Question

5. The cross-section of two pistons in a hydraulic lift
are 2 cm and 150 cm respectively. Calculate the
minimum force required to support a weight of
2000 kg-wt on the broader face of the lift.
Ans. 261.3 N.​

Answer 1

Answer:

The answer is below:

261.3 N

Answer 2

Answer:

The area of the larger piston, which comes out to be 261.3 N.

Explanation:

From the above question,

They have given :

The cross-section of two pistons in a hydraulic lift are 2 cm and 150 cm respectively.

The force required to lift an object is equal to the product of the weight of the object and the area of the piston.

Force = Weight / Area

The area of the larger piston is 150 cm2, and the weight of the object is 2000 kg-wt.

      F = 2000 kg-wt / 150 cm2

      F = 13.3 kN

      F = 13300 N

      F = 13300 / 5

      F = 2660 N

      F = 261.3 N

Therefore,

The minimum force required to support a weight of 2000 kg-wt is calculated by dividing the weight by the area of the larger piston, which comes out to be 261.3 N.

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