Question

The small piston of a hydraulic lift has a diameter of 8.0 cm, and its large piston has a diameter of 40cm. the lift raises a load of 15,000N.
a) Determine the force that must be applied to the small piston.
b) Determine the pressure applied to the fluid in the lift.

Answer 1

a) Pressure on both sides must be equal, that is:
\[\frac{F_1}{A_1}=\frac{F_2}{A_2}\]
here load is bigger area, and applied force is on small area. remember area of the circular objects
\[\pi r^2=\pi (d/2)^2\]
thus
\[\frac{15000}{\pi 20^2}=\frac{F}{\pi 8^2}\]
solve for F
b) pressure is either side of the equality, but you may need to convert the units to get (N/m^2)
\[P=\frac{F}{A}=\frac{150000/\pi(0.2^2)}]

Answer 2

\[P=\frac{F}{A}=\frac{150000}{pi(0.2^2)}\]

Answer 3

Thanks, but i think you meant 40 instead of 20 for part a
\[\frac{ 15000}{ 40^2 }=\frac{ F }{ 8^2 }\]
and I got 600N.

b)I have the answer but not sure about the solution.
the answer is 120000 Pa

Answer 4

40 is diameter, you need to use radius (r=d/2), so 8 must be 4 for part a, you have that in both sides, so they cancel out. your solution should give the correct result.
for part b, I have 150 000 instead of 15 000, if you correct that, you will get:
119426.75N/m^2 you can round it to 120 000Pa