9

Solids and Fluids

PROBLEM SOLUTIONS

 9.3 Four acrobats of mass75.0 kg, 68.0 kg, 62.0 kg and 55.0kg for a human tower,

with each acrobat standing on the shoulders of another acrobat. The 75.0kg

acrobats is at the bottom of the tower. (a) What is the normal force acting on

the 75—kg acrobat? (b) If the area of each of the 75.0 kg acrobat’s show is

425 cm2, what average pressure ( not including atmospheric pressure) does

the column of acrobats exert on the floor? (c) Will the pressure be the same if

a different acrobat is on the bottom?

9.3, 9.5, 9.7, 9.9, 9.11, 9.13,9.15, 9.17, 9.23,9. 25, 9.33,9.39, 9.63, 9.88

Weight of the four Acrobats is

= A1 + A2 +A3 +A4 = -2548N (Earthward

)

A4= 75.0kg x 9.8 m/s2=735 N

A3= 68.0kg x 9.8 m/s2=666 N

A2 = 62.0kg x 9.8 m/s2=608 N

A 1= 55.0kg x 9.8 m/s2=539 N

A) The Normal

force is equal

and opposite to the force of gravity and is perpendicular to the surface of

the floor. =

2548 Skyward

B) Average Pressure P on the floorP = Force/AreaP = W/ AP = 2548 N/2(.04M)P = 31.8 x 103 Pa

C) If the area of either of the feet of another acrobats is different than that of the 75.0kg acrobat, then the pressure will also be different. If the area of the feet of another acrobat is the same are the 75.0kg acrobat, then the pressure will be the same.

 9.5 The nucleus of an atom can be modeled as several protons and neutrons closely packed together.Each particle has a mass of 1.67 x 10-27 kg and radius on the order of 10-15 m. (a) Use this model and thedata provided to estimate the density of the nucleus of an atom. (b) Compare your result with the density ofa material such as iron. What do your result and comparison suggest about the structure of matter?

(a) If the particles in the nucleus are closely packed with negligible space between them, the average nuclear

density should be approximately that of a proton or neutron. That is

  

(b) The density of iron is and the densities of other solids and liquids are on the

order of . Thus, the nuclear density is about times greater than that of common solids and

liquids, which suggests that atoms must be mostly empty space. Solids and liquids, as well as gases, are

mostly empty space.

9.3, 9.5, 9.7, 9.9, 9.11, 9.13,9.15, 9.17, 9.23,9. 25, 9.33,9.39, 9.63, 9.88

9.3, 9.5, 9.7, 9.9, 9.11, 9.13,9.15, 9.17, 9.23,9. 25, 9.33,9.39, 9.63, 9.88

 9.7 Suppose a distant world with surface gravity of 7.44 m/s2 has an atmospheric pressure of 8.04x 104 Pa at the surface. (a) What force is exerted by the atmosphere on a disk-shaped region 2.00 m in radius at the surface of a methane ocean? (b) What is the weight of a 10.0-m deep cylindrical column of methane with radius 2.00 m? (c) Calculate the pressure at a depth of 10.0 m in the methane ocean. Note: The density of liquid methane is 415 kg/m3.

(a)

(b)

(c) Now, consider the thin disk-shaped region 2.00 m in radius at the bottom end of the column of methane.

The total downward force on it is the weight of the 10.0-meter tall column of methane plus the

downward force exerted on the upper end of the column by the atmosphere. Thus, the pressure (force per

unit area) on the disk-shaped region located 10.0 meters below the ocean surface is

  

9.3, 9.5, 9.7, 9.9, 9.11, 9.13,9.15, 9.17, 9.23,9. 25, 9.33,9.39, 9.63, 9.88

 9.9 A 200-kg load is hung on a wire of length 4.00 m, cross-sectional area 0.200 x 10-

4m2, and Young’s modulus 8.00 x 1010 N/m2. What is its increase in length?

