Calculating Force Applied to Input Piston in Hydraulic System

[mike2007]

The load bearing piston in a hydraulic system as an area 50 times as the the input piston. If the larger piston supports a load of 6000N, how large a force must be applied to the input piston?

Since the area is 50 times as large i used a ratio 10:50

Using the formula F2/F1 = A2/A1

6000/F1 = 50/10
6000/F1 = 5
F1 = 6000/5
= 1200N
I have no idea if i am anywhere close!

 

[G01]

The ratio should be 50:1, not 50:10. Otherwise, that would be how to find the force required on the input piston to balance out the 6000N on the output. If this was what you were looking for then this seems correct, except for the ratio mistake. I'm assuming you are looking for this equilibrium state. (The load is not supposed to be accelerated correct?)

 

[mike2007]

Thats correct the load is not supposed to be accelerated and it should be in equilibrium state.
thank you

 

[G01]

Thats correct the load is not supposed to be accelerated and it should be in equilibrium state.
thank you

Please notice that I didn't notice the mistake in you ratio the first time around. Check my edited first post again to make sure you see the mistake. Sorry for missing that. I should have been more observant. It's getting late here!

 

[mike2007]

I see where i made the mistake but if the ratio is now 50:1, won't that change the answer to F1 = 6000/50 = 120N
?

 

[Doc Al]

...won't that change the answer to F1 = 6000/50 = 120N
?

That's correct.