Pulleys moving down with constant speed

In summary: So if the mass is at y=L*cos(Θ), and dL/dt = -u, then the velocity with the mass at y=L*cos(Θ) is u/cosθ. If you replace AC with DC in the second equation it will be right.
  • #1
amal
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0

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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  • #2
Why don't you explain your reasoning for the 'obvious' result, u*cos(θ)?
 
  • #3
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
gneill said:
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
ehild said:
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
amal said:
But then is my interpretation of decreasing vertical velocity wrong?

It is wrong. And you did not prove it.

ehild
 
  • #7
amal said:
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
amal said:
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
amal said:
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
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
amal said:
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
Yes, thank you.
 

1. How do pulleys move down with constant speed?

Pulleys move down with constant speed due to the law of conservation of energy. The force applied on the pulley by the weight of the object is equal to the force required to lift the object, resulting in a constant speed.

2. What factors affect the speed of pulleys moving down?

The speed of pulleys moving down can be affected by the weight of the object, the friction between the pulley and the rope, and the angle of the rope. In addition, the number of pulleys and their arrangement can also impact the speed.

3. Can the speed of pulleys moving down change?

Yes, the speed of pulleys moving down can change depending on the forces acting on the pulley system. If the weight of the object or the angle of the rope changes, the speed of the pulleys moving down will also change.

4. How does the number of pulleys affect the speed of pulleys moving down?

The more pulleys there are in a system, the more the weight of the object is distributed between them. This results in a decrease in the force required to lift the object, thus increasing the speed of the pulleys moving down.

5. Are there any real-life applications of pulleys moving down with constant speed?

Yes, pulleys moving down with constant speed are commonly used in elevators, cranes, and other lifting systems. They are also used in exercise machines, such as weightlifting machines, to provide a constant resistance throughout the movement.

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