Question

hydraulic press with the larger piston of diameter 35cm at a height of 1.5m relative to the smaller piston of diameter 10cm the mass on the smaller piston is 20kg. What is the force exerted on the load by the larger piston. the density of oil in the press is 750kg/m^3​

Answer 1

Explanation:

Given Hydraulic press with the larger piston of diameter 35 cm at a height of 1.5 m relative to the smaller piston of diameter 10 cm the mass on the smaller piston is 20 kg. What is the force exerted on the load by the larger piston. the density of oil in the press is 750 kg/m^3

Given h = 1.5 m, d1 = 35 m , d2 = 10 cm

So mass of piston = 20 kg

Force is exerted and so pressure will be Pa = Pb

So Pa + ρgh = Pb

  Pa = Pb – ρgh

Now F1 / A1 = 20 x 10 x 4 / π x 10^2 x 10^-4 – 750 x 10 x 1.5

                   = (20 x 10 x 4 / π x 10^-2 – 750 x 10 x 1.5) π x (35 x 10^-2)^2 / 4

So F1 = 1.34 x 10^3 N

Answer 2

1340 N

Explanation:

Given that we have a hydraulic press with larger piston of 35 cm at a height of 1.5m relative to the smaller piston of diameter 10cm.

It is also given that there is a mass of 20 kg lying on the small piston.

This can easily solved by equating the pressures on both the sides.

Let us first calculate the pressure on the right hand side.

Initial pressure at the top is ,(assuming the force acting on the piston to be F)

$p=\frac{F}{A}$

At 1.5m below we get the pressure as ,

$P=\frac{F}{A}+750*10*1.5$

A in the abow equation is area of the 35 cm cylinder.

Coming to left side with smaller piston,

$P=\frac{20*10}{A1}$

where A1 is area of the smaller cylinder.

$A=\frac{\pi *35^2}{10000*4}\\\\A1=\frac{\pi *10^2}{10000*4}$

Substituting all and equating the pressures on both sides we get F as 1340 N.