Physics 102 Homework solutions, problem set 2 Chapter 13: Exercises: 20: No. there would be no buoyancy force. The buoyancy force comes from having a larger pressure pushing up on the bottom of the object than pushing down on the top of the object. If the pressures are the same than there is no net upwards force and no buoyancy. 28: Deeper. If you put more mass in a ship, it will need to displace more water in order to float, and thus it will ride deeper. (Since you are not changing the size of the ship.) 38: Ice cubes will float lower in a mixed drink than in water since adding alcohol will decrease the density of the liquid. Thus the ice cube will have to displace more volume in order to displace the same weight of fluid in order to have the same buoyancy force. If the ice cubes are at the bottom of the glass, then the drink is more than 50% alcohol (since the density of the water plus alcohol is less than the density of ice). Problem: 6: For the 2 kg object to only 'mass' 1.5 kg when suspended in water, it must displace 0.5 kg of water. Since the density of water is 1 kg per liter, the volume of the displaced water is 0.5 liter. Thus, the density of the object is 2 kg / 0.5 liter = 4 kg/liter (or four times as much as water). 4 kg/liter = 4 g /cm^3 = 4000 kg / m^3. Note that you can also solve the problem a little faster by observing that the 2 kg object displaces 0.5 kg of water. Thus, it must be four times the density of water. --------------- Chapter 14: Exercises: 4: The ridges are on the outside of the funnel. Liquid flows down into the bottle through the funnel. Therefore, the air in the bottle must be able to leave somehow to make room for the liquid. The ridges on the funnel provide the space for the air to leave the bottle. 10: The air pressure at 10,000 m (30,000 feet) is much less than the air pressure at sea level. The airplanes are pressurized to increase the pressure inside the plane, but it is still much less than at sea level. Thus, there is much less pressure in the airplane so the gas in the package expands, puffing up the package. 22: You are unharmed by air pressure because you have compensating pressure inside you pushing out. When the elephant steps on you, it drastically increases the force on you without any compensating internal pressure increase. 34: Assuming that neither balloon pops, the expanding balloon will float higher. This is because the buoyant force is due to the weight of the displaced air. As the balloons rise, the air pressure decreases and hence the air density decreases. This causes the weightof the displaced air to decrease and hence the buoyant force decreases. The rigid balloon always has the same volume so it always displaces the same volume of air. The expanding balloon expands as it rises (since the pressure decreases) and thus displaces more air as it rises. Thus, at any altitude, the expanding balloon will have a greater buoyant force. Problems: 2: Since pressure time volume is constant (at constant temperature), if you decrease the volume by a factor of ten, you increase the pressure by a factor of ten. 4: To get the buoyant force, I need to determine my volume. To get this, I can use my mass and my density. I know that my density is very close to 1 (since I barely float in water). My mass is about 170 pounds or 77 kg. Since mass = density * volume, my volume is V = M/d = 77 kg / 1000 kg/m^3 = 0.077 m^3. The density if air (from the table on page 270) is 1.2 kg/m^3 at room temperature so the mass of the displaced air is 0.077 m^3 * 1.2 kg/m^3 = 0.093 kg. The weight of the displaced air is 0.093 kg * 10 m/s^2 = 0.93 N or a little less than 1 newton. Very little actually. ---------------------------------- Estimation: My calf is about 1 m below my heart. The extra pressure due to the 1 m column of blood is p = dhg = 1000 kg/m^3 * 1 m * 10 m/s^2 = 1 * 10^4 Pa. Now I need to convert this to tor. Since 10^5 Pa = 760 tor, we know that 10^4 Pa = 76 tor. This difference is a lot compared to a typical reading (which is why they are so careful to measure blood pressure at the same height as the heart).