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?
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.