Question

two pistons of a hydraulic lift have diameters of 60cm and 5cm. what is the force exerted by the larger piston when 50N is placed on the smaller piston?

a) 720N
b) 50N
c) 500N
d) 7200N​

Answer 1

The force exerted by the larger piston when 50N is placed on the smaller piston is 7200 N

Explanation:

Given that,

Two pistons of a hydraulic lift having diameters of 60 cm and 5 cm.

As diameter of the pistons are given in question thus calculate the radius of the piston,   $r = \frac{d}{2}$

Radius of small piston = 5/2 = 2.5

Radius of large piston =60/2 = 30

Area of small piston,  A1 = $\pi (\frac{5}{2})^2$ = $\pi (2.5)^2$

Area of large piston,  A2 = $\pi (\frac{60}{2})^2$ = $\pi (30)^2$

Pressure p2 is felt at the other piston that is equal to p1

Thus p1 = p2

$p1 = \frac{F1}{A1}$

$p2 = \frac{F2}{A2}$

$\frac{F1}{A1} = \frac{F2}{A2}$

$F2=\frac{A2}{A1} \times F1 = 50\times (\frac{30}{2.5})^2$

$= 7200N$

Thus, the force exerted by the larger piston when 50N is placed on the smaller piston is 7200 N

Answer 2

Answer:

d 7200

Explanation:

because F1/A1=F2/A2

A=R