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Question: A boy lifts up a stone of \( 1000kg \) using a hydraulic lift as shown below! ![](https://www.veda...

A boy lifts up a stone of 1000kg1000kg using a hydraulic lift as shown below!

If the larger piston area has 250250 times the smaller piston area, by ignoring the difference in height of fluid between the two pistons, find the minimum force to lift up the stone.

Explanation

Solution

We can answer the above question using Pascal’s law. Hydraulic lift is said to be an application of Pascal's law. Pascal’s law states that the external pressure applied on a confined liquid is distributed evenly throughout the liquid in all directions. Hydraulic lift systems always use incompressible liquids such as oil or water.

Complete Step By Step Answer:
Let the area of the smaller piston can be represented as AA . And the area of the larger piston is 250250 times the area of the smaller piston. Therefore the area of the larger piston is 250A250A .
Since from pascal’s law, we know that the pressure at one point in the system is the same throughout the liquid.
P1=P2{P_1} = {P_2}
We know the pressure formula is given by force divided by area. Therefore,
P=FAP = \dfrac{F}{A}
Therefore substituting this in the above equation.
F1A1=F2A2\dfrac{{{F_1}}}{{{A_1}}} = \dfrac{{{F_2}}}{{{A_2}}} …… (1)
We need to find the force required to lift the mass of 1000kg1000kg the stone. So the weight of this stone can be calculated using the weight formula,
W=mgW = mg
Here, mm is the mass of the stone and gg is the gravitational acceleration.
Therefore substituting the mass of the stone in the weight of the stone and also we can substitute for the gravitational acceleration 10m/s210m/{s^2} . Therefore,
W=1000×10W = 1000 \times 10
W=10000kg\Rightarrow W = 10000kg
We know that, F=mgF = mg
Therefore we can substitute this as Force1Forc{e_1} in the equation (1)
Rearranging equation (1) to get Force2Forc{e_2} we get,
A1×F1A2=F2\dfrac{{{A_1} \times {F_1}}}{{{A_2}}} = {F_2}
A×10000250A=F2\Rightarrow \dfrac{{A \times 10000}}{{250A}} = {F_2}
1000g250=F2\Rightarrow \dfrac{{1000g}}{{250}} = {F_2}
F2=40N{F_2} = 40N
Correct Answer: Therefore the minimum force required to lift the stone of mass 1000kg1000kg is given by 40N40N .

Note:
We know that hydraulic lift is one of the applications of Pascal's law. There are also other applications of Pascal's law. They are hydraulic jack, hydraulic brakes, hydraulic pumps, and aircraft hydraulic systems. This law was first given by a French mathematician, physicist, and philosopher Blaise Pascal in the year 16471647 .