A hydraulic press is an ingenious device that allows the...

A hydraulic press is an ingenious device that allows the lifting of heavy objects by the application of a small force to a small piston. A schematic of a hydraulic press is shown below: Figure can't copy A force $\mathrm{F}_1$ is applied to Piston 1. This creates a pressure $P_1$ on Piston 1, which increases the pressure uniformly throughout the much larger oil reservoir and results in a force $\mathrm{F}_2$ on Piston 2. A pressure $\mathrm{P}_2$ is therefore applied to Piston 2. In the following problems, assume that the radius of Piston 2, $r_2$, is five times that of the radius of Piston $1, r_1$. If Piston 1 is moved a distance $x_1$, by how much will Piston 2 move? A. $x_2=5 x_1$ B. $x_2=10 x_1$ C. $x_2=\frac{x_1}{5}$ D. $x_2=\frac{x_1}{25}$

A hydraulic press is an ingenious device that allows the lifting of heavy objects by the application of a small force to a small piston. A schematic of a hydraulic press is shown below:
Figure can't copy
A force $\mathrm{F}_1$ is applied to Piston 1. This creates a pressure $P_1$ on Piston 1, which increases the pressure uniformly throughout the much larger oil reservoir and results in a force $\mathrm{F}_2$ on Piston 2. A pressure $\mathrm{P}_2$ is therefore applied to Piston 2.

In the following problems, assume that the radius of Piston 2, $r_2$, is five times that of the radius of Piston $1, r_1$.

If Piston 1 is moved a distance $x_1$, by how much will Piston 2 move?
A. $x_2=5 x_1$
B. $x_2=10 x_1$
C. $x_2=\frac{x_1}{5}$
D. $x_2=\frac{x_1}{25}$

Transcript

00:01 So in this example, we're told that we have two pistons, and they're connected by a small tube or small pipe, that there is some liquid in their cylinders.

00:12 And what part a asks is what force magnitude with a larger piston sustain without moving? so because it's not moving, that tells us that both pistons are at equilibrium, which means that the force on the small piston divided by the area of the small piston, has to be equal to the force in the large piston divided by the area of the large piston.

00:39 We can rearrange this to solve for the force on the large piston.

00:43 That tells us that the force in the large piston will be equal to the area of that piston divided by the area of the small piston times the force on the small piston.

00:58 In part b, they actually give us values.

01:04 And what they want to know is what force magnitude on the small piston will be.

01:09 Balance out a 20 -kilonuton force on the large piston.

01:17 So they tell us that the diameter of the small piston is 3 .8 centimeters because we know the radius is half the diameter.

01:24 It tells us the radius is 1 .9 centimeters.

01:27 They tell us that the diameter of the large piston is 53 centimeters, which means its radius is 26 .5 centimeters.

01:33 And of course, they tell us that the force on a large piston is 20 kilonetons.

01:40 So they want to know what force on the small piston will equal that out, right, so that they'll still be at equilibrium...

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