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A block with mass 40 kg in placed on a block of 60kg which lies on a frictionless surface. The coefficient of friction between 40kg mass and 60kg mass is 0.2. Now, a force of 100 N is applied on the upper block. We have to find the force of friction between the blocks.

ATTEMPT ON SOLUTION

I attempted the problem in the following way.

The normal force on the 40kg block would be 400N. Since, the coefficient of friction is 0.2, the maximum friction that can act is 400 * 0.2 = 80N.


The answer came out to be wrong, i need to know here i am thinking wrong.

I also know that that the friciton would act in the second block (60Kg) and it will start to move with 80N of force. The remaining N would be left with the upper block. But, that shouldn't be the problem because over here, friction wouldn't depend on relative acceleration. Help me approach the problem.

The block on block mass in problem

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closed as off-topic by John Rennie, ACuriousMind, HDE 226868, Qmechanic Aug 31 '15 at 7:13

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  • $\begingroup$ Is it the coefficient of static friction or of kinetic friction that you have? $\endgroup$ – Steeven Aug 29 '15 at 15:43
  • $\begingroup$ I don't know either, the question doesn't tell anything about that. By default, we consider to be static friction though. $\endgroup$ – Mrinal Gautam Aug 29 '15 at 15:49
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    $\begingroup$ Hi Mrinal and welcome to the Physics SE! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. $\endgroup$ – John Rennie Aug 29 '15 at 18:44
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If it is static friction, then the two blocks are stuck together and both have the same acceleration. In that case, the top block has a net force of 40 N (100 N pull - 60 N friction) and the bottom block has 60 N (just from the friction). Since 60N < 80N max friction, then the ansatz that this is static friction is consistent.

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  • $\begingroup$ When there's static friction, then 2 blocks are stuck.. i dont get it. I mean, we're applying 100 N on upper block, and the max friciton between both the blocks can be 80 N. So, the upper black is able to overcome it's friciton? And, the same amount of friciton would act on the lower block! $\endgroup$ – Mrinal Gautam Aug 29 '15 at 16:12
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    $\begingroup$ Static friction = stuck by definition. The upper block has +100 N from you and -60 N from contact with the lower block (this is the friction) for a net acceleration of (100 N-60 N)/40 kg =1 m/s^2. The lower block has +60 N from contact with the upper block (equal and opposite the -60 N on the upper block) for a net acceleration of 60 N/ 60 kg = 1 m/s^2. Same acceleration for the two, so they stay stuck. They stay stuck until you apply ~133 N. $\endgroup$ – Jason A Aug 29 '15 at 16:23
  • $\begingroup$ Oh! I get it. will the total normal force on the above block be 100 N? $\endgroup$ – Mrinal Gautam Aug 29 '15 at 16:27
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    $\begingroup$ You calculated the normal force correctly (assuming g=10 m/s^2): 400 N. The coefficient of static friction gives you the maximum friction force possible. Once you exceed that force, the two blocks will slip and you switch to using the coefficient of kinetic friction to calculate the friction force (which is less than the maximum static force). $\endgroup$ – Jason A Aug 29 '15 at 16:35
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The problem is that you have not solved the question yet. What you have found is not the friction between the boxes. It is something else. As you actually state yourself, you have instead found the maximum [static] friction. This is just the maximum possible value and not at all necessarily equal to the actual friction. Static friction can be anything from $0$ to this maximum limit of $80\:\mathrm{N}$ that you found.

In math-terms you have looked at static friction $f_s$ with this expression:

$$f_s\leq n \mu_s$$

There is only an equal sign here, if you are looking for maximum static friction. If you are just looking for static friction, you cannot use this. Always, when you have a force that you don't have a formula for, then use Newtons laws to find it:

Newtons 2nd law on the top block:

$$\sum F_x=m_{top}a\quad\Leftrightarrow\quad F-f_s=m_{top}a\quad\Leftrightarrow\quad f_s=F-m_{top}a$$

Newtons 2nd law on the bottom block:

$$\sum F_x=m_{bottom}a\quad\Leftrightarrow\quad f_s=m_{bottom}a \quad\Leftrightarrow\quad a=\frac{f_s}{m_{bottom}}$$

The accelerations $a$ are equal if this is static friction (if they are kept together). This is two equations with only two unknowns. If you put in numbers and solve them you get the right result.


You have in your question shown that the maximum static friction is $80\:\mathrm{N}$, so if our result here is larger, then we know that we do not have static but rather kinetic friction between them. Then you could easily have found the result without Newtons laws. Because in that case it must be kinetic friction $f_k$ between the boxes (as this is the only other possibility) since they slide over each other. And you do have a clear formula for kinetic friction:

$$f_k=n\mu_k$$

Note that this is another $\mu$ than before (typically $\mu_k$ is smaller than $\mu_s$)

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    $\begingroup$ A! i get it, thanks! I was making the novice mistake of not actually writing the equations but, solving them broadly. $\endgroup$ – Mrinal Gautam Aug 29 '15 at 16:41

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