A 500-N weight sits on the small piston of a hydraulic machine. The small piston has an area of $2.0 \mathrm {~cm} ^ { 2 }$ If the large piston has an area of $40 \mathrm {~cm} ^ { 2 }$ how much weight can the large piston support? Assume the pistons each have negligible weight.
A) 10000 N
B) 40000 N
C) 500 N
D) 25 N

Question

A  500-N weight sits on the small piston of a hydraulic machine. The small piston has an area of $2.0 \mathrm {~cm} ^ { 2 }$ If the large piston has an area of $40 \mathrm {~cm} ^ { 2 }$ how much weight can the large piston support? Assume the pistons each have negligible weight.
A)  10000 N
B)  40000 N 
C)  500 N
D)  25 N

Quizplus · Accepted Answer

The principle of hydraulic machines states that the pressure applied at one point is transmitted equally throughout the liquid. Pressure is calculated as force (weight, in this case) divided by area. Thus, the pressure applied by the small piston is $500 \, \text{N} / 2.0 \, \text{cm}^2 = 250 \, \text{N/cm}^2$. The large piston, with an area of $40 \, \text{cm}^2$, will support a weight that generates the same pressure, so the weight it can support is $250 \, \text{N/cm}^2 \times 40 \, \text{cm}^2 = 10000 \, \text{N}$.

A 500-N Weight Sits on the Small Piston of a Hydraulic