Question

a certain hydraulic jack is used for lifting load. a force of 250N is applied on a small piston with a diameter of 80cm while the diameter of the larger piston is 2.0m.the applied force moves through a distance of 25cm.calculate the maximum mass of a load that can be lifted by the jack and the distance through which the load is lifted.(take g=9.81ms^-2)​

Answer 1

The maximum mass of a load that can be lifted by the jack and the distance covered are:

m = 160.2 Kg

h = 25 cm

Given that a certain hydraulic jack is used for lifting load. a force of 250N is applied on a small piston with a diameter of 80cm while the diameter of the larger piston is 2.0m.

The parameters given are

$F_{1}$ = 250

$A_{1}$ = Area of the small piston = π$r^{2}$

$A_{1}$ = 22/7 x $0.4^{2}$

$A_{1}$ = 0.5 $m^{2}$

$F_{2}$ = ?

$A_{2}$ = Area of the large piston = π$r^{2}$

$A_{2}$ = π x 1

$A_{2}$ = 3.14 $m^{2}$

To calculate the force on the large piston, we will use the below formula

$F_{1}$/ $A_{1}$ = $F_{2}$ / $A_{2}$

Substitute all the parameters into the equation

250/0.5 =  $F_{2}$/3.14

$F_{2}$ = 1570 N

To calculate the maximum mass of a load that can be lifted by the jack, let us apply Newton second law

F = mg

1570 = 9.8m

m = 1570/9.8

m = 160.2 Kg

.(take g=9.81ms^-2)​

If the applied force moves through a distance of 25cm, the distance through which the load is lifted will be

$F_{1}$/ 0.25$A_{1}$ = $F_{2}$ / $A_{2}$h

250/0.125 = 1570/3.14h

make h the subject of the formula

6280h = 1570

h = 1570/6280

h = 0.25 m

Therefore, the distance through which the load is lifted is 25 cm

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