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

Sm = small piston
la = large piston

P=F/A
P=15000/(20^2)π

F of sm = PA
= (75/2π)•((8^2)π)
= (75•64)/2
= 4800/2
= 2400N
We already know the pressure but giving it in an approximate decimal form, to two significant figures (since that's what your supplied precision is at):

a) 12 Pa
b) 2400 N

Answer 2

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