When Friction the block m2 start to move?

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SUMMARY

The discussion focuses on the conditions under which block m2 begins to move when subjected to a force F=k*t, where k is a constant and t is time. The key conclusion is that block m2 will start moving when the applied force exceeds the sum of the static friction forces acting on it, specifically fr1 and fr2. The static friction forces are defined by the coefficients of friction μ1 and μ2, and the normal forces related to the weights of the blocks m1 and m2. A critical correction is noted regarding the maximum static friction, which is μsF_N, emphasizing that static friction can vary between zero and this maximum value.

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  • Understanding of Newton's second law (F=ma)
  • Knowledge of static and kinetic friction concepts
  • Familiarity with coefficients of friction (μ1 and μ2)
  • Ability to analyze free body diagrams (FBD)
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navneet9431
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Homework Statement


Screenshot_2018-07-29-09-57-49-598_com.hashlearn.now.jpg

A block m2 is placed on a surface where the coefficient of friction is μ2 and another block m1 placed over the block m2 where the coefficient of frictio is μ1.
A force F=k*t(Force as a function of time),is acting on the block m2.

Now my question is that when would the block m2 start moving?

Homework Equations


F=ma

Static Friction=Coefficient Of Friction*Normal Force

The Attempt at a Solution


I tried solving this by drawing the FBD of the objects.
IMG_20180729_100910.jpg

See two forces fr1 and fr2 will be acting on the block m2.
So m2 would start to move when F>fr1+fr2.
Or simply we can say k*t=fr1+ft2.
So the block m2 would start to move just after time "t".
Is my result correct?
 

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Yes it is, just be careful on how you going to equate forces fr1 and fr2 with relation to the weights ##m_1g ## and ##m_2g ## and the static friction coefficients ##\mu_1## and ##\mu_2##.
 
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Delta² said:
Yes it is, just be careful on how you going to equate forces fr1 and fr2 with relation to the weights ##m_1g ## and ##m_2g ## and the static friction coefficients ##\mu_1## and ##\mu_2##.
Thanks for the reply!
So my result was correct.
See the pic I have uploaded.
Have I proceeded correctly?
1532841119367.jpeg
 

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Yes your result seems correct to me. Actually a minor correction, it should be ##t_0\geq…## not just >.
 
Last edited:
navneet9431 said:
Static Friction=Coefficient Of Friction*Normal Force
There is a common blunder in that equation, and it is causing you to get the wrong answer.
For kinetic friction it works: ##F_k=\mu_kF_N##, but for static friction there is a key difference.
 
haruspex said:
There is a common blunder in that equation, and it is causing you to get the wrong answer.
For kinetic friction it works: ##F_k=\mu_kF_N##, but for static friction there is a key difference.
You mean to say that my result is wrong?
 
navneet9431 said:
You mean to say that my result is wrong?
Yes.
 
haruspex said:
Yes.
How?Please explain.
And,also mention the correct result!
 
navneet9431 said:
How?Please explain.
And,also mention the correct result!
Consider the situation when the applied force is zero. What are the frictional forces?
 
  • #10
Zero
haruspex said:
Consider the situation when the applied force is zero. What are the frictional forces?
 
  • #11
navneet9431 said:
Zero
Right, so what is wrong with ##F_s=\mu_sF_N##?
 
  • #12
Ah, I think @haruspex is right, it is the maximum static friction that is equal to ##\mu_sF_N##. Static friction can take any value between zero and this maximum value.

When the box ##m_2## is about to move then the static friction between ##m_2## and the surface is indeed as you calculate ##fr_2=\mu_2(m_1+m_2)g## but box ##m_1## doesn't move with respect to box ##m_2## so the kinetic friction between m1 and m2 is just zero and also the static friction ##fr_1## is zero at the start of movement, later it gets bigger as the external force gets bigger.
 
Last edited:

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