Question

the ratio of the area of the larger piston to the smaller piston Ina hydraulic lift is 25. if the force on the smaller piston is 75N, how much load can we lift using it​

Answer 1

Assuming that the hydraulic lift is ideal and frictionless, the force exerted on the larger piston is directly proportional to the area of the larger piston and inversely proportional to the area of the smaller piston.

Let A1 be the area of the smaller piston, A2 be the area of the larger piston, and F1 be the force exerted on the smaller piston. Then, the force F2 exerted on the larger piston is given by:

F2 = (A2/A1) * F1

We are given that the ratio of the areas A2/A1 is 25, so we can write:

F2 = 25 * F1

Substituting F1 = 75N, we get:

F2 = 25 * 75N = 1875N

Therefore, the hydraulic lift can lift a load of up to 1875N using the smaller piston.

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