Question

A mechanic uses a hydraulic lift to raise a car above her head. The car weighs 22,500N, and is raised by a piston with a 30cm diameter. What is the diameter of the smaller piston, if the force acting on it is 1500 N

Answer 1

First, take into account that the pressure on each piston must be equal:

$P_1=P_2$

then, by conisdering that P=F/A, you have:

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

furthermore, consider that the area of the piston (circular shape) is:

$A=\pi r^2$

then, in order to determine the diameter of the smaller piston, replace the previous expression into the equation for F/A, and solve it for r2, just as follow:

$\begin{gathered} \frac{F_1}{\pi r^2_1}=\frac{F_2}{\pi r^2_2} \\ r_2=\sqrt[]{\frac{F_2}{F_1}r^2_1}^{} \end{gathered}$

take into account that the radius of the bigger piston is

r1=d1/2=30cm/2 = 15cm,

then by replacing F1 = 22,500N and F2=1,500N, you obtain for r2:

$\begin{gathered} r_2=\sqrt[]{\frac{(1500N)(15cm)^2}{22500N}} \\ r_2=3.87\operatorname{cm} \end{gathered}$

the diameter of the smaller piston is twice its radius, then, you have:

d2 = 2*r2 = 2*(3.87 cm) = 7.74 cm