A certain hydraulic system is designed to exert a force 100 times as large as the one put into it. (a) What must be the ratio of the area of the slave cylinder to the area of the master cylinder? (b) What must be the ratio of their diameters? (c) By what factor is the distance through which the output force moves reduced relative to the distance through which the input force moves? Assume

A) 100

B) 10

C) 0.01

Explanation:

Let

Fs = force on slave cylinder

Fm = force on master cylinder

Given that the designed to exert a force 100 times as large as the one put into it. That is

Fs = 100 Fm

Using Pascal's law

Fs/As = Fm/Am

100Fm/As = Fm/Am

100/As = 1/Am

Cross multiply

As= 100Am

As/Am = 100

b) What must be the ratio of their diameters

Using their areas ratio, area of a cylinder is  πr^2

A = πr^2 =  π(D/2)^2

 π(Ds/2)^2 ÷  π(Dm/2)^2 = 100

(Ds/2)^2 ÷ (Dm/2)^2 = 100

(Ds/Dm)^2 = 100

Ds/Dm = 10

(c) By what factor is the distance through which the output force moves reduced relative to the distance through which the input force moves? Assume no losses to friction

Let us consider the workdone at the input and at the output.

Work done = force × distance

Fs × Hs = Fm × Hm

100Fm × Hs = Fm × Hm

100Hs = Hm

100Hs/Hm = 1

Hs/Hm = 1/100

Hs/Hm = 0.01

the ima (higher than 1m) will be greater than the ima with the height of 1m.