. A hydraulic lift consists of two cylindrical pistons, one with a radius of 1.5 m and the other with a radius of 8.0 cm. What force must be applied to the smaller piston if
a crate with a mass of 1.5x103 kg is to be raised on the larger piston?
F=42N,

The force that must be applied to the smaller piston if the crate is to be raised by the larger piston is approximately 42 N

Explanation:

The given parameters of the hydraulic lift are;

The diameter of one of the pistons = 1.5 m

The diameter of the other pistons = 8.0 cm = 0.08 m

The mass of the crate to be lifted with the larger piston, m = 1.5 × 10³ kg

The weight of the mass to be lifted by the larger piston, W₁ = m × g

Where;

g = The acceleration due to gravity = 9.8 m/s²

Therefore;

W₁ = 1.5 × 10³ kg × 9.8 m/s² = 14,700 N

The area of the surface of the circular piston, A = π × r²

Where;

r = The piston's radius

∴ The area of the larger piston with 1.5 m radius, A₁ = π × (1.5 m)² = 7.06868347058 m² ≈ 7.07 m²

A₁ ≈ 7.07 m²

The area of the smaller piston with 0.08 m (8 cm) radius, A₂ = π × (0.08 m)² = 02010619298 m² ≈ 0.02 m²

A₂ ≈ 0.02 m²

The pressure in the fluid of the hydraulic lift is constant, therefore, we have;

The pressure applied to the larger piston with 1.5 m radius, P₁ = The pressure applied to the smaller piston with 0.08 m radius, P₂

∴ P₁ = P₂

Where, F₂, represents the force that must be applied to the smaller piston if the crate is to be raised by the larger piston;

Given that we have the ratio of two areas which have π as a factor, we have;

The force that must be applied to the smaller piston if the crate is to be raised by the larger piston, F₂ = 41.81 N ≈ 42 N.

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