Question

example 3
the large piston in a hydraulic lift has a radius of 20 cm. what force must be applied to the small piston of radius 2 cm to raise a car of mass 1,500 kg?
solution:
pascals law gives, ( p_s = p_l ),
( \frac{f}{a}=\frac{f}{a} )
( f=\frac{a}{a}f=\frac{pi r^{2}}{pi r^{2}}mg )
to lift, ( f ) must be equal to the weight of the car
( mg=(1,500 kg)(9.81 n/kg)=1.47\times10^{4}n )
( f=\frac{(0.02 m)^{2}}{(0.200 m)^{2}}(1.47\times10^{4} n)=147 n )

Explanation:

Step1: Determine the weight of the car

The weight of the car \(F = mg\), where \(m = 1500\space kg\) and \(g=9.81\space N/kg\).

$$F=(1500\space kg)(9.81\space N/kg)=1.47\times 10^{4}\space N$$

Step2: Apply Pascal's law

Pascal's law states \(\frac{f}{a}=\frac{F}{A}\), and since \(a = \pi r^{2}\) and \(A=\pi R^{2}\), then \(f=\frac{\pi r^{2}}{\pi R^{2}}F=\frac{r^{2}}{R^{2}}F\).
Given \(r = 0.02\space m\) and \(R = 0.2\space m\), \(F = 1.47\times 10^{4}\space N\)

$$f=\frac{(0.02\space m)^{2}}{(0.2\space m)^{2}}\times(1.47\times 10^{4}\space N)$$

$$f=\frac{0.0004}{0.04}\times(1.47\times 10^{4}\space N)$$

$$f = 0.01\times(1.47\times 10^{4}\space N)=147\space N$$

Answer:

\(147\space N\)