Physics Problem?

Question

A hydraulic system uses hydraulic fluid with a density of 1.6 grams/cm3 . An extra force of 220 Newtons is exerted on the small cylinder (diameter of 2 cm). Calculate the force available on the large cylinder of the system (diameter of 30 cm).

Answer 1

Force on the small Cylinder = 220 N.

Diameter of the small cylinder = 2 cm.

Radius of the small cylinder = 1 cm.

Diameter of the larger cylinder = 30 cm.

Radius of the larger cylinder = 30 / 2 cm = 15cm.

Area of the small cylinder = πr² = π * (1)² = 3.14 cm² = 3.14 * 10^-4 m²

Area of the larger cylinder = πr² = π * (15)² = 706.8 cm² = 706.8 * 10^-4 m²

We know that In Hydraulic Jack system, Pressure on the Smaller Cylinder = Pressure on the Larger Cylinder.

Force on the Smaller Cylinder/ Area of the Smaller Cylinder = Force on the Larger Cylinder/ Area of the Larger Cylinder

220 /( 3.14 *10^-4)= F / (706.8 *10^-4)

F = 49500 N.

Therefore the Force on the larger Cylinder is 49500 N.