Physics 1114: Unit 6 Homework: Answers

Problem set 1

1. A rod 4.2m long and 0.50cm$^{2}$ in cross-sectional area is stretched 0.20cm under a tension of 12,000N.

   a) The stress is the Force ( $1.2\times 10^{4}$N) per Area ( $\rm0.5\,cm^{2} = 5.0 \times 10^{-5}\,m^{2}$), so $\rm Stress = 2.4\times 10^{8}\,N/m^{2}$.

   b) The strain is $\Delta L/L_{0} = \rm (2.0\times10^{-3}\,m)/(4.2\,m)=4.8\times10^{-4}$ (unitless ratio)

   c) Young's Modulus = stress/strain = $5\times10^{11}$N/m$^{2}$

2. A typical value for systolic blood pressure is 120mmHg. Converted to other units, this is (a) $1.6\times10^{4}$N/m$^{2}$, (b) 0.16atm, (c) 2.32lb/in$^{2}$.

3. The arm of a record player exerts a force of 0.0098N (the weight of 1 gram) on a record. If the diameter of the stylus is 0.0013cm find the pressure on the record groove in N/m$^{2}$ and in atmospheres. (Assume a circular cross-section.)

   Pressure = Force/Area, so $p = \rm (9.8\times10^{-3}\,N)/(1.327\times10^{-10}\,m^{2})=7.38\times10^{7}\,N/m^{2}=730\,atm$

4. (a) The downward force (in N) of the atmosphere acting on the top of a table that measures 3.2m by 1.2m is calculated from $p=F/A$ or $F = pA =\rm (1.013\times10^{5}\,N/m^{2})(3.84\,m^{2})=3.9\times10^{5}\,N$.

   (b) The total (pressure) force acting upward on the underside of the table is the same magnitude as the downward force. (Actually it is very slightly larger, providing a very small buoyant force on the table. This is because the underside is at a deeper level in the atmosphere, so the pressure is slightly larger.)

5. On a day when the atmospheric pressure is 14.8psi a tire is inflated to 32.1psi (psi = lb/in$^{2}$).

   (a) The absolute pressure is the gauge pressure plus atmospheric pressure:

   $p = p_{\rm gauge} + p_{\rm atm} = \rm (14.8 + 32.1)\,psi=46.9\,psi$

   (b) On the following day, when the atmospheric pressure has fallen to 14.6psi, the pressure gauge read $p_{\rm gauge} = p - p_{atm}$, where $p$ is the absolute pressure. If the air in the tire has the same volume and temperature as before, then the absolute pressure inside does not change. The gauge pressure, then, is $p_{\rm gauge} = \rm (46.9 - 14.6)\,psi = 32.3\,psi$. (Note that the net outward force on the walls of the tire does increase because the outside pressure is lower but the inside pressure stays the same).

6. The contraction of the left ventricle (chamber) of the heart pumps blood into the body. Assuming that the inner surface of the left ventricle has an area of $\rm 85\,cm^{2}$ ( $A=8.5\times10^{-3}\rm\,m^{2}$) and the maximum pressure in the blood is 120mmHg ( $p=\rm\,1.6\times10^{4}$N/m$^{2}$), the total force exerted by the ventricle at the point of maximum pressure is $F = pA = 136$N.

7. The difference in blood pressure between the top of the head and bottom of the feet of a 1.60m tall person standing vertically can be calculated using the pressure change in a depth of fluid $\Delta p = \rho g h$. Here, the `depth' is the person's height and $\rho$ is the density of blood ( $\rho =\rm 1030\,kg/m^{3}$). Thus, $\Delta p = \rm (1030\,kg/m^{3}) (9.8\,m/s^{2})(1.60\,m) = 1.6\times10^{4}\,N/m^{2}$. Note that from Problem 2 you found that $\rm 1.6\times10^{4}\,N/m^{2} = 120\,mm\,Hg$.
8. For water to come out of a faucet, there has to be enough pressure at the incoming pipe to lift the weight of the water column plus pushing against air pressure. In other words, the (absolute) pressure has to be greater than $p_{\rm atm} + \rho g h = \rm 1.013\times10^{5}\,N/m^{2} + (1000\,kg/m^{3})(9.8\,m/s^{2})(40\,m)=4.93\times10^{5}\,N/m^{2}$.

