Question

The piston of a hydraulic elevator used for lifting trucks has a 0.3m radius. What pressure is required to lift a truck of 2500 kg mass? What force was applied to the small piston if it has a radius of 3cm?

Answer 1

Answer:

1. 88370.45 N/$m^{2}$

2. 2500 N.

Explanation:

Pressure, P, is define as the force, F, per unit area, A, applied on/ by an object.

i.e P = $\frac{F}{A}$

1. The area, A, of the piston of the hydraulic elevator can be determined by;

A = $\pi$$r^{2}$

where r is the radius of the piston.

A = $\frac{22}{7}$ × $(0.3)^{2}$

  = $\frac{22}{7}$ × 0.09

  = 0.2829 $m^{2}$

Pressure required to lift a truck of 2500 kg mass can be determined as;

P = $\frac{F}{A}$

F = W = mg

  = 2500 × 10

  = 25 000 N

So that,

P = $\frac{25000}{0.2829}$

  = 88370.45 N/$m^{2}$

The pressure required is 88370.45 N/$m^{2}$.

2. $\frac{F_{1} }{A_{1} }$ = $\frac{F_{2} }{A_{2} }$

 Area of small piston = $\pi$$r^{2}$

                                   = $\frac{22}{7}$ × $(0.03)^{2}$

                                   = 0.02829 $m^{2}$

So that,

$\frac{25000}{0.2829}$ = $\frac{F_{2} }{0.02829}$

$F_{2}$ = 2500 N

The force applied to the small piston is 2500 N.