What magnitude force must be
exerted on the master cylinder of a hydraulic lift to support the weight of a car
with a mass of 2500 kilograms resting on a second cylinder? The master cylinder’s diameter is
2.50 centimeters and the second cylinder’s diameter is 25.0 centimeters.
From the information given, we can
label the mass of the car 2500 kilograms as 𝑚, the diameter of the master cylinder
2.50 centimeters as 𝑑 sub 𝑚, and the diameter of the second cylinder 25.0
centimeters we’ll call 𝑑 sub two. We want to solve for the force
magnitude that must be exerted on the master cylinder to accomplish this lift. We’ll call this force capital 𝐹
and start on our solution by drawing a diagram of the situation. In this scenario, we have a master
cylinder where we apply a force and a secondary cylinder that does the lifting of
the object we’re interested in elevating. In our case, that object is a car
of mass 𝑚 of 2500 kilograms. We apply a force of magnitude 𝐹 to
our master cylinder creating a pressure in our hydraulic system. That pressure is transmitted
through this system and then acts to keep the car elevated.
To solve for the magnitude of that
applied force 𝐹, we can recall that pressure is equal to force spread over an area
𝐴. And we also recall that in a
hydraulic system such as we have here pressure is constant. That means if we choose any two
points in our system, we’ll choose the input master cylinder and the output or
secondary cylinder. The pressure at these two points,
we can call them 𝑃 one and 𝑃 two, are the same. They’re equal to one another. This means that if we look at the
area, where 𝑃 one is applied and we call that area 𝐴 sub 𝑚, then we look at the
area, where 𝑃 two is applied and we call that area 𝐴 sub two, by the equality of
𝑃 one and 𝑃 two, we can write 𝐹 over 𝐴 sub 𝑚 equals the force applied by the
car, its mass times gravity, divided by 𝐴 sub two.
Rearranging this expression to
solve for the force 𝐹, we see it’s equal to 𝐴 sub 𝑚, the area of our master
cylinder in cross section, divided by 𝐴 sub two, the cross-sectional area of our
secondary lift, all multiplied by the weight force of the car 𝑚 times 𝑔. We can treat 𝑔, the constant, as
exactly 9.8 meters per second squared. And to solve for 𝐴 sub 𝑚 and 𝐴
sub two, we remember that the cross sections of these cylinders are circles. We recall that the area of the
circle is equal to 𝜋 times its diameter squared divided by four. The diameter of the master cylinder
and the diameter of the secondary cylinder are given.
So we can rewrite our expression
for force as 𝜋 over four 𝑑 sub 𝑚 squared over 𝜋 over four 𝑑 sub two squared all
multiplied by 𝑚𝑔. And we see the factors of 𝜋 over
four cancel out. Knowing 𝑑 sub 𝑚, 𝑑 sub two, 𝑚,
and the acceleration due to gravity 𝑔, we’re ready to plug in and solve for 𝐹. When we do, entering these values
on our calculator, we find that 𝐹 is 245 newtons. That’s the force we would need to
apply to lift up a car of mass 2500 kilograms.