The diameters of the two pistons of a hydraulic pressare 0.1 m and 1 m. The lengths of the two arms of a first class lever are 0.15 m and 0.9 m. A 50 N force is applied at the end of the lever. Determine the total thrust developed on the larger piston.

Explanation

We know that the thrust developed on the larger piston,
`F_(2)=W*y/x*beta/alpha`
Here, force applied at the end of the lever arm, W = 50N
Area of the larger piston, `beta=pi(0.5)^(2)=0.25pim^(2)`
Area of the smaller piston, `alpha=pi(0.05)^(2)=0.0025pim^(2)`
Length of the longer arm of the lever, y = 0.9 m
Length of the smaller arm of the lever,
x = 0.15 m
`therefore" "F_(2)=50xx0.9/0.15xx(0.25pi)/(0.0025pi)=3xx10^(4)N`.