Ahydraulic 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 : 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 :

---------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 :

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

Area = meter square (m²)

On plugging value of pressure in equation (1) :

------ 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) :

Multiplying both side by 0.0725 m²

F2 = 11,745 N

the answer would be c