Question

A floor jack constists of a small piston of cross-sectional area 2.0 cm2 connected to a large 2 piston of cross-sectional area 100 cm . What is the force on the small piston necessary to lift a 10,000 N car that rests on the large piston

Answer 1

force acting on small piston is 200 N

Step-by-step explanation:

given data

cross-sectional area a = 2.0 cm² = 2 ×
`10^(-4)` m²

cross-sectional area a = 100 cm²= 100 ×
`10^(-4)` =
`10^(-2)`m²

force f = 10,000 N

to find out

force acting on small piston

solution

we know that here pressure will be equal at all level in horizontal

so pressure at p1 = pressure p2

and pressure =
`(force)/(area)`

so
`(force1)/(area1)` =
`(force2)/(area2)`

put here value we get

`(force1)/(2*10^(-4))` =
`(10000)/(10^(-2))`

solve it we get

force 1 = 200 N

so force acting on small piston is 200 N