Question

To inspect a 12,500 N car, it is raised with a hydraulic lift. If the radius of the small piston is 5.0 cm, and the radius of the large piston (the car is sitting on the large piston) is 50 cm, find the force that must be exerted on the small piston to lift the car. N (precision to the nearest whole number)

Answer 1

Answer:

The force will be "125 N".

Explanation:

The given values are:

$F_1=12500 \ N$

$R_1 = 50 \ cm$

$R_2=5 \ cm$

As we know,

⇒  $A=\pi(H)^2$

then,

⇒  $A_2=\pi(5)^2$

⇒  $A_1=\pi(50)^2$

Since,

The pressure on both the pistons are equal, then

⇒  $\frac{F_1}{A_1} =\frac{F_2}{A_2}$

or,

⇒  $\frac{F_2}{F_1} =\frac{A_2}{A_1}$

By substituting the values, we get

⇒  $\frac{F_2}{12500} =\frac{\pi(5)^2}{\pi(50)^2}$

⇒  $\frac{F_2}{12500} =\frac{\pi(25)}{\pi(2500)}$

⇒      $F_2=\frac{25}{2500}\times 12500$

⇒           $=0.01\times 12500$

⇒           $=125 \ N$