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Sagot :
Answer:
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
[tex]Pressure = \dfrac{Force}{Area}[/tex]
[tex]P_1 = \dfrac{W_1}{A_1} = \ P_2 = \dfrac{F_2}{A_2}[/tex]
[tex]\therefore \dfrac{W_1}{A_1} = \ \dfrac{F_2}{A_2}[/tex]
Where, F₂, represents the force that must be applied to the smaller piston if the crate is to be raised by the larger piston;
[tex]\therefore F_2 = A_2 \times \dfrac{W_1}{A_1}[/tex]
Given that we have the ratio of two areas which have π as a factor, we have;
[tex]\therefore F_2 = A_2 \times \dfrac{W_1}{A_1} = 0.08^2 \times \pi \times \dfrac{14,700}{1.5^2 \times \pi } = 41.81 \overline 3[/tex]
The force that must be applied to the smaller piston if the crate is to be raised by the larger piston, F₂ = 41.81[tex]\overline 3[/tex] N ≈ 42 N.
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