Figure E-6

-In the hydraulic device shown, when the small piston is pushed down 10 cm with a force of 10 N, the large piston is raised 1 cm. This means that the large piston is capable of lifting
A)10 N,
B)100 N,
C)1000 N,
D)any amount of weight, Which illustrates
E)that impulse is equal to a change in momentum.
F)(F × d)input = (F × d)output.
G)Newton's second law.
H)that the stored energy that a body possesses may be transformed into other forms of energy.

Question

Figure E-6
 
-In the hydraulic device shown, when the small piston is pushed down 10 cm with a force of 10 N, the large piston is raised 1 cm. This means that the large piston is capable of lifting
A)10 N, 
B)100 N, 
C)1000 N, 
D)any amount of weight, Which illustrates 
E)that impulse is equal to a change in momentum. 
F)(F × d)input = (F × d)output. 
G)Newton's second law. 
H)that the stored energy that a body possesses may be transformed into other forms of energy.

Quizplus · Accepted Answer

The principle illustrated is (F × d)input = (F × d)output, which is a statement of the conservation of energy in a hydraulic system. Given that the small piston is pushed down with a force of 10 N over a distance of 10 cm, and the large piston moves up 1 cm, the force exerted by the large piston can be calculated using the principle of hydraulic advantage. The work done on both pistons is equal, so (10 N × 10 cm) = (F × 1 cm), solving for F gives 100 N. This demonstrates how a hydraulic system can multiply force.

Figure E-6 -In the Hydraulic Device Shown, When the Small Piston Is