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coefficient of friction between two blocks is 0.5 and lower surace is smooth.

I solved it like this:

$$F(\text{st max})=5\text{ N}$$

For the top block,

$$\begin{align} 6\text{ N} - 5\text{ N} &= 1a \\ a &= 1\ \mathrm{m/s^2} \end{align}$$

For the lower block, the driving force will be the frictional force, so

$$\begin{align} 2a &= 5\text{ N} \\ a &= \frac{5}{2} = 2.5\ \mathrm{m/s^2} \end{align}$$

I am confused as to how the lower block could have acceleration greater than the upper block, since the force is acting on the top block.

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  • $\begingroup$ I thought lower block never get more acc than above in this case because if it does then friction force would act on it towards left. $\endgroup$
    – imrran
    Oct 21, 2013 at 20:35
  • $\begingroup$ This question is related, and possibly even a duplicate. $\endgroup$
    – David Z
    Oct 21, 2013 at 21:06
  • $\begingroup$ I got the answer 5N is not critical or max force because it is also depend upon mass of lower block. $\endgroup$
    – imrran
    Oct 21, 2013 at 21:30
  • $\begingroup$ Solution: F-f=m1a and f=ma2 after solving it we get F=((m1+m2)/m2)*f so your F(max) will be 7.5N not 5N $\endgroup$
    – imrran
    Oct 21, 2013 at 21:30
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    $\begingroup$ No, that's still a comment. Look further down, to the box just above the "Post Your Answer" button. $\endgroup$
    – David Z
    Oct 21, 2013 at 22:25

3 Answers 3

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Your calculations are wrong.

The basic assumption that friction = u x N u = coefficient of friction N = Normal force (in this case the weight of the block) Above assumption is valid only if there is relative motion between the two blocks i.e a case of sliding motion, but before we consider that sliding occurs we should verify whether the block are moving relative to each other of not i.e. checking for static friction.

Now the maximum value static friction can reach is uN i.e. Sliding/kinetic friction but can also be lesser than that. Taking that into account and assuming friction to be f (a variable) and no relative motion between the blocks. No relative motion means that both blocks will have same acceleration.

Calculations :

6N− f =a m/s2 (for small block) f = 2a m/s2 (for big block)

substituting f=2a for small block

6N - 2a = a m/s2

6N = 3a m/s2

2 m/s2 = a

both blocks having same acceleration, hence no relative motion.

Value of friction in this condition is 2 x 2 = 4N which is less than uxN = 5N

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You have got the right answers, just are interpreting it a little bit wrong;

The net force on upper block is $6N - 5N = 1N$ and it's accelerations is $1 ms^{-2}$ while the force on bottom block is $5N$ and it's acceleration is $2.5ms^{-2}$ but you are missing the fact that the acceleration of the bottom block is with respect to ground while that of upper block is with respect to lower block. Therefore, if you see from ground bottom block accelerates with $2.5ms^{-2} $ while upper block moves with $3.5ms^{-2} $.

I apologise for misunderstanding and posting the wrong answer earlier

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  • $\begingroup$ This is incorrect, accelerations are always measured and calculated relative to an inertial frame (ie one undergoing zero acceleration (in Newtonian mechanics)) Thus assuming the acceleration of the upper block is relative to the lower block is incorrect. $\endgroup$
    – Rick
    Aug 18, 2015 at 12:58
  • $\begingroup$ @Rick: With utmost respect please try and understand what I have done properly before saying that the answer is incorrect. In this particular solution the force "1N" is calculated with respect to the lower block and hence the acceleration is also with respect to the lower block. If you understand now, kindly withdraw your downvote! $\endgroup$ Aug 18, 2015 at 15:42
  • $\begingroup$ The 1 N is force acting on the upper block. It doesn't matter where the force came from, it could be 1N of gravity from the moon, in calculating the acceleration via F=ma, the sources of the forces that add to the net force do not matter. The acceleration is always calculated with respect to an inertial frame. The lower block is not an inertial frame as it's accelerating. The ground is an inertial frame and thus it a valid reference frame for the acceleration. Please read potato's answer for an explanation to the question. I'd gladly upvote your answer after you correct it. $\endgroup$
    – Rick
    Aug 18, 2015 at 15:53
  • $\begingroup$ Looking back at your edit history, your first and second versions were actually much better. Your third version however, is incorrect. $\endgroup$
    – Rick
    Aug 18, 2015 at 16:03
  • $\begingroup$ @Rick: LOL, did you not read the question properly or what? the actual applied force is 6N and with respect to lower block it becomes 1N, get it? $\endgroup$ Aug 18, 2015 at 16:13
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This the result of not keeping track of what your calculation or measurement is relative to. The small block is not accelerating very much relative to the large block, this is correct:

5N-6N=-1(kg)*1(m/s^2)

The large block is accelerating much relative to the surface, this is correct:

2(kg)*(5/2)(m/s^2)=5N

But how is the small block moving relative to the surface? It is accelerating much faster than the large block:

(5/2)(m/s^2)+1(m/s^2)=(7/2)(m/s^2) or as you say (7/2)a

This is the acceleration of the small block you are missing out on.

Yes relative to the large block the small block is only accelerating at "a" or 1(m/s^2). But to find this relative to the surface you must add the acceleration of the large block.

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  • $\begingroup$ All the numbers imrran gave looked correct to me assuming a gravitational field of 10(m/s^2). did David make it so? $\endgroup$ Oct 22, 2013 at 4:57
  • $\begingroup$ This is incorrect, accelerations are always measured and calculated relative to an inertial frame (ie one undergoing zero acceleration (in Newtonian mechanics)) Thus assuming the acceleration of the upper block is relative to the lower block is incorrect. $\endgroup$
    – Rick
    Aug 18, 2015 at 12:57
  • $\begingroup$ "assuming the acceleration of the upper block is relative to the lower block is incorrect." ~ 5N was subtracted from the constantly provided 6N, why? How is it that an object in acceleration of the same direction can provide a force of 5N opposite to its acceleration, would not an equal and opposite force be present? What you are trying to say is that the 5N provided by the lower block is not relative to the lower block but relative to the friction-less surface, and I must disagree. The lower block is being accelerated not standing still. $\endgroup$ Jan 30, 2016 at 16:06
  • $\begingroup$ The 5N is incorrect. That is the maximum friction. However, in this case there is less friction as the blocks are not moving relative to each other. In this case the friction is 4N, giving each block an acceleration of 2m/s/s (relative to an inertial frame). Forces are on objects and are not relative quantities. Changing from one frame to another does not change the forces between objects. So you don't need to specify a reference frame when referring to forces between objects. So specifying what the 5N of force is relative to doesn't make sense. $\endgroup$
    – Rick
    Jan 31, 2016 at 16:39
  • $\begingroup$ "you don't need to specify a reference frame when referring to forces between objects" -> this is why I stated "I must disagree." But now I am to understand that you are trying to say the "u=0.5" is a coefficient of static friction. While I am inclined to believe this is true it is not made clear in the question. $\endgroup$ Feb 4, 2016 at 3:22

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