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Asked by Basanta · 2 years ago

By using a hydraulic lift a load of 2500 N needs to be lifted. If cross sectional area of the small piston is $\rm \frac {1}{10}$ of the big piston, calculate the force required to apply on small piston.

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Arpit Sundas Arpit Sundas · 1 year ago
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Solution

Here, Given:

  • Load to be lifted(F)= 2500N
  • Let the cross sectional area of big piston(A) be ‘x’ then, according to question cross sectional area of small piston(a) will be  '$\frac{x}{10}$'
  • Force required(applied) on small piston(f) =?

We have,

$\frac{A}{a}$ = $\frac{F}{f}$

or, $\frac{x}{\frac{x}{10}}$ = $\frac{2500}{f}$

or, $\frac{10}{1}$ = $\frac{2500}{f}$

or, f= $\frac{2500}{10}$

or, f= 250 N

Hence, 250 N of force should be applied on small piston to lift the load of 2500 N.
 

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