PHYSICS 121 - LANGENBERG

HOMEWORK 11
SOLUTIONS


1. Hanging blocks

A. It is less than. The pressure in the water at the top surface of A is equal to the atmospheric pressure at the surface of the water plus the weight (mass x acceleration of gravity) of the water in a column of unit area and height equal to the depth below the surface of the top surface of A. Similarly with B. B hangs deeper than A, therefore there is a greater depth of water above it, and the water pressure on its top surface is greater than it is for A. Because the two blocks are identical, the areas of their top surfaces are equal, and the force exerted by the water (pressure x area) is less for A than it is for B.

B. It is equal to. Blocks A and C hang at the same depth. Hence by the argument above, the forces on their top surfaces are equal.

C. By N3 the force exerted on the water by each block is equal in magnitude to the buoyant force exerted by the water on the block. The buoyant force on a block is equal to the weight of the water displaced by it. Because both blocks are identical in volume, the forces exerted by the water on each block are equal.

D. Because the blocks have equal volumes, they displace equal weights of water and the buoyant forces on them are all equal.

So what effects do the differences in masses of the blocks have? They cause the tensions in the strings supporting the blocks to be different. The string tension for A is equal to that for B, and the tension C is less.


2. The Magdeburg hemispheres

Each team of horses will pull on one of the rings shown in the diagram, one toward the left and one toward the right. In order to separate the two hemispheres, each team would have to exert a force greater than the net force exerted by the atmosphere on the hemisphere it is attached to. How would we calculate that net force?

You will notice that the force exerted by the atmosphere on a hemisphere varies with direction over its curved surface. The atmospheric force components perpendicular to the axis along which the horses are pulling cancel out around the sphere. To determine the net force parallel to the axis requires a little bit of elementary calculus. We can avoid that and get an answer that is probably pretty close if we think of a cube instead of a sphere, split in the middle. Then the net force is just that of the atmosphere on the square end surface of the cube. If the cube is about a foot on a side, then the net force is just F = PA, where P is atmospheric pressure and A is the area of the cube face. Plugging in P = 15 pounds/square inch (yes, I know its really 14.7 pounds/square inch) and A = 144 square inches yields F = about 2200 pounds, or about a ton!

That's a big number, but it does seem that eight horses ought to be able to exert a force that big. That's only about 250 pounds per horse, and many humans can bench press way more than that. Following my own advice always to contemplate the reasonableness of a numerical result, I am led to wonder what might be awry here. I suppose horses might have been smaller and weaker back then. On the other hand, if you look at the contemporary engraving on p. 302 of the textbook you will see that the sphere pictured is rather larger than a basketball. It looks like it is closer to the size of a horse's haunches, i.e., two to three times the diameter of a basketball. That means that the surface area of each hemisphere is probably something on the order of five times that of a basketball hemisphere, so the force required goes up to something like five tons. That's more like it. Poor horses!

Curiosity question: Where's Magdeburg?


3. Pushing iron

a. less than (<) This answer assumes that what is meant by the question is the normal force directly exerted by the bucket on the iron chunk at its contact with the bottom of the bucket. The remainder of the upward vertical force required to balance the (unchanged) gravitational force on the iron is the buoyant force of the water on the iron.

b. same as (=) See answer above. The total upward force on the iron when it is in the bucket (contact force by the bucket plus buoyant force of the water) remains equal in magnitude to the downward gravitational force on the iron. The three forces cancel, as they must to satisfy N3.

c. greater than (>) The evacuation of the bell jar removes the (very small) buoyant force experienced by the iron in air, so the contact force exerted on the block by the table must increase (very slightly).

d. less than (<) Again, the buoyant force will reduce the normal upward contact force exerted by the scale, and thus the apparent weight. However, assuming the scale would read zero with nothing sitting on it, even under water, is a bit of a stretch. Many types of scales would detect the pressure force of the water at whatever depth they sat.


4. Balloon in a car

First, a so-called "Gedanken Experiment": (That's German for "thought experiment," and is a commonly used term and technique in physics.) Think about a car as approximated by a closed spherical container partially filled with water. At rest, the surface of the water will be in a plane perpendicular to the upward force exerted by the container on the water, which balances the downward gravitational force on the water, i.e., the surface will be horizontal. Now imagine adding an external force that accelerates the container in the "forward" direction parallel to the ground. The container must in turn exert a force on the water in order to accelerate it so that it remains inside the container. The total force exerted by the container on the water is then the vector sum of the initial upward force and the new "forward" force and will have a direction that tilts forward from the vertical by an amount that increases as the external acceleration increases. From the perspective of the water, the "up" direction effectively tilts forward, and its surface will rotate away from the horizontal toward the forward direction. If there's anything floating in the water, it will stay on the tilted surface and thus move toward the forward side of the container.

Now imagine replacing the closed spherical container with a closed automobile and the water with air in which there floats a helium-filled balloon. The behavior of the floating object (balloon) in the fluid (air) contained in the accelerating container (car) will be qualitatively just as described in the idealized example above.


5. The 3 vase puzzle

This one's actually rather tricky, so don't worry if you had a difficult time with it.

Yes, it is true that the water pressure is the same at the bottom of each vessel. And it is true that the force exerted by the water on the bottom of each vessel is the same, because the areas of the bottoms of the vessels are equal. But the total force exerted by the water on the vessels is not the same for all three vessels. Remember that a fluid in static equilibrium in a container exerts a normal force at each point of a container surface proportional to the pressure in the fluid at that point (N2 again). So we must really consider the pressures and forces exerted by the fluid on all parts of a container surface, not just those on its bottom surface. Now look at the container at the right in the diagram. Its walls are vertical, so the normal forces exerted by the water on those vertical walls have no vertical components. Furthermore, the horizontal components of the forces due to water pressure sum to zero, so the container doesn't move horizontally. Now consider the middle container in the diagram. The normal forces exerted by the water on its side walls do have vertical components, and they're upward everywhere around the container. This net upward force is opposite in direction to the force exerted by the water on the bottom of the container, and thus reduces the "weight" of the full container as measured by a scale. For the container at the left the opposite is true. The normal forces exerted by the water on the container side walls have a downward vertical component that adds to the force exerted by the water on the bottom. The result is that this container weighs more.

You could, if you wished, calculate the side-wall forces for the three idealized containers considered here, being careful to take into account the fact that the pressure due to the water varies with its depth. What you would find is that they are just the forces necessary to provide different measurable weights for the containers that are exactly what you would expect from the different volumes of water they contain. The remarkable thing is that this result is totally independent of the shape of the container. It might, for example, be artistically shaped like a glass swan. Would you call this intuitively obvious or weirdly counter-intuitive?