### 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 πΉ sub two β is 50.0
times greater than the input force β we can call it πΉ 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
π΄ sub two, the area at location two, to π΄ 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 π one and the distance that the
secondary cylinder ascends π two, then we want to solve for the ratio of those
distances π one to π two. We can start solving for these
three ratios by recalling the fact that pressure π is equal to force divided by
area and that all throughout this closed hydraulic system the pressure π is a
constant, which means we can write that the pressure at point one is equal to the
pressure at point two or πΉ one over π΄ one equals πΉ two over π΄ two.

We can rearrange this expression to
read π΄ two over π΄ one equals πΉ two over πΉ one. And by our given design constraint,
we know that πΉ two is equal to 50 times πΉ one. So in this ratio, πΉ one cancels
out. And we find that π΄ two over π΄ one
is 50.0. Thatβs 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 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
over four in numerator and denominator cancel.

From our earlier work, we know that
this ratio π two over π 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 π two to π
one. To three significant figures, this
is 7.07. Thatβs the ratio of the diameter of
the secondary cylinder to that of the master cylinder. Finally, we move on to solving for
π one over π two, the ratio of the distance the master cylinder descends to the
distance the secondary cylinder ascends. To solve for this ratio, what weβll
consider is the volume of fluid thatβs being moved at each of these points one and
two.

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