Young’s modulus is defined as . Thus, the elongation of the wire is

  

9.3, 9.5, 9.7, 9.9, 9.11, 9.13,9.15, 9.17, 9.23,9. 25, 9.33,9.39, 9.63, 9.88

9.11 A plank 2.00cm thick and 15.0cm wide is firmly attached to the railing of a ship by clamps so that the

rest of the board extends 2.00 m horizontally over the sea below. A man of mass 80.00 kg is forces to

stand on the very end. If the end of the board drops by 5.00cm because of the man’s weight, find the

shear modulus of the wood

Two cross-sectional areas in the plank, with one directly above the rail and one at the outer end of the plank, separated

by distance 2.00 mh and each with area, 22.00 cm 15.0 cm 30.0 cmA move a distance

25.00 10 mx parallel to each other. The force causing this shearing effect in the plank is the weight

of the man F mg applied perpendicular to the length of the plank at its outer end. Since the shear modulus

S is given by

shear stress F A Fh

Sshear strain x h x A

we have

27

2 2 2 4 2

80.0 kg 9.80 m s 2.00 m1.05 10 Pa

5.00 10 m 30.0 cm 1 m 10 cmS

9.3, 9.5, 9.7, 9.9, 9.11, 9.13,9.15, 9.17, 9.23,9. 25, 9.33,9.39, 9.63, 9.88

2.00cm x 15.0cm x 2.00 m

F=80.00kgx 9.8

9.13 For safety in climbing, a mountaineer uses a nylon rope that is 50 m long and 1.0

cm in diameter. When supporting a 90-kg climber, the rope elongates 1.6 m. Find

its Young’s modulus

Using with and , we get

  

9.3, 9.5, 9.7, 9.9, 9.11, 9.13,9.15, 9.17, 9.23,9. 25, 9.33,9.39, 9.63, 9.88

9.15 Bone has a Young’s modulus of 18 x 109 Pa. Under compression, it can withstand a stress ofabout 160 x 106 Pa before breaking. Assume that a femur (thigh bone) is 0.50 m long, and calculate theamount of compression this bone can withstand before breaking.

From , the maximum compression the femur can withstand before breaking is

  

9.3, 9.5, 9.7, 9.9, 9.11, 9.13,9.15, 9.17, 9.23,9. 25, 9.33,9.39, 9.63, 9.88

9.17 A walkway suspended across a hotel lobby is supported at numerous points along its edges by a vertical cable above each point and a vertical column underneath. The steel cable is 1.27 cm in diameter and is 5.75 m long before loading. The aluminum column is a hollow cylinder with an inside diameter of 16.14 cm, an outside diameter of 16.24 cm, and unloaded length of 3.25 m. When the walkway exerts a load force of 8 500 N on one of the support points, how much does the point move down?

The upward force supporting the load is the sum of the compression force exerted by the column and the

tension force exerted by the cable.

  

Since the magnitude of the change in length is the same for the column and the cable, this becomes

  

yielding  

or

9.3, 9.5, 9.7, 9.9, 9.11, 9.13,9.15, 9.17, 9.23,9. 25, 9.33,9.39, 9.63, 9.88

9.23 A collapsible plastic bag contains a glusoe solution. If the average gauge

pressure in the vein is 1.33 x 103 Pa, what must be the minimum height h of

the bag in ore vein? Assume the specific gravity of the solution is 1.2.

The density of the solution is . The gauge

pressure of the fluid at the level of the needle must equal the gauge

pressure in the vein, so , and

9.3, 9.5, 9.7, 9.9, 9.11, 9.13,9.15, 9.17, 9.23,9. 25, 9.33,9.39, 9.63, 9.88

9.25 A container is filled to a depth of 20.0 cm with water. On top of the water floats a 30.0-cm-thick layer of oil with specific gravity 0.700. What is the absolute pressure at the bottom of the container?

We first find the absolute pressure at the interface between oil and water.

  

This is the pressure at the top of the water. To find the absolute pressure at the bottom, we use

, or

  

9.3, 9.5, 9.7, 9.9, 9.11, 9.13,9.15, 9.17, 9.23,9. 25, 9.33,9.39, 9.63, 9.88

9.33 A wooden block of volume 5.24 3 10-4 m3 floats in water, and a small steel object of mass m is placed on top of the block. When m 5 0.310 kg, the system is in equilibrium, and the top of the wooden block is at the level of the water. (a) What is the density of the wood? (b) What happens to the block when the steel object is replaced by a second steel object with a mass less than 0.310 kg? What happens to the block when the steel object is replaced by yet another steel object with a mass greater than 0.310 kg?

