Question

A car of mass 1.5 x 103 kg is hoisted on the large cylinder of a hydraulic press. The area of the large piston is 0.20 m2, and the area of the small piston is 0.015 m2.

a.Calculate the magnitude of the force of the small piston needed to raise the car with slow speed on the large piston.
b.Calculate the pressure, in pascals and kilopascals, in this hydraulic press.

Answer 1

@Michele_Laino

Answer 2

Hey there how are you?

Answer 3

I hope all is well with you;)

Answer 4

yes! I'm fine! thanks! :)

Answer 5

here you have to keep in mind the principle of Pascal, namely the pressure of the fluid is the same on both pistons

Answer 6

Where water is used as multiplier of force in call cases.

Answer 7

Since for question a I know the area of small piston

Answer 8

I guess all I need is divide the mass to be lifted by the area of the piston?

Answer 9

more precisely are used special oils, instead of water.
the pressure exerted by the fluid on larger piston is:

\[\Large p = \frac{{Mg}}{S} = \frac{{1500 \cdot 9.81}}{{0.2}} = ...Pascals\]

Answer 10

Right... I needed to calculate the normal force acting on the object divided by the area of the piston.

Answer 11

that's right!

Answer 12

According to my calculator the pascals required on the piston is 73500 pascals

Answer 13

So that's considering if we only use the large piston right?

Answer 14

I got p = 73.6 kPa approximately

Answer 15

On the other hand if I use the smaller piston the answer would be slightly greater I think

Answer 16

I got p=981000 on my side for the smaller piston.

Answer 17

no, the pressure is the same through all the fluid.
The force which is acting on the smaller piston, is perpendicular with respect to that piston, and its magnitude is:
\[\Large F = p \times s = 73575 \cdot 0.015 = ...Newtons\]

Answer 18

Oh ok.. but why are we only using the pressure applied to the large piston? Is it because of the structure of hydraulic press?

Answer 19

no, we are using the pression of the fluid which is transmitted by the fluid itself, namely your hydraulic press can be represented as below:

Answer 20

what changes are the magnitude of the forces applied to both pistons

Answer 21

magnitudes*

Answer 22

So basically a force is applied to one end and that results in other end

Answer 23

yes! we apply a little force and the hydraulic press multiply this little force

Answer 24

multiplies*

Answer 25

73575 Pa on the large piston.

Answer 26

I understand until that point

Answer 27

But then, isn't it like by 0.2/0.015 that the pressure is multiplied?

Answer 28

that pressure is acting inside all the fluid, so the pressure of 73575 is acting on both pistons

Answer 29

the pressure is the same!

Answer 30

Are you saying that fluid supplied from the inside the hydraulic press?

Answer 31

in your exercise, we apply a force whose magnitude is 1,104 Newtons, and we can rise a car whose weight is: 14,715 Newtons

Answer 32

yes! the fluid is working inside the hydraulic press

Answer 33

The force the small piston needs to lift the car right? so I need to apply force on the small piston to lift the car to the larger piston and force multiplies

Answer 34

correct! we can see a hydraulic press as a multiplier of force

Answer 35

So it's correct to say 73575 Pa* 0.015= pressure require to lift the car? The only thing I don't understand is why we multiply by 0.015 for a pressure calculated using 0.2. Isn't it like we have to multiply by 0.015/0.2 ?

Answer 36

pressure is 73575 Pa, and 73575*0.015 is a force

Answer 37

likewise 73575 Pa*0.2 is the force I need to apply to larger piston to do the same trick?

Answer 38

better is:
73575*0.2 is the force which is acting on the larger piston

Answer 39

wait a moment it's taking long so

Answer 40

that's my thanks for being patient with me XD

Answer 41

thanks! :)

Answer 42

I am trying my best to understand

Answer 43

ok! :)

Answer 44

You previously calculated the pressure applied to the large piston using the formula
Mg/s=pascals. What if you used 0.015 instead of the 0.2? The overall pressure in the hydraulic press would then increase no?

Answer 45

According to your reasoning that pressure is the same across all points in a hydraulic press, then the pressure calculated using 0.015 instead of 0.2 would be greater...

Answer 46

I have used 0.2 since 0.2 is the area of the piston on which the weight of the car is acting on

Answer 47

I can not use 0.015, since it will be an error