Friction and kinematics problem

AI Thread Summary
To determine the minimum time required to move a 50 kg weight over 10 meters with a maximum tension of 20 kgf, the net force must be calculated considering friction and the applied force. The initial force of 10 kgf is sufficient to overcome static friction, but the kinetic friction value is unclear. The discussion highlights the need to clarify whether the weight must come to rest at the end of the distance, as this affects the calculations. It is suggested to use the maximum acceleration for the entire distance to minimize time. The conversation emphasizes the importance of correctly identifying the forces and conditions involved in the problem.
stackptr
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Homework Statement


What minimum time is required to move a weight of mass 50 kg over a distance of 10 meters along a horizontal floor if the rope used to pull it breaks when the tension exceeds 20kgf, while a force of 10kgf is sufficient to start the weight moving or move it uniformly, overcoming the force of friction?

Homework Equations



$$v = \frac{x}{t} = at = \frac{\Sigma F}{m}t$$

Where ##v## is the velocity of the object, ##x## is the horizontal distance covered, ##t## is the time in motion (what we are trying to solve for), ##a## is acceleration of the object and ##\Sigma F## is the net force on the object.

The Attempt at a Solution


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I rearranged the equation for ##t##, giving me

$$t = \sqrt{\frac{mx}{\Sigma F}}$$

However, I am having trouble figuring out the net force. I know the tension is acting to the right with a force of 10kgf, and I know ##f_s\leq10 kgf##. What would the kinetic friction be though? Using this, I can do ##T-f_k## to find the net force on the object.
 
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stackptr said:
Relevant equations

$$v = \frac{x}{t} = at = \frac{\Sigma F}{m}t$$
No. v=at is correct for constant acceleration, but x=vt is only correct for constant velocity, so will not apply here.
 
stackptr said:
the tension is acting to the right with a force of 10kgf
Horizontally?
 
haruspex said:
Horizontally?
Yes
 
haruspex said:
No. v=at is correct for constant acceleration, but x=vt is only correct for constant velocity, so will not apply here.
So what equations would we be able to use here?
 
stackptr said:
Yes
It doesn't say so.
stackptr said:
So what equations would we be able to use here?
Suppose the force is constant for at least part of the time. What does that say about the acceleration? What equations do you know for that condition?

One thing is not clear... does the weight have to finish at rest at the desired position?
 
haruspex said:
does the weight have to finish at rest at the desired position?

I guess so, this is the whole problem I copied word for word from a textbook.

Anyway we could use a = F/m to find acceleration, if it is constant
 
stackptr said:
I guess so, this is the whole problem I copied word for word from a textbook.
It is a lot tougher if it has to finish at rest, so for now I recommend you do not require that. We can address that later if you wish.
stackptr said:
Anyway we could use a = F/m to find acceleration, if it is constant
You are making two assumptions. First, that the force is horizontal; you are not told it is horizontal, and it will turn out that the minimum time is not with a horizontal force.
Second, that the force is constant; since we are not requiring the mass to arrive at rest that is reasonable - you might as well use the maximum acceleration all the way.
 

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