(a) While the system floats, , or

  

When , , giving

  

(b) If the total weight of the block+steel system is reduced, by having , a smaller buoyant

force is needed to allow the system to float in equilibrium. Thus, the block will displace a smaller

volume of water and will be only partially submerged. The block is fully submerged when

. The mass of the steel object can increase slightly above this value without causing it

and the block to sink to the bottom. As the mass of the steel object is gradually increased above 0.310 kg,

the steel object begins to submerge, displacing additional water, and providing a slight increase in the

buoyant force. With a density of about eight times that of water, the steel object will be able to displace

approximately of additional water before it becomes fully submerged. At this

point, the steel object will have a mass of about 0.349 kg and will be unable to displace any additional

water. Any further increase in the mass of the object causes it and the block to sink to the bottom. In

9.3, 9.5, 9.7, 9.9, 9.11, 9.13,9.15, 9.17, 9.23,9. 25, 9.33,9.39, 9.63, 9.88

conclusion,

9.39 A cube of wood having an edge dimension of 20.0 cm and a density of 650 kg/m3

floats on water.(a) What is the distance from the horizontal top surface of the cube to the water level? (b) What mass oflead should be placed on the cube so that the top of the cube will be just level with the water surface?

(a) The wooden block sinks until the buoyant force (weight of the displaced water) equals the weight of the block.

That is, when equilibrium is reached,

, giving

  

or  

(b) When the upper surface of the block is level with the water surface, the buoyant force is

  

This must equal the weight of the block plus the weight of the added lead, or , and

  

9.3, 9.5, 9.7, 9.9, 9.11, 9.13,9.15, 9.17, 9.23,9. 25, 9.33,9.39, 9.63, 9.88

giving  

9.3, 9.5, 9.7, 9.9, 9.11, 9.13,9.15, 9.17, 9.23,9. 25, 9.33,9.39, 9.63, 9.88

9.63 The block of ice (temperature 0°C) shown in Figure P9.63 is drawn over a level surface lubricated by a layer of water 0.10 mm thick. Determine the magnitude of the force F S needed to pull the block with a constant speed of 0.50 m/s. At 0°C, the viscosity of water has the value h 5 1.79 x 1023 N ? s/m2

From the definition of the coefficient of viscosity, , the required force is

  

9.3, 9.5, 9.7, 9.9, 9.11, 9.13,9.15, 9.17, 9.23,9. 25, 9.33,9.39, 9.63, 9.88

9.88 A U-tube open at both ends is partially filled with water (Fig. P9.88a). Oil (r = 750 kg/m3) is then poured into the right arm and forms a column L = 5.00 cm high (Fig. P9.88b). (a) Determine the difference h in theheights of the two liquid surfaces. (b) The right arm is then shielded from any air motion while air is blown across the top of the left arm until the surfaces of the two liquids are at the same height (Fig. P9.88c). Determine the speed of the air being blown across the left arm. Assume the density of air is 1.29 kg/m3.

(a) Consider the pressure at points A and B in part (b) of the figure by applying . Looking at

the left tube gives , and looking at the tube on the right, .

Pascal’s principle says that . Therefore, , giving

  

(b) Consider part (c) of the diagram showing the situation when the air flow over the left tube equalizes the

fluid levels in the two tubes. First, apply Bernoulli’s equation to points A and B. This gives

. Since , , and , this reduces to

9.3, 9.5, 9.7, 9.9, 9.11, 9.13,9.15, 9.17, 9.23,9. 25, 9.33,9.39, 9.63, 9.88

   [1]

Now use to find the pressure at points C and D, both at the level of the oil–water interface

in the right tube. From the left tube, , and from the right tube, .

Pascal’s principle says that , and equating these two gives or

   [2]

Combining Equations [1] and [2] yields

  

9.3, 9.5, 9.7, 9.9, 9.11, 9.13,9.15, 9.17, 9.23,9. 25, 9.33,9.39, 9.63, 9.88