Problem set 2

1. The gauge pressure of the liquid column weight comes from $p_{\rm g}=\rho g h$. We want $p_{\rm g}=\rm 60\,mm\,Hg = 8.0\times10^{3}\,N/m^{2}$, so
   
   $${ \rm 8.0\times10^{3}\,N/m^{2} = (1000\,kg/m^{3})(9.8\,m/s^{2})}h$$
   
   
   so that $h = 0.81$m.

   The blood pressure in the vein ( $\rm 18\,mm\,Hg = 2.4\times10^{3}\,N/m^{2}$) is $gauge$ pressure ($p_{\rm g}$), so the IV bottle has to be raised so that the fluid gauge pressure ($\rho g h$) matches blood pressure: $p_{\rm g}=\rho g h$, so
   

   $${ \rm 2.4\times10^{3}\,N/m^{2} = (1000\,kg/m^{3})(9.8\,m/s^{2})}h$$
   
   
   so that $h = 0.245$m.

2. The force from the lift is $F=pA$, where $p$ is the gauge pressure (maximum of 16atm). (If the interior pressure were given as the absolute pressure, then atmospheric pressure would have to be subtracted out first.) The maximum force, then, is $F_{\rm max} = p_{\rm max} (\pi r^{2})$ for a circular cross-section. $F_{\rm max} = \rm (1.62\times10^{6}\,N/m^{2})(2.27\times10^{-2}\,m^{2}) = 3.68\times10^{4}\,N$. Setting $F_{\rm max}=mg$ (the weight), we get $m=3.75\times10^{3}$kg.

3. A 3.0N force is applied to the plunger of a hypodermic needle. The diameter of the plunger is 1.0cm and that of the needle 0.20mm.

   (a) To find the force ($F_{2}$) at the needle end, we can use the fact that the pressure is the same at both ends: $p_{1}=p_{2}$. Then $F_{1}/A_{1} = F_{2}/A_{2}$, where $A=\pi(d/2)^{2}$.
   

   $$F_{2}=\frac{A_{2}}{A_{1}}F_{1}= \frac{\pi(d_{2}/2)^{2}}{\pi(d_{1}/2)^{2}}F_{1}= \frac{d_{2}^{2}}{d_{1}^{2}}F_{1}$$
   
   
   
   
   $$F_{2} = \rm\frac{(0.2\,mm)^{2}}{(10\,mm)^{2}}(3.0\,N) = 1.2\times10^{-3}\,N$$
   
   
   (Since the length units cancel out, we don't have to put the diameters into meters, but they do have to be the same units.)

   (b) To push fluid into a vein where the (gauge) pressure is 18mmHg ( $2.4\times10^{3}$N/m$^{2}$), we find the plunger force as $F_{1}=p A = \rm (2.4\times10^{3}\,N/m^{2})(\pi (5.0\times10^{-3})^{2}\,m^{2}) = 0.19\,N$

4. The force required to raise a 1000kg block of concrete to the surface of a freshwater lake is less than $mg$ because of the buoyant force ($F_{B}$) from being submerged in water. $F_{B} = \rho_{f}V$, where $\rho_{f}$ is the fluid density (1000kg/m$^{3}$) and $V$ is the displaced volume (the volume of the concrete block: $V_{b} = m_{b}/\rho_{b} = \rm (1000\,kg)/(2.3\times 10^{3}\,kg/m^{3})= 0.435\,m^{3}$. The lifting force, $F_{\rm lift}$ is the weight of the block minus the buoyant force:
   
   $$F_{\rm lift}=mg - \rho_{f}V_{b} =\rm 5.5\times10^{3}\,N$$
   
   

   As the block comes out of the water, the buoyant force decreases (less water displaced), becoming zero when the block is completely out of the water. So the force to lift the block becomes equal to its weight ($m_{b}g=9800$N).

