Question

The press plungers of a braman (hydraulic) press is $40c{m}^2$ in cross-section and is used to lift a load of mass $800kg$. What minimum force will be required to be applied on the pump plunger if its cross-sectional area is $0.02m^2$?

Answer 1

Pressure at Plunger is $P=\frac{weight}{area}$

$Weight=800\times 9.8=7840N$

Area $A=40c{m}^2=40\times {10}^{-4}{m}^2$

$P=\frac{7840}{40\times {10}^{-4}}=196\times {10}^{4}N{m}^{-2}$

Now at pump plunger required pressure is $P=196\times {10}^{4}N{m}^{-2}$

Let the force require be F.

Area, $a=0.02m^2$

$P=\frac{F}{a}$

$F=p\times a=196\times {10}^4\times 0.02=39200N$