We can use the formula:
force=pressure × area
The pressure is the same throughout the hydraulic system, so we can set the pressures on the small and large pistons equal to each other:
pressure on small piston = pressure on large piston
We can rearrange the formula to solve for the force on the large piston:
force on large piston = pressure on large piston × area of large piston
We can substitute the area of the large piston in terms of its diameter:
area of large piston = π(diameter/2)² = π(200/2)² cm² = 31416 cm²
We can also substitute the area of the small piston:
area of small piston = 5 cm²
We can solve for the pressure on both pistons by dividing the force by the area:
pressure on small piston = 400N ÷ 5 cm² = 80 N/cm²
pressure on large piston = 80 N/cm²
Finally, we can substitute the pressure and area of the large piston into the formula to solve for the force:
force on large piston = 80 N/cm² × 31416 cm² = 2,513,280 N
Therefore, the resulting force exerted on the large piston is 2,513,280 N.
the diameter of the large piston of hydraulic press us 200 cm and the cross sectional area of the small piston is 5cm² if the force of 400N is applied to the small piston what is the resulting force exerted on the large piston?
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