• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Master and Slave Cylinder at different heights

  • Thread starter nmsurobert
  • Start date
246
26
Homework Statement
Suppose the master cylinder in a hydraulic system is at a greater height than the slave cylinder. Explain how this will affect the force produced at the slave cylinder.
Homework Equations
P = F/A
F[SUB]1[/SUB]/A[SUB]1[/SUB] = F[SUB]2[/SUB]/A[SUB]2[/SUB]
P = ρgh
I think I've made sense of this.

m: master, s:slave

Fm/Am = Fs/As

Fs = As(Fm/Am)

Fs = (As)ρghm

So with that said, if the height of the master cylinder increases then the F produced by the slave will also increase. Anyone see any problems here?

My concern is that P = ρgh isn't applicable here. Thanks.
 
660
249
Problem Statement: Suppose the master cylinder in a hydraulic system is at a greater height than the slave cylinder. Explain how this will affect the force produced at the slave cylinder.
Relevant Equations: P = F/A
F1/A1 = F2/A2
P = ρgh

I think I've made sense of this.

m: master, s:slave

Fm/Am = Fs/As

Fs = As(Fm/Am)

Fs = (As)ρghm

So with that said, if the height of the master cylinder increases then the F produced by the slave will also increase. Anyone see any problems here?

My concern is that P = ρgh isn't applicable here. Thanks.
Your concern is correct and the equations need to reflect this. If Fm is zero there is nonzero Fs from the gravitational head. If Fm is nonzero then it adds to it.
 
246
26
Your concern is correct and the equations need to reflect this. If Fm is zero there is nonzero Fs from the gravitational head. If Fm is nonzero then it adds to it.
I think I understand what you’re saying. Can I compensate for that by just doing this?
Fs = (As)ρghm + Fm

Or something like this. Or add mg?
 
Last edited:
660
249
I think I understand what you’re saying. Can I compensate for that by just doing this?
Fs = (As)ρghm + Fm

Or something like this. Or add mg?
Close but Its the pressure that gets added to.
In particular in the limit of ρ=zero you should recover the previous result.
 
246
26
Close but Its the pressure that gets added to.
In particular in the limit of ρ=zero you should recover the previous result.
Fs = (As)Fm/Am + ρghm

This way the force applied to the master cylinder can be zero but the force applied to the slave cannot be zero because of the pressure created by the difference in height.
 
660
249
Fs = (As)Fm/Am + ρghm

This way the force applied to the master cylinder can be zero but the force applied to the slave cannot be zero because of the pressure created by the difference in height.
Your idea is correct but ρghm is a pressure. You now have it equated to a force.
 
246
26
Your idea is correct but ρghm is a pressure. You now have it equated to a force.
Ahh that makes sense. The only thing that makes sense to me is to use weight. If the fluid is on top of the slave cylinder then it’s weight applies a constant force to the slave cylinder always keeping that force nonzero.
 

Want to reply to this thread?

"Master and Slave Cylinder at different heights" You must log in or register to reply here.

Related Threads for: Master and Slave Cylinder at different heights

Replies
3
Views
264
Replies
13
Views
3K
  • Posted
Replies
1
Views
3K
  • Posted
Replies
2
Views
4K
Replies
2
Views
1K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top