# When Friction the block m2 start to move?

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## Homework Statement 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. 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?

#### Attachments

• Delta2

## Answers and Replies

Delta2
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Yes it is, just be careful on how you gonna 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##.

• navneet9431
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Yes it is, just be careful on how you gonna 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? #### Attachments

Delta2
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Yes your result seems correct to me. Actually a minor correction, it should be ##t_0\geq…## not just >.

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haruspex
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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.

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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?

haruspex
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You mean to say that my result is wrong?
Yes.

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Yes.
How?Please explain.
And,also mention the correct result!

haruspex
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How?Please explain.
And,also mention the correct result!
Consider the situation when the applied force is zero. What are the frictional forces?

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Zero
Consider the situation when the applied force is zero. What are the frictional forces?

haruspex
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Zero
Right, so what is wrong with ##F_s=\mu_sF_N##?

Delta2
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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.

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