Friction problem - Relation between coefficient of friction

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  • #1
Saitama
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Homework Statement


A man of mass m is applying a horizontal force to slide a box of mass m' on a rough horizontal surface. It is known that the man does not slide. The coefficient of friction between the shoes of the man and the floor is ##\mu## and between the box and the floor is ##\mu'##. In which of the following cases it is certainly not possible to slide the box?
A) μ>μ', m<m'
B) μ<μ', m<m'
C) μ<μ', m>m'
D) μ>μ', m>m'


Homework Equations





The Attempt at a Solution


If man applies a force F on box, due to Newton's third law, he also experiences a force F in the opposite direction. According to the question, F<=μmg and F<=μ'm'g. Also, since it is given that man does not slide, F has to be less than or equal to μmg. The box can slide if μ'm'g<F<μmg. The block never slides if μmg<μ'm'g. But I don't see how to find relation between the coefficient of friction and masses by considering the above inequalities.

Any help is appreciated. Thanks!
 

Answers and Replies

  • #2
Doc Al
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The box can slide if μ'm'g<F<μmg. The block never slides if μmg<μ'm'g.
OK.

But I don't see how to find relation between the coefficient of friction and masses by considering the above inequalities.
Go through the set of choices and see which ones make that second inequality true.
 
  • #3
Saitama
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Go through the set of choices and see which ones make that second inequality true.

Call f as the friction force on man and f' on box. For the first case,
Let μ=0.6, μ'=0.5, m=3 kg and m'=5kg. Then f=18N and f'=25N. Here f<f' so A is the answer but the given answer is B. :confused:
 
  • #4
Doc Al
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Call f as the friction force on man and f' on box. For the first case,
Let μ=0.6, μ'=0.5, m=3 kg and m'=5kg. Then f=18N and f'=25N. Here f<f' so A is the answer but the given answer is B. :confused:
Don't plug in specific numbers. You want an answer that will make it impossible to slide for any number in the given range.

What's an easy way to ensure that μmg<μ'm'g? One of the given choices ensures that will inequality will be met.
 
  • #5
Saitama
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What's an easy way to ensure that μmg<μ'm'g? One of the given choices ensures that will inequality will be met.

I am still confused. The inequality can be rearranged as m'>(μ/μ')m. For the first case, μ/μ'>1 so I guess (μ/μ')m may become greater than m' so first is false. Is this a correct way to put it?
 
  • #6
Doc Al
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The inequality can be rearranged as m'>(μ/μ')m. For the first case, μ/μ'>1 so I guess (μ/μ')m may become greater than m' so first is false. Is this a correct way to put it?
Yes. That shows that choice A won't work.

Now look at choice B.
 
  • #7
Saitama
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Now look at choice B.

In this case μ<μ' and m<m'. Rearranging as before, m'>(μ/μ')m. Already, m < m' and since (μ/μ')<1, the product on the RHS becomes more smaller and hence it is always true. Correct?
 
  • #8
Doc Al
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In this case μ<μ' and m<m'. Rearranging as before, m'>(μ/μ')m. Already, m < m' and since (μ/μ')<1, the product on the RHS becomes more smaller and hence it is always true. Correct?
Yes.
 
  • #9
Saitama
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