Question

A hydraulic press has one piston of diameter 4.0 cm and the other piston of diameter 8.0 cm. What force must be applied to the smaller piston to obtain a force of 1600 N at the larger piston

Answer 1

Answer:

The  force is $F_1 = 400.8 \ N$

Explanation:

From the question we are told that

   The first  diameter is  $d_1 = 4.0 \ cm = 0.04 \ m$

   The second diameter is  $d_2 = 8.0 \ cm = 0.08 \ m$

   

Generally the first area is  

         $A_1 = \pi * \frac{d^2_1 }{4}$

=>      $A_1 = 3.142 * \frac{0.04^2}{4}$

=>       $A_1 = 0.00126 \ m^2$

The  second area is  

     $A_2 = \pi * \frac{d^2_2 }{4}$

     $A_2 = 3.142 * \frac{0.08^2}{4}$

     $A_2 = 0.00503 \ m^2$

For a hydraulic press the pressure at both end must be equal .

Generally  pressure is mathematically represented as

    $P = \frac{F}{A}$

=>  

   $\frac{F_1}{A_1 } = \frac{F_2}{A_2 }$

=>   $F_1 = \frac{1600}{0.00503} * 0.00126$

=>    $F_1 = 400.8 \ N$