Dynamics. Inclined plane with friction coefficient

  • Thread starter Pepealej
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I've got the following exercise, which I've done 3 times already and I think I haven't made any mistake:



A body is slid upwards through an inclined plane of angle [tex]\theta=15\degree[/tex]. Determine the friction coefficient of the body with the surface of the plane if the time it takes to go upwards [tex]t_a[/tex] is double the time it takes to go downwards [tex]t_b[/tex] ([tex]t_a=2t_b[/tex]).

Mi procedure is as follows:

The normal force: [tex]N=mg\cos(\theta)[/tex].

1. Upwards:

During the ascent the x componen of the weight and the friction force act in the same direction, therefore: [tex]a_a=-g\sin(\theta)-\mu g\cos(\theta)=-g(\sin(\theta)+\mu\cos(\theta))[/tex].

The ecuacion of space as a function of time is: [tex]x=2v_0 t_b -2g(\sin(\theta)+\mu\cos(\theta))t_b^2[/tex] (1).
The ecuation of speed as a function of time is: [tex]0=v_0 -2g(\sin(\theta)+\mu\cos(\theta))t_b[/tex] (2).

2. Downwards

During the descent the x component of the weight and the friction force act in opposite directions, therefore: [tex]a_a=-g\sin(\theta)+\mu g\cos(\theta)=-g(\sin(\theta)-\mu\cos(\theta))[/tex].

The ecuation of space as a function of time is: [tex]0=x-\frac{1}{2}g(\sin(\theta)-\mu\cos(\theta))t_b^2[/tex] (3).


Combining (1), (2) and (3) we obtain:

[tex]\frac{1}{2}g(\sin(\theta)-\mu\cos(\theta))t_b^2=4g(\sin(\theta)+\mu\cos(\theta))t_b^2-2g(\sin(\theta)+\mu\cos(\theta))t_b^2[/tex]
[tex]\sin(\theta)-\mu\cos(\theta)=4\sin(\theta)+4\mu\cos(\theta)[/tex]
[tex]\mu=-\frac{3}{5}\tan(\theta)[/tex]

From here I deduce that a value for [tex]\mu[/tex], for with the conditions are met, doesn't exist. Is it correct?

Thanks! :)
 

Answers and Replies

  • #2
haruspex
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From the calculation, I deduce that the object is given an initial push up the plane, then left to the mercies of gravity and friction. That was not clear from the description.
If so, it seems obvious to me that the descent will last longer than the ascent. So maybe the problem description is wrong and it should be tb = 2ta.
 
  • #3
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From the calculation, I deduce that the object is given an initial push up the plane, then left to the mercies of gravity and friction. That was not clear from the description.
If so, it seems obvious to me that the descent will last longer than the ascent. So maybe the problem description is wrong and it should be tb = 2ta.

Adding to that, the relation between ascent time and descent time depends on the push velocity given to the object, which is not specified. Is that the complete problem statement?

Also, there are several algebraic errors in the solution.
 
  • #4
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From the calculation, I deduce that the object is given an initial push up the plane, then left to the mercies of gravity and friction. That was not clear from the description.
If so, it seems obvious to me that the descent will last longer than the ascent. So maybe the problem description is wrong and it should be tb = 2ta.

Oh my.. That might be the mistake.. U_U Didn't notice that (the exercise is wirtten wrongly) Thanks. I'll see if that was the problem.
 
  • #5
haruspex
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Adding to that, the relation between ascent time and descent time depends on the push velocity given to the object, which is not specified.
No, that doesn't affect the ratio.
Also, there are several algebraic errors in the solution.
Yes, but I suspect these were in transcribing it for the post. They all seem to cancel out later.
 
  • #6
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Thanks for your replies. The problem was the question was wrongly written (tb=2ta, not ta=2tb). That was it :)
 

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