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.

ADetermine the force that must be applied to the small piston.
BDetermine the pressure applied to the fluid in the lift.

Verified Answer :

Answer:

Part a)

F = 600 N

Part b)

P = 119366.20 Pascal

Explanation:

Part a)

As we know that by Pascal's law pressure must be transmitted through out the liquid without any change

So pressure at bigger piston = pressure at smaller piston

\(\frac{F_1}{A_1} = \frac{F_2}{A_2}\)

\(\frac{F_1}{\pi(0.04)^2} = \frac{15000}{\pi(0.20)^2}\)

now by solving above equation we have

\(F_1 = \frac{16}{400} (15000)\)

\(F_1 = 600 N\)

Part b)

Pressure applied at fluid is given as

\(P = \frac{F}{A}\)

\(P = \frac{15000}{\pi(0.20)^2}\)

P = 119366.20 Pascal