Question

12) The small piston of a hydraulic lift has a diameter of 8.0 cm, and its large piston has a diameter of 40 cm. The lift raises a load of 15,000 N. (a) Determine the force that must be applied to the small piston. (b) Determine the pressure applied to the fluid in the lift.

Answer 1

600 N

119366.20731 Pa

Step-by-step explanation:

`F_1` = Force on the smaller piston

`A_1` = Area of the smaller piston =
`\pi 0.04^2`

`F_2` = Force on the larger piston = 15000 N

`A_2` = Area of the larger piston =
`\pi 0.2^2`

From Pascal's law we have

`(F_1)/(A_1)=(F_2)/(A_2)\\\Rightarrow F_1=(F_2A_1)/(A_2)\\\Rightarrow F_1=(15000\pi 0.04^2)/(\pi 0.2^2)\\\Rightarrow F_1=600\ N`

The force that must be applied to the small piston is 600 N

Pressure would be

`P=(F)/(A)\\\Rightarrow P=(600)/(\pi * 0.04^2)\\\Rightarrow P=119366.20731\ Pa`

The pressure applied to the fluid in the lift is 119366.20731 Pa