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Hydraulic lift

  • Thread starter Mowgli
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  • #1
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Homework Statement



In a hydraulic lift the radii of the pistons are 1.75cm and 10.5cm. A car weighing 18.5kN is to be lifted by the force of the large piston.

a.) what force must be applied to the small piston?

b.) when the small piston is pushed 1.00cm, how far is the car lifted?

c.) find the mechanical advantage of the lift (this is the ratio of the large force to the small force)


The Attempt at a Solution



a.) F= (r/R)^2 (F)
= (1.75 cm/10.5 cm)^2 (18.5 kN)
= .514 or 514 N

b.) Win(f)= Wout(F)= fh=FH=H= (f/F)h= (pi)r^2/(piR^2=(r/R)^2
= (1m/R)^2 = (100cm/10.5cm)^2
= 90.7cm

c.) MA= F/f= 18.5kN/1.75cm
= 10.57


I have no clue if I am doing this right.. can anyone help?
 
Last edited:

Answers and Replies

  • #2
Doc Al
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a.) F= (r/R)^2 (F)
= (1.25 cm/15.0 cm)^2 (14.5 kN)
= .1007 or .101 N
Good, but careful with units. (Those are kN, not N.)

b.) Win(f)= Wout(F)= fh=FH=H= (f/F)h= (pi)r^2/(piR^2=(r/R)^2
= (1m/R)^2 = (100cm/10.5cm)^2
= 90.7cm
FH = fh is correct. What's the ratio of the forces?

c.) MA= F/f= 18.5kN/1.75cm
= 10.57
What's the ratio of the forces?
 
  • #3
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I'm not sure how to find that?
 
  • #4
Doc Al
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I'm not sure how to find that?
Use what you found in part a.
 
  • #5
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is it 1/10?
 
  • #6
Doc Al
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is it 1/10?
Nope. You were given one force and for part a you found the other force. What's their ratio?
 
  • #7
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14.5/101
 
  • #8
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I just realized my problem had the wrong numbers- so I edited the problem...

For part b.) is my ratio now 18.5/514????
 
  • #9
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For part b.) is my ratio now 18.5/514????
(18.5/514)^2 ??

and then I think c.) is MA=F/f = 18.5/.514
= 35.9

Can anyone help?
 
  • #10
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Trying to figure this out:

b.) (r/R)^2
so, therefore it should be (1.75/10.5)^2
which = .0278

c.) MA= F/f = 18.5/.514 = 35.9

????
 
  • #11
Doc Al
Mentor
44,867
1,114
Trying to figure this out:

b.) (r/R)^2
so, therefore it should be (1.75/10.5)^2
which = .0278

c.) MA= F/f = 18.5/.514 = 35.9

????
That's fine. The ratio of forces is given by (R/r)^2 = (10.5/1.75)^2 = 6^2 = 36.

So for b, the large piston will raise by 1/36 of the distance the small piston is lowered.

And for c, the mechanical advantage is just that ratio of forces.
 

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