5. A freight ship has a horizontal cross-sectional area of 3100m$^{2}$ at the water line. When loaded, the ship drops 6.1m. The amount of drop times the cross-sectional area of the ship gives the increase in the volume of the ship that is below the water line, i.e., the volume of extra seawater that is displaced. The weight of the displaced seawater equals the weight of the load (floating). So the mass of the load is the mass of the displaced water: $m=\rho_{f}V=\rho_{f}Ah=\rm (1025\,kg/m^{3})(3100\,m^{2})(6.1\,m) = 1.94\times10^{7}\,kg$.
6. To find the maximum load, consider what the maximum buoyant force would be. Since the raft floats when there is no load ( $F_{B}=m_{\rm raft}g$), as weight is added the raft will be pushed farther down into the water. The most it can be pushed is to when it is just submerged and the volume of displaced water equals the volume of the raft. At that point the buoyant force is $F_{B}= \rho_{f}V_{\rm raft}g$ ( $\rho_{f}=1025\rm\,kg/m^{3}$) which balances the weight of the raft and its load: $F_{g}=m_{\rm raft}g + m_{\rm load}g$.
   
   $$F_{B} = \mbox{weight of raft and load}$$
   
   
   
   
   $$\rho_{f}V_{\rm raft}g = m_{\rm raft}g + m_{\rm load}g$$
   
   
   The weight of the load, then, is
   
   $$m_{\rm load}g = \rho_{f}V_{\rm raft}g - m_{\rm raft}g$$
   
   

   The raft has a volume of 7.49m$^{3}$. The weight of the raft is $mg=\rho_{\rm wood}Vg = \rm (130\,kg/m^{3})(7.49\,m^{3})(9.8\,m/s^{2})=9.54\times10^{3}\,N$. Plug these values in above and get the weight of the load as $6.57\times10^{4}$N, or a mass of 6700kg.

7. A uniform cube of wood is 0.25m on each edge. It is floating in pure water with its upper surface 0.07m above the surface of the water. To determine the density of the wood, we must find the mass of the block. (We know that the volume is $V = l^{3} = 1.56\times10^{-2}$m$^{3}$, and $\rho = m/V$.) Since the block is floating, $F_{B}=F_{g}$:
   
   $$\rho_{f}Vg = m_{\rm block}g$$
   
   
   
   
   $$m_{\rm block} = \rho_{f}V$$
   
   
   The mass of the block equals the mass of the displaced water. The volume of water is the volume of the submerged part of the block: $V =\rm (0.25\,m)(0.25\,m)(0.25 - 0.07\,m) = 1.125\times10^{-2}\,m^{3}$. Since $\rho_{f}=\rm 1000\,kg/m^{3}$, then $m_{\rm block} = 11.25\rm\,kg$ and $\rho_{\rm block}=\rm 720\,kg/m^{3}$. (Less than water, obviously!)

Problem set 3

1. a) The surface temperature of the sun: $\rm 5160\,K = 4887^{\circ}C = 8828^{\circ}$F.

   b) The ``night-time'' surface temperature of the moon: $\rm -173^{\circ}C = -279.4^{\circ}F = 100.15\,K$.

   c) Normal body temperature: $\rm 98.6^{\circ}F = 37^{\circ}C = 310.15\,K$.

