A mass sliding down the inclined plane -- Will it stop?

AI Thread Summary
The discussion centers on the dynamics of a mass sliding down an inclined plane and the calculations involving friction and normal force. Participants debate the relationship between the normal force and the weight of the mass, with some asserting that the normal force equals the weight at all angles, while others argue this is incorrect. The correct expression for the normal force is identified as F_N = mg cos φ, leading to the conclusion that the coefficient of friction μ can be expressed as μ = 2 tan φ. The conversation also touches on the work done by friction and gravity, confirming they are equal in magnitude but opposite in sign. Ultimately, the thread clarifies that the problem involves determining how the coefficient of friction varies with the angle of the ramp.
rudransh verma
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Homework Statement
The upper half of an inclined plane with inclination ##\phi## is perfectly smooth,while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom, if the coefficient of friction for the lower half is
Relevant Equations
##Fnet=ma##
##v^2=u^2+2as##
For lower half ,$$Fnet=-\mu F_N+mg\sin \phi$$
For upper half,
$$v^2=u^2+2as$$ (s is half of the total slant distance)
$$v^2=0+2\frac{mg\sin \phi}ms$$
$$v=\sqrt{2g\sin \phi s}$$
again for lower half,
$$v^2=u^2+2as$$
$$0=2g\sin \phi s+2\frac{-\mu F_N+mg\sin \phi}ms$$
$$\mu=\frac{2gm\sin \phi}{F_N}$$
$$\mu=2\sin \phi$$
 
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rudransh verma said:
$$0=2g\sin \phi s+2\frac{-\mu F_N+mg\sin \phi}ms$$
$$\mu=\frac{2gm\sin \phi}{F_N}$$
$$\mu=2\sin \phi$$
You cannot get rid of ##F_N## in the denominator by setting it equal to the weight. What is ##F_N##?
 
Last edited:
kuruman said:
You cannot get rid of ##F_N## in the denominator by setting it equal to the weight. What is ##F_N##?
Normal force which will be equal to weight mg. I don't think i have done anything wrong here.
 
Last edited:
rudransh verma said:
Normal force which will be equal to weight mg. I don't think i have done anything wrong here.
Why will it be equal to weight? Have you checked the limiting cases such as a vertical wall?
 
rudransh verma said:
Normal force which will be equal to weight mg. I don't think i have done anything wrong here.
If it makes sense to you that the normal force is equal to the weight no matter what the angle of the incline, leave it at that but it is wrong. I will not suggest a method for finding the correct normal force because you have ignored such advice by numerous people on numerous occasions in other threads. I am out of this particular thread.
 
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rudransh verma said:
Normal force which will be equal to weight mg. I don't think i have done anything wrong here.
Only on level ground.
 
rudransh verma said:
Normal force which will be equal to weight mg.
What do you think the word "normal" means here? Have you tried solving for the normal force?
 
rudransh verma said:
Normal force which will be equal to weight mg. I don't think i have done anything wrong here.
You have an impressive rate of posting to this forum. Unfortunately it does not seem to be leaving you time to think in between.
 
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haruspex said:
You have an impressive rate of posting to this forum. Unfortunately it does not seem to be leaving you time to think in between.
##F_N=mg\cos \phi##. By mistake I did that. ##\mu= 2\tan \phi## is coming.
 
  • #10
Would the work done by friction be equal to the work done by gravity?
 
  • #11
valenumr said:
Would the work done by friction be equal to the work done by gravity?
Equal in magnitude, opposite sign.
 
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  • #12
haruspex said:
Yes.
Thanks. I would assume that would be the way to solve it, but not enough information given.
 
  • #13
valenumr said:
Thanks. I would assume that would be the way to solve it, but not enough information given.
Really? Looks enough to me.
 
  • #14
haruspex said:
Really? Looks enough to me.
Oh... I misunderstood the question. It is trying to solve for the coefficient of friction vs the angle of the ramp, given that the coefficient changes from zero to some value after half the length of the ramp?
 
  • #15
valenumr said:
Oh... I misunderstood the question. It is trying to solve for the coefficient of friction vs the angle of the ramp, given that the coefficient changes from zero to some value after half the length of the ramp?
Yes.
 
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