Question

in a hydraulic garage lift, the small piston has a radius of 5.0 cm and the large piston has a radius of 15 cm. what force in newton must be applied on the small piston in order to lift a car weighing 20,000 n on the large piston? assume the pistons each have negligible weight.

Answer 1

To lift a car weighing 20,000 N on a large piston using a hydraulic lift with a small piston of radius 5.0 cm and a large piston of radius 15 cm, a force of approximately 1,111 N must be applied on the small piston.

Step-by-step explanation:

The force that must be applied on the small piston to lift a car weighing 20,000 N on the large piston can be calculated using Pascal's principle:

From Pascal's principle, the force needed to lift the car is less than the weight of the car:

F1/A1 = F2/A2

Where F1 is the force to be applied on the small piston, A1 is the cross-sectional area of the small piston, F2 is the weight of the car (20,000 N), and A2 is the cross-sectional area of the large piston.

Given that the radius of the small piston is 5.0 cm and the radius of the large piston is 15 cm, we can calculate the force required:

F1 = (A1/A2) * F2

F1 = (5.0 cm/15 cm)^2 * 20,000 N

Using this equation, we find that the force required to lift the car is approximately 1,111 N.