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

a) F = 4.9 10⁴ N, b) F₁ = 122.5 N

Step-by-step 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

`(F_1)/(A_1) = (F_2)/(A_2)`

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

F₁ =
`(A_1)/(A_2) \ F_2`

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

F₁ =
`( (d_1)/(d_2) )^2\ m g`

F₁ = (0.05 / 1) ² 5000 9.8

F₁ = 122.5 N