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The small and large pistons of a hydraulic press
have areas of 2 m2 and 4 m2. If the load on the large piston in 3200 N, what is the input force (effort) that must be applied on the small piston

Respuesta :

asuwafohjames

The Force the must be applied on the small piston is 1600 N.

To solve the problem above, we apply Pascal's principle.

Pascal's Principle: The pressure applied to an enclosed fluid in a container is transmitted undiminished to every part of the container.

Mathematically this principle is expressed as

⇒ Formula:

  • F/A = f/a..................... Equation 1

⇒Where

  • F = Force on the Large piston
  • f = force on the small piston
  • A = Area of the large piston
  • a = are of the small piston.

⇒Make f the subject of the equation

  • f = F×a/A............... Equation 2

From the question,

⇒Given:

  • F = 3200 N
  • A = 4 m²
  • a = 2 m²

⇒Substitute these values into equation 2

  • f = 3200×2/4
  • f = 1600 N

Hence, The Force the must be applied on the small piston is 1600 N

Learn more about pascal's principle here:  https://brainly.com/question/4262025

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