Question

A hydraulic system is designed to lift cars for inspection in a service station. The narrow end of the system has a surface area of 5.00 cm2, and the lift platform (the wide end) has a surface area of 725 cm2. If a force of 81.0 newtons is applied to the narrow end, how much upward lift force will be exerted at the wide end?
Use the following formula:


0.559 N
11,745 N
44.75 N
293,625 N

Answer 1

Given:

Surface area at the narrow end, A1 = 5.00 cm2

Force applied at the narrow end, F1 = 81.0 N

Surface area at the wide end, A2 = 725 cm2

To determine:

Force F2 applied at the wide end

Explanation:

Use the relation

F1/A1 = F2/A2

F2 = F1*A2/A1 = 81.0 N * 725 cm2/5.00 cm2 = 11,745 N

Ans: (b)

The force applied at the wide end = 11,745 N

Answer 2

Answer : The correct answer for force = 11.745 N .

Hydraulic system is a system which is operated by moving liquid in a definite area which is under pressure . It uses the force which is produced due to difference in pressure at two ends of a system .

It basically uses oil or water ( incompressible liquids ) , which can relocate the force from one area to other in the fluid .

The force around different areas can be calculated using Pascal's Principle.

Pascal's Principle : It states that if there in increase in pressure in any area in fluid , there will be equal to increase in pressure at other area . It can be expressed as :

$P1 = P2$ ---------Equation (1)

, where P1 = pressure in one area P2 = pressure in other area .

Pressure is defined as force acting per unit area . Hence , Pressure and force can be related as :

$Pressure(P) = \frac{Force }{area }$

Where unit of pressure= pascals(p) Force= newton(N)

Area = meter square (m²)

On plugging value of pressure in equation (1) :

$\frac{F1}{A1} = \frac{F2}{A2}$ ------ Equation (2)

Where , F1 = force at one area A1 = area of one area

F2 = force at other area A2 = area of other area

Given : F 1 = 81 N A1 = 5 cm² or 0.0005 m² ( 1 m²= 10000 cm²)

F2 = ? A2 = 725 cm² or 0.0725 m²

Plugging these value in equation (2) :

$\frac{81 N}{0.0005 m^2} = \frac{F2}{0.0725 m^2}$

Multiplying both side by 0.0725 m²

$\frac{81 N}{0.0005 m^2} * 0.0725 m^2 = \frac{F2 }{0.0725 m^2} * 0.0725 m^2$

F2 = 11,745 N