2. An aluminum wire 1.4km long is strung between two towers for the transmission of electricity.

   a) How much does its length change when the temperature goes from $-$10$^{\circ}$C to 40$^{\circ}$C? ( $\alpha = 22.2\times 10^{-6}$$^{\circ}$C$^{-1}$).
   

   $$\Delta L = \alpha L_{0} \Delta T =\rm 1.55\,m$$
   
   
   b) If the wire is 1.4 km long when the temperature is $-$10$^{\circ}$C, how long will it be when the temperature reaches 20$^{\circ}$C?
   
   $$\Delta L = \alpha L_{0} \Delta T =\rm0.93\,m$$
   
   
   So that $L = 1400.9$m.
3. An automobile tire is filled to a gauge pressure of $2\times 10^{5}$N/m$^{2}$ at 10$^{\circ}$C. After driving 100km, the temperature within the tire rises to 40$^{\circ}$C. What is the pressure within the tire now? Atmospheric pressure is $1.01\times 10^{5}$N/m$^{2}$.

   The original absolute pressure in the tire is $p_{1} = p_{\rm g} + p_{atm} = \rm (2\times 10^{5} + 1.01\times 10^{5})\,N/m^{2} = 3.01\times10^{5}\,N/m^{2}$. The starting temperature is $T_{1}=283.15$K, and the final temperature is $T_{2}=313.15$K. We assume that the volume of the tire is constant, i.e., $V_{1}=V_{2}$. Using the ideal gas law:
   

   $$\frac{p_{1}V_{1}}{T_{1}} = \frac{p_{2}V_{2}}{T_{2}}$$
   
   
   Divide out the volume ($V_{1}=V_{2}$), and solve for $p_{2}$:
   
   $$p_{2}= p_{1}\frac{T_{2}}{T_{1}} = \rm 3.33\times10^{5}\,N/m^{2}$$
   
   
   So the new gauge pressure is $p_{\rm g}= p - p_{atm} = 2.3\times10^{5}\rm\,N/m^{2}$

4. STP means Standard Temperature and Pressure ( $\rm0^{\circ}C=273\,K$ and 1atm). If 5.00m$^{3}$ of gas initially at STP is placed under a pressure of 4.0atm and the temperature of the gas rises to 25$^{\circ}$C (298K), then to find the final volume we again use the ideal gas law:
   
   $$\frac{p_{1}V_{1}}{T_{1}} = \frac{p_{2}V_{2}}{T_{2}}$$
   
   
   And solve for $V_{2}$:
   
   $$V_{2} = V_{1}\left(\frac{T_{2}}{T_{1}}\right) \left(\frac{p_{1}}{p_{2}}\right) = 1.36\rm\,m^{3}$$
   
   
   Note that because of the ratio you don't have to convert pressure to N/m$^{2}$, but the temperature must be in Kelvin (absolute temperature).

5. If 50.0L of oxygen at 10$^{\circ}$C (283K) and an absolute pressure of 1.88atm are compressed to 36.6L and at the same time the temperature is raised to 80$^{\circ}$C (353K), the new pressure is
   
   $$p_{2} = p_{1}\left(\frac{T_{2}}{T_{1}}\right) \left(\frac{V_{1}}{V_{2}}\right) = 3.2\rm\,atm$$
   
   
   Note that because of the ratio you don't have to convert volume to m$^{3}$.
6. A sample of gas occupies $2.0 \times 10^{-3}$m$^{3}$ at an absolute pressure of 1.0atm and a temperature of 0$^{\circ}$C (273K).

   a) Its volume at the same pressure and a temperature of 200$^{\circ}$C (473K): $V_{2} = 3.5\times10^{-3}\rm\,m^{3}$

   b) Its pressure at a volume of $5.0 \times 10^{-4}$m$^{3}$ and a temperature of 200$^{\circ}$C: $p_{2} = 6.9\rm\,atm$

7. Describe what happens when the temperature of a gas is decreased while the pressure remains constant. Explain this change based on the Kinetic-Molecular Theory:

   The volume decreases as the temperature decreases at constant pressure, thus the density increases (the so-called Charles' Law). So the pressure stays the same because although the average impulse is smaller from each molecule colliding with the walls, more molecules per time hit the walls because of the higher density.