Student Neighbour Problem

[kuiperbelt]

A 1000 kg car is placed on the large piston of an hydraulic system whose radius is 5 metres.
a.)What is the minimum force exerted by the large piston to lift the car?
b.)If the mechanical advantage of the hydraulic system is 10, what is the minimum downward force required to lift the car?
c.)What is the radius of the master cylinder piston?
d.)What is the pressure applied to the slave piston?
e.)How far should master piston travel to lift the car by 10cm?
Show your working and take acceleration due to gravity as 10m/s^2

[kuiperbelt]

See diagram fig 1

https://ibb.co/b6zDzBn

[ZebuSanctuary]

a.) Force (weight) of car acting downwards = mg
Which is 1000kg x 10m/s² = 10kN

The large piston has to exert a force on the car a tadge over 10kN

b.) Mechanical Advantage MA = Fout/Fin

Therefore 10 = 10kN/Fin

Minimum downward force required to lift car is a smidgen over 1kN

c.) From Pascal's Law;

Pressure in (master piston acting on fluid) = Presure out (Hydraulic fluid acting on slave piston)

Utilising the formula P = F/A

We obtain (Master) 1kN/Ain = (Slave) 10kN/Aout

Fout/Fin = Aout/Ain = MA = 10 thus;

1000/Ain = 10,000/Aout

Aout/Ain = 10,000/1000 = 10

The area of the master cylinder Ain is ten times smaller than the slave cylinder Aout

Slave piston radius = 5 metres

Then Area Slave Piston = πr² = π5² = 25π m²

As master piston Area is ten times less Ain = 2.5π m²

Ain = 2.5π = πr²

radius of master piston = √(2.5 = 1.58 metres

d.)Pressure exerted on the slave piston by the hydraulic oil ;

P = F/A ---> 10,000/25π = 127.3 N/m² or Pascals

e.) Work in = Work out i.e., Fin x Distance = Fout x Distance

1000 x Din = 10,000 x 10cm

Din = 100,000/1000 = 100cm

The master piston must travel 1 metre to lift the car 10 cm.

Note the ratio 1m/10cm = MA = 10