What is the force equation for power from a trolley motor?

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Homework Help Overview

The discussion revolves around the force equation related to power output from a trolley motor, specifically examining the relationship between force, velocity, and power in a mechanical context.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the implications of the equation Fv = av(b-v^2) and question the validity of assuming F = ab when v = 0. They discuss the continuity of functions and the limits involved in this context. Some express confusion about the nature of force when the object is at rest and how power relates to force and motion.

Discussion Status

The conversation is active, with participants raising questions about the assumptions made in the original post and exploring different interpretations of the equations involved. There is a focus on the mathematical and physical implications of force at zero velocity, with some participants suggesting that deeper concepts may need to be considered.

Contextual Notes

Participants note the constraints of the problem, including the assumption of continuous functions and the implications of power being defined in terms of average over time intervals. There is also mention of the physical scenario where the trolley is not moving, prompting questions about the nature of force in that state.

walking
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Homework Statement
For part (c) author says F=ab when v=0. I don't see how since Fv=av(b-v^2) doesn't necessarily mean F=ab at v=0. In fact F can take any value?
Relevant Equations
Fv=av(b-v^2)
e4vf9.png


Author solution:

9yum3.png
 
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walking said:
Homework Statement:: For part (c) author says F=ab when v=0. I don't see how since Fv=av(b-v^2) doesn't necessarily mean F=ab at v=0. In fact F can take any value?
Relevant Equations:: Fv=av(b-v^2)

View attachment 279836

Author solution:

View attachment 279837
From a strictly mathematical perspective you are right, but it is reasonable to assume that F is a continuous function of v, so you can take the limit as v tends to zero.
 
haruspex said:
From a strictly mathematical perspective you are right, but it is reasonable to assume that F is a continuous function of v, so you can take the limit as v tends to zero.
Does this only work for continuous functions? If so why?
 
I do not understand how an author got that answer. That result, if possible at all, is possible only in one way when you make such equation: $$Fv = av(b-v^2)$$ Then you divide by v and get this formula $$F= \frac {av(b-v^2)} {v}.$$ You can, obviously simplify it and you get your $$F = a(b-v^2)= ab, v=0$$ But if the object is not moving the power output is zero (also written in a task). But I'm wondering where the force equation comes from. And if this force exists what is this force at all, what it exerted to, and what it is equilibrated by? (the case when object does not move)

IMHO, this questions is smth like: you sit on a chair, your velocity equals zero, does it mean you cannot stand up? If there is a deeper idea in it, tell me please.
 
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walking said:
Does this only work for continuous functions? If so why?
add314 said:
You can, obviously simplify it and you get
We are told P(v), not F(v). If we do not assume F is continuous then it could be
0 when v=0
a(b-v2) otherwise.
add314 said:
If there is a deeper idea in it, tell me please.
It's akin to Xeno's paradox.
How can you start an object moving from rest if the power is given by Fv and v is zero?
We can maybe explain it in terms of limits: over any nonzero time interval dt there is work done and a nonzero average power. If we say that power is not a fundamental concept and is only defined in terms of an average over a time interval then the problem goes away.
Or perhaps, in the real world, we should be considering the Planck length, etc.?
 
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add314 said:
... But I'm wondering where the force equation comes from. And if this force exists what is this force at all, what it exerted to, and what it is equilibrated by? (the case when object does not move)
It seems to me that the force of reference is the result of the initial torque from the electric motor, applied to the wheel(s), just before the trolley starts moving forward.
That would be a tangential force at the contact point of the wheel with the rolling surface, transferring momentum to the accelerating mass of the whole trolley.
Just like the instantaneous situation of a pendulum at maximum swing position, there is a force and an acceleration, but its velocity is zero.

3Pbfz.gif
 

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