KnowledgeBoat Logo

Physics

The areas of pistons in a hydraulic machine are 5 cm 2 and 625 cm 2. What force on the smaller piston will support a load of 1250 N on the larger piston ? State any assumption which you make in your calculation.

Fluids Pressure

45 Likes

Answer

As we know,

Pressure (P) = Thrust (F)Area (A)\dfrac{\text {Thrust (F)}}{\text {Area (A)}}

and by the principle of hydraulic machine

Pressure on narrow piston = pressure on broader piston

Hence,
F1A1=F2A2\dfrac{F{1}}{A{1}} = \dfrac{F{2}}{A{2}} \\[0.5em]

Given,

Area of narrow piston (A1) = 5 cm 2

Area of wider piston (A2) = 625 cm 2

force (F2) = 1250 N

Substituting the values in the formula above we get,

F15=1250625F1=1250×5625⇒F1=10N\dfrac{F{1}}{5} = \dfrac{1250}{625} \\[0.5em] F{1} = \dfrac{1250 \times 5 }{625} \\[0.5em] \Rightarrow F_{1} = 10 N \\[0.5em]

Hence, force acting on the smaller piston = 10 N

Assumption — There is no friction and no leakage of liquid.

Answered By

30 Likes

Related Questions