Answer:
the force applied to the smaller piston is 600 N
Explanation:
Given;
weight of the car, F = 15,000 N
radius of the lager piston, R = 0.2 m
radius of the smaller piston, r = 0.04 m
let the force applied to the smaller piston = f
The pressure applied on both piston is constant;
[tex]P = \frac{F}{A} = \frac{f}{a} \\\\\frac{F}{R^2} = \frac{f}{r^2} \\\\f = \frac{F\times r^2}{R^2} = \frac{15,000 \times (0.04)^2}{(0.2)^2} = 600 \ N[/tex]
Therefore, the force applied to the smaller piston is 600 N
