In a hydraulic system, small piston area A1 = 0.02 m^2 experiences F1 = 800 N. What is the output force F2 on a large piston with area A2 = 0.10 m^2?

In a hydraulic system, small piston area A1 = 0.02 m^2 experiences F1 = 800 N. What is the output force F2 on a large piston with area A2 = 0.10 m^2?

• A. 2000 N
• B. 1000 N
• C. 400 N
• D. 4000 N

Answer: (D)

Pascal's principle says that pressure in an enclosed fluid is transmitted equally in all directions, so the same pressure that acts on the small piston also acts on the large piston. The pressure created by the input is p = F1/A1. That same pressure pushes on the large piston, giving F2 = p × A2. Compute the numbers: p = 800 N / 0.02 m² = 40,000 Pa. Then F2 = 40,000 Pa × 0.10 m² = 4,000 N. So the output force is five times the input force, since F2/F1 = A2/A1 = 0.10/0.02 = 5. In an ideal, lossless system, energy is conserved, so the larger piston moves less to balance the work done on the smaller piston.

Pascal's principle says that pressure in an enclosed fluid is transmitted equally in all directions, so the same pressure that acts on the small piston also acts on the large piston. The pressure created by the input is p = F1/A1. That same pressure pushes on the large piston, giving F2 = p × A2.

Compute the numbers: p = 800 N / 0.02 m² = 40,000 Pa. Then F2 = 40,000 Pa × 0.10 m² = 4,000 N.

So the output force is five times the input force, since F2/F1 = A2/A1 = 0.10/0.02 = 5. In an ideal, lossless system, energy is conserved, so the larger piston moves less to balance the work done on the smaller piston.