# Confusing problem on friction -- pushing the bottom block in a stack of blocks

teo11
Homework Statement:
Hey!
Suppose there are n blocks for instance 3 stacked on top of one another and a force F is applied on the bottom one. Let the blocks from bottom to top be 1,2,3,…,n. Let the friction coefficients be u1, u2,…un. Now I want to know on which surfaces will slipping occur.
Relevant Equations:
Newtons laws of motion
So far I have reached a pretty promising equation that if uk * g >= F/(m1+m2+…+mn) then kth block will atleast surely not slide wrt the lower one. But the converse to this that if uk * g < … then it will “surely” slide is wrong!!

Stuck on this from many days!
Thanks (:

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Hello and !

From your description I understand that F is applied to the lowest block. Your pretty promising equation seems to apply to the whole stack !

##\ ##

Homework Helper
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Welcome, @teo11 !

What uk stands for?
Why?

Homework Helper
Something like ##\mu## for block k ?

Let the friction coefficients be u1, u2,…un.

##\ ##

• teo11
teo11
Welcome, @teo11 !

What uk stands for?
Why?
edited the question
Homework Statement:: Hey!
Suppose there are n blocks for instance 3 stacked on top of one another and a force F is applied on the bottom one. Let the blocks from bottom to top be 1,2,3,…,n. Let the friction coefficients be u1, u2,…un. Now I want to know on which surfaces will slipping occur.
Relevant Equations:: Newtons laws of motion

So far I have reached a pretty promising equation that if uk * g >= F/(m1+m2+…+mn) then kth block will atleast surely not slide wrt the lower one. But the converse to this that if uk * g < … then it will “surely” slide is wrong!!

Stuck on this from many days!
Thanks (:
Welcome, @teo11 !

What uk stands for?
Why?
Uk stands for friction coefficient of kth block with the lower one, and I reached that eqn by considering common acc as if no sliding was there in whole system because that is the maximum possible acc in any case for a combined system of blocks above kth block and it.

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edited the question

Uk stands for friction coefficient of kth block with the lower one, and I reached that eqn by considering common acc as if no sliding was there in whole system because that is the maximum possible acc in any case for a combined system of blocks above kth block and it.
Normally, ##\mu_k## is the symbol used for the kinetic coefficient of friction.
In our problem, the blocks are not sliding respect to each other initially.
Therefore, we should consider the static coefficient of friction instead, which symbol is ##\mu_s##.

Sorry, what is the meaning of "acc"?

Have you tried a free body diagram of the blocks?

• BvU
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But the converse to this that if uk * g < … then it will “surely” slide is wrong!!
What if you assume none of the other interfaces are sliding? Would it be true then?

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Normally, ##\mu_k## is the symbol used for the kinetic coefficient of friction.
Quite so, but the notation was explained in post #1. ##u_r## would have been less confusing.
what is the meaning of "acc"?
acceleration

• Lnewqban
teo11
What if you assume none of the other interfaces are sliding? Would it be true then?
Yes it would be true then.

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Yes it would be true then.
So you need to figure which will slip first. Then, maybe, which will slip next as F is increased?

• teo11
teo11
I think I have found it. It involves using the concept of finding max. friction required at any surface which can be done by using an FBD, considering a system of either the upper of lower blocks depending upon net forces on this system and taking its acc to be Ac that is (net force on all blocks/ total mass). Now I see the weakest interface (which will break first by considering an imaginary force increasing), which in our simple case of n blocks with one force at bottom for first break would turn out to be finding min coefficient of static friction among all. Now repeat with more complicated expressions for the first surface which will break as there are #2 forces (one of top of first break) and the 2nd on the bottom one and we have a solution.

teo11
Can anyone put up some examples with answers for 4 or such blocks to test?

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But the static friction for each interface not only depends on the coefficient, but on the normal force and reaction.
It seems to me that there is one block that has the highest normal force and one block that has the minimum one.
Could you tell which is which?

You can analize the system, regarding which one would slide first as the pulling or pushing force reaches certain value, assuming two extreme conditions:
1) All the blocks have the same mass and each pair has a different coefficient of static friction.
2) All the blocks have the same coefficient, but each has a different mass.

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the static friction for each interface not only depends on the coefficient, but on the normal force and reaction.
Are you disagreeing with the formula in post #1?

teo11
So you need to figure which will slip first. Then, maybe, which will slip next as F is increased?
But the static friction for each interface not only depends on the coefficient, but on the normal force and reaction.
It seems to me that there is one block that has the highest normal force and one block that has the minimum one.
Could you tell which is which?

You can analize the system, regarding which one would slide first as the pulling or pushing force reaches certain value, assuming two extreme conditions:
1) All the blocks have the same mass and each pair has a different coefficient of static friction.
2) All the blocks have the same coefficient, but each has a different mass.
I actually did the same thing. For rth block and blocks above it I can write
Fr required <= F*(mr + m(r+1)+ … + mn)/(m1 + m2 +…+ mn)
So ur * g should be >= F/(m1 + m2 +…+ mn) as Fr= Normal reaction *g = (mr + m(r+1)+ … + mn)*g

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Are you disagreeing with the formula in post #1?
No, if applicable to the lowest block only.
Yes, otherwise.

I may still be confused about the n's, the k's and the r's shown above.
A FBD would greatly help.

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Yes, otherwise.
Draw the FBD for the top k blocks and consider the frictional force required at its base to achieve the desired acceleration.

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Draw the FBD for the top k blocks and consider the frictional force required at its base to achieve the desired acceleration.
As I see it, there is no force reaching the top blocks until the bottom block starts accelerating right after sliding.

Perhaps the problem should include a frictionless flat surface-bottom block, or a stack of blocks sitting on a wheeled cart?

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Your pretty promising equation seems to apply to the whole stack !
No, I believe it is general.

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