Question

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?

Answer 1

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.