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Pulleys moving down with constant speed

  • Thread starter amal
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  • #1
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Homework Statement


Please refer diagram attached.
The pulleys are moving down with constant speed u each. Pulleys are light , string ideal. What is the speed with the mass moves up?






Homework Equations





The Attempt at a Solution


Now, I got the answer as ucosΘ, obviously. But the answer is u/cosΘ.
the solution too is given. But I feel that the latter is wrong as the velocity goes to to infinity as the mass goes upwards. Please tell me which is correct.
 

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Answers and Replies

  • #2
gneill
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Why don't you explain your reasoning for the 'obvious' result, u*cos(θ)?
 
  • #3
ehild
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If the half-length of the rope between the pulleys is L, the mass is at depth y=L*cos(Θ), and dL/dt = -u. The rate of change of y is dy/dt=d(L*cos(Θ))/dt. Both L and Θ are changing as the rope is pulled while x = LsinΘ stays constant. If you take these into account you will get the correct result, that the mass raises wit u/cosΘ.

ehild
 
  • #4
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Why don't you explain your reasoning for the 'obvious' result, u*cos(θ)?
Well, the string section 'r' should move with speed u only as it is a light pulley. Then the vertical component is ucosθ.
 
  • #5
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If the half-length of the rope between the pulleys is L, the mass is at depth y=L*cos(Θ), and dL/dt = -u. The rate of change of y is dy/dt=d(L*cos(Θ))/dt. Both L and Θ are changing as the rope is pulled while x = LsinΘ stays constant. If you take these into account you will get the correct result, that the mass raises wit u/cosΘ.
But then is my interpretation of decreasing vertical velocity wrong?
 
  • #6
ehild
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But then is my interpretation of decreasing vertical velocity wrong?
It is wrong. And you did not prove it.

ehild
 
  • #7
gneill
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Well, the string section 'r' should move with speed u only as it is a light pulley. Then the vertical component is ucosθ.
The problem with this analysis (as tempting as it is) is that the rope end at D will NOT move along the line DA (or DB), as it is constrained by its mirror-image partner. While your length r may be getting shorter, point D has no velocity along r. So trying to treat dr/dt = u as a velocity component for point D is not right.
 
  • #8
gneill
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But then is my interpretation of decreasing vertical velocity wrong?
The vertical velocity will decrease, but not for the reason you think! As the mass rises and the angle θ approaches 90°, the tension in the rope required to maintain the rope speed u will head to infinity. IF the rope speed could be maintained at u (or any positive real value greater than zero) then the velocity of the mass would go infinite, but rope speed cannot be so maintained by any real apparatus.
 
  • #9
Doc Al
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Well, the string section 'r' should move with speed u only as it is a light pulley. Then the vertical component is ucosθ.
The rate at which the length of segment DA changes will equal u. You need to relate the rate of change of segment DC to the rate of change of DA. How are those lengths related?
 
  • #10
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OK, I see now. Using Pythagoras theorem we can relate the legths as
[itex]DA^{2}[/itex]=[itex]DC^{2}[/itex]+[itex]AC^{2}[/itex]
then differentiating,
[itex]2DA\frac{dDA}{dt}[/itex]=[itex]0+2AC\frac{dDC}{dt}[/itex]
And finally the answer will come u/cosθ. Right?
 
Last edited:
  • #11
ehild
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OK, I see now. Using Pythagoras theorem we can relate the legths as
[itex]DA^{2}[/itex]=[itex]DC^{2}[/itex]+[itex]AC^{2}[/itex]
then differentiating,
[itex]2DA\frac{dDA}{dt}[/itex]=[itex]0+2AC\frac{dDC}{dt}[/itex]
And finally the answer will come u/cosθ. Right?
If you replace AC with DC in the second equation it will be right.

ehild
 
  • #12
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Yes, thank you.
 

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