Video: Applying Pascal’s Principle to a Hydraulic System

A certain hydraulic system is designed to exert an output force 50.0 times as large as its input force. The system consists of two cylinders, the master cylinder and the second cylinder. What must the ratio of the areas of the second cylinder and master cylinder be? What must the ratio of the diameters of the second cylinder and master cylinder be? By what factor is the distance through which the input force moves relative to the distance through which the output force moves? Assume no losses du

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Video Transcript

A certain hydraulic system is designed to exert an output force 50.0 times as large as its input force. The system consists of two cylinders, the master cylinder and the second cylinder. What must the ratio of the areas of the second cylinder and master cylinder be? What must the ratio of the diameters of the second cylinder and master cylinder be? By what factor is the distance through which the input force moves relative to the distance through which the output force moves? Assume no losses due to friction.

In this scenario, we have a hydraulic system with an input or master cylinder and an output or secondary cylinder. The system is designed to meet a particular specification that the output force β€” we can call it 𝐹 sub two β€” is 50.0 times greater than the input force β€” we can call it 𝐹 sub one. If we call the location in the system where the first force is applied, location one, and the place in the system where the second is applied, location two, we want to solve for three pieces of information. First, we want to find the ratio of 𝐴 sub two, the area at location two, to 𝐴 sub one, the area at location one.

Next, we want to find the ratio of the diameters of the two circular cylinders at those same two spots. And finally, if we call the distance that the first or master cylinder descends 𝑠 one and the distance that the secondary cylinder ascends 𝑠 two, then we want to solve for the ratio of those distances 𝑠 one to 𝑠 two. We can start solving for these three ratios by recalling the fact that pressure 𝑃 is equal to force divided by area and that all throughout this closed hydraulic system the pressure 𝑃 is a constant, which means we can write that the pressure at point one is equal to the pressure at point two or 𝐹 one over 𝐴 one equals 𝐹 two over 𝐴 two.

We can rearrange this expression to read 𝐴 two over 𝐴 one equals 𝐹 two over 𝐹 one. And by our given design constraint, we know that 𝐹 two is equal to 50 times 𝐹 one. So in this ratio, 𝐹 one cancels out. And we find that 𝐴 two over 𝐴 one is 50.0. That’s the ratio of the area of the second cylinder to that of the primary or master cylinder. Next, we want to solve for the ratios, not the areas of these cylinders, but their diameters. And to solve for this, we can build off of our result for part one. Recalling that the area of a circle is equal to πœ‹ times its diameter squared divided by four, we can rewrite our area ratio in terms of the diameters. And we see that the factor of πœ‹ over four in numerator and denominator cancel.

From our earlier work, we know that this ratio 𝑑 two over 𝑑 one quantity squared is equal to 50.0. So if we take the square root of both sides, at that point then, we have an expression for the ratio 𝑑 two to 𝑑 one. To three significant figures, this is 7.07. That’s the ratio of the diameter of the secondary cylinder to that of the master cylinder. Finally, we move on to solving for 𝑠 one over 𝑠 two, the ratio of the distance the master cylinder descends to the distance the secondary cylinder ascends. To solve for this ratio, what we’ll consider is the volume of fluid that’s being moved at each of these points one and two.

Since the fluid in this system is incompressible, that means that the volume 𝑉 one, which is the amount of fluid moved by 𝐹 one pressing down on the master cylinder, is equal to the volume 𝑉 two. Since this hydraulic system is a closed system, no fluid can escape and the fluid is incompressible. Since 𝑠 one and 𝑠 two are distances, if we multiply those distances by the cross-sectional areas of their respective points, writing 𝑉 one equals 𝑉 two is equivalent to writing 𝑠 one 𝐴 one equals 𝑠 two 𝐴 two. When we cross-multiply, we find that 𝑠 one over 𝑠 two is equal to 𝐴 two over 𝐴 one, which we solved for earlier. It’s equal to 50.0. That’s the ratio of the master cylinder’s distance moved to the secondary cylinder’s distance moved.

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