A thin, rectangular-prism-shaped piston has sides 0.01 meters and 0.015 meters in length. The piston is pushed a horizontal distance of 0.25 meters along a tube attached to a larger water tank, as shown in the diagram. The tube has the same cross section as the piston. The water tank is a rectangular prism with a cross section that has sides of lengths of 0.05 meters and 0.075 meters. Floating on top of the water in the tank is a thin, wooden, rectangular-prism-shaped plank that has the same cross section as the water tank. Assume that the friction between the piston and the tube and that between the plank and the tank is negligible. What is the distance 𝑑 that the plank rises?
Looking at our diagram, we see this rectangular-prism-shaped piston. The piston is located in a tube filled with water that’s connected to this larger tank. Resting on top of the water in the tank is this thin board. In this scenario, we’re told that our piston moves a distance of 0.25 meters in along the tube. This causes the water in the tube and then the water in the tank to move.
Water is an example of an incompressible fluid. This means the volume of the water in this system before the piston started to move and the volume after it moved is the same. Because of the piston’s motion, the surface of water in the tank moved up some distance 𝑑. This is the same as the distance that the plank on top of the water rose. We want to solve for that distance 𝑑. And to do it, let’s recall that the volume of water in this system is the same at all times. Therefore, when the moving piston clears out this volume of water we’ve marked in pink, that leads to an increase in the volume of water in the tank marked out here.
If we call the volume of water cleared out by the moving piston 𝑉 one and that volume added to the tank 𝑉 two, we can simply write that 𝑉 one equals 𝑉 two. When it comes to the volume 𝑉 one, we have all the dimensions of this rectangular prism that make up that volume marked out. The volume 𝑉 one has a height of 0.015 meters, a depth of 0.01 meters, and a width of 0.25 meters. When it comes to the volume 𝑉 two, this volume has a height of distance 𝑑, a depth of 0.05 meters, and a width of 0.075 meters.
In this equation, we know everything except the distance 𝑑. If we divide both sides of this equation by the width and depth of the volume 𝑉 two, then on the right-hand side both that depth and that width cancel out. The distance 𝑑 is equal to this fraction. Before we calculate this result, notice that both of the factors of meters in the denominator cancel out with the factor of meters in the numerator. We’ll be left with a final answer with units of meters.
Calculating 𝑑, we find that it’s exactly 0.01 meters. Or since there are 100 centimeters in a meter, this distance is one centimeter. Due to the motion of the piston in the tube, the plank rises through a distance of one centimeter.
Let’s look now at part two of this question.
The work done on the piston to move it is five joules. What is the average force applied to the plank?
Let’s begin on our answer by recalling that the work done on an object equals the force acting on that object multiplied by the distance the object travels. In the case of our plank, we’re told that the work done on it is five joules, and we solved for the distance it moves in the previous part of our question. What we’ll do then is rearrange this equation so that 𝐹 is the subject. We can do that by dividing both sides by the distance 𝑑 so that it cancels on the right and then using the resulting relationship to solve for the average force 𝐹.
Note that when we solve for 𝐹, this is indeed an average force. If the piston happens to move at a constant speed, then the water will exert a constant force on the plank. If the piston doesn’t move at a constant speed though, sometimes moving faster and sometimes slower, then the upward-acting force on the plank will vary over time. Any variation like that, though, doesn’t come into account when we’re solving for the average force applied to the plank. Now we’ve seen that the work done on the plank is five joules and the distance through which it travels is one centimeter.
Before we calculate 𝐹, though, we’d like to change this distance into the SI base unit of meters. We recall that 100 centimeters equals one meter, which means that one centimeter is one one hundredth of a meter. That is, it’s equal to 0.01 meters. Now, when we calculate this fraction, we’ll get a result in units of newtons. Five divided by 0.01 is 500, so the average force applied to the plank is 500 newtons.