Question

1. A hydraulic lift is to be used to lift a truck masses 5000 kg. If the diameter of the
small piston and large piston of the lift is 5 cm and 1 m respectively,
a. What gauge pressure in Pa must be applied to the oil?
b. What is the magnitude of the force required on the small piston to lift the truck?​

Answer 1

Answer:

a)   F = 4.9 10⁴ N,  b)   F₁ = 122.5 N

Explanation:

To solve this problem we use that the pressure is transmitted throughout the entire fluid, being the same for the same height

1) pressure is defined by the relation

           P = F / A

to lift the weight of the truck the force of the piston must be equal to the weight of the truck

          ∑F = 0

          F-W = 0

          F = W = mg

          F = 5000 9.8

          F = 4.9 10⁴ N

the area of ​​the pisto is

          A = pi r²

          A = pi d² / 4

          A = pi 1 ^ 2/4

          A = 0.7854 m²

pressure is

          P = 4.9 104 / 0.7854

          P = 3.85 104 Pa

2) Let's find a point with the same height on the two pistons, the pressure is the same

          $\frac{F_1}{A_1} = \frac{F_2}{A_2}$

where subscript 1 is for the small piston and subscript 2 is for the large piston

          F₁ = $\frac{A_1}{A_2} \ F_2$

the force applied must be equal to the weight of the truck

          F₁ = $( \frac{d_1}{d_2} )^2\ m g$

          F₁ = (0.05 / 1) ² 5000 9.8

          F₁ = 122.5 N