that must support
the weight of the auto, mg = 1500 kg9.81 m/s2 = 14715 N. Since the pressure is the force per unit area, P = F/A,
we find that Atot = 0.074 m2, which means that each tire provides a contact area of A = 0.074 m2 /4 = 0.0184 m2.
19.
A force of 147 N is applied to a small piston to lift a car with weight 14,700 N. Show that the law of
conservation of mechanical energy is not violated.
The work done on the small piston is equal to the exerted force times the distance through which it
works, W1 = Fh = 147h N-m. The work done on the car is equal to the change in potential energy of the car, W2
= U = mgh. The distance moved by the car, h, is found by noting that the amount of fluid is constant, so the
volume change created by the movement of the small
piston must be equal to that created by the motion of the
large piston, A1h = A2h, which gives h = (A1/A2)h. In the example referred to in the problem, the area ration is
1/100, which means that h = 0.01h. Thus we have W2 = 14700N0.01h = 147h N-m, so the two are equal.
23.
A rock of mass M with a density twice that of water is sitting on the bottom of an aquarium tank filled
with water. The normal force exerted on the rock by the bottom of the tank is (c) Mg/2. To determine this, draw
a free-body diagram for the rock, replacing M by rockV = 2wV.
28.
**A 20-cm3 block of lead and a 20-cm3 block of copper rest at the bottom of an aquarium filled with
water.
(c) The buoyant force is the same on both blocks, since the buoyant force is equal to the weight of the
displaced water, which depends on the volume displaced. Both blocks have the same volume.
35.
**A large piece of cork weighs 0.285 N in air. When held submerged underwater by a spring scale,
the spring scale reads 0.855N. Find the density of the cork.
The force read by the scale (as shown in Fig. 13-27) is given by the difference between the buoyant force
and the weight, with the former being greater. Thus FB  Fg  0.855 N , and we know that Fg = 0.285 N. The
buoyant force is therefore FB = 1.140 N and is equal to the weight of the displaced water, FB   wVg . The
weight in air is equal to Fg   corkVg , so we can put these together to find
 cork 0.285 N

 0.25 which leads
 w 1.140 N
to cork = 0.25 g/cm3 = 250 kg/m3.
44.
A horizontal pipe narrows from a diameter of 10 cm to 5 cm. For a fluid flowing from the larger
diameter to the smaller, (b) the velocity increases and the pressure decreases. See Bernoulli’s Equation
for the special case given in Eqn. 13-18.
49.
**Blood flows in an aorta of radius 9 mm at 30 cm/s. (a) Calculate the volume flow rate in liters per
minute. (b) If all the blood form the aorta flows into capillaries and the speed of flow through
the capillaries is
1.0 mm/s, calculate the total cross-sectional area of the capillaries.
cm
(a) The volume flow rate is IV  Av   (0.9cm)2 (30 cm
. One liter is 1000 cm3, so to
s )  76.3 s
3
convert to liters per minute we have IV = 76.3cm3/s60s/min(1 L/1000 cm3) = 4.6 L/min. (b) The speed of flow
in the capillaries is 1.0 mm/s = 0.1 cm/s. Since the volume flow rate is constant, we have Acap vcap  Aaorta vaorta
and we can then solve for the total cross-sectional area Acap = 763 cm2.
51.
**Referring to Fig. 13-30, the outlet for water is at a height h below the surface. How far does the water
spray out?
Using Bernoulli’s equation, P   gh 
1 2
 v  constant , with the pressure at the top equal to Patm and
2
the pressure at the outlet being Patm as well, we can write (assuming as in Example 13-8 that the speed of water in
the tank as the tank empties is negligible,  gH   g  H  h  
1 2
 v . The speed of the water leaving the tank
2
is therefore v  2 gh . Now we can return to kinematics equations to find that the time it takes the water to
reach the ground is found from  H  h  
1 2
at , or t 
2
2  H  h
and the distance traveled in that time is
g
x  vt  2 h  H  h  .
60.
The buoyant force on an object depends on the shape of the object. False.
87.
Two communicating vessels contain a liquid of density 0. The cross-sectional areas of the vessels are A
and 3A. Find the change in elevation of the liquid level if an object of mass m and density  = 0.80 is put into
one of the vessels.
The new object displaces a certain amount of fluid. The level in both of the containers must rise by the
same amount, so that the level remains the same in each. The increase in volume is due to the volume of the
object is not the total volume V = m/, but the volume of displaced fluid for a floating object V = m/0, and the
change of volume of fluid is V = (3A + A) h = 4Ah. We set these equal and rearrange to find h 
m
.
4 A0
93.
**A balloon can just lift a load of 750 N. The skin of the balloon has a mass of 1.5 kg. a) What is the
volume of the balloon? b) If the volume of the balloon is twice what you found in part a), what is the initial
acceleration of the balloon if the load is now 900N?
The forces acting on the balloon are 1) the weight of the load plus balloon downward, 2) the weight of the
helium downward, and 3) the buoyant force upward. To just float, these must sum to zero. Thus we have
airVg  HeVg   750N  1.5kgg   0 . Using the densities for air and helium, air = 1.293 kg/m3 and He =
0.1786 kg/m3 respectively, we find V = 70 m3. For a volume of V2 = 140 m3, the buoyant force would be larger,
leading to a net upward force and therefore an acceleration. The net force would be given by the same equation as
before, but with the new load and an acceleration term, airVg  HeVg   900N  1.5kgg   M tot a . Plugging
in the values, we end up with a = 5.2 m/s2.