Question

a effort of 100n can raise a load of 2000n in a hydraulic press. calculate the cross-sectional area of a small piston in it. The cross-sectional area of a large piston is 4m^s​

Answer 1

Answer:

$A_{1}$ = 0.2 $m^{2}$

Explanation:

The pressure on the pistons is given as;

Pressure = $\frac{Force}{Area}$

So that,

Pressure on the small piston = $\frac{F_{1} }{A_{1} }$ and Pressure on the large piston = $\frac{F_{2} }{A_{2} }$

Thus,

$\frac{F_{1} }{A_{1} }$ = $\frac{F_{2} }{A_{2} }$

Given that: $F_{1}$ = 100 N, $F_{2}$ = 2000 N, $A_{2}$ = 4 $m^{2}$.

$\frac{100}{A_{1} }$ = $\frac{2000}{4}$

$A_{1}$ = $\frac{100*4}{2000}$

    = $\frac{400}{2000}$

    = 0.2

$A_{1}$ = 0.2 $m^{2}$

The area of the small piston is 0.2 $m^{2}$.