A 10 N force is applied on the 3 kg block. What are the accelerations of the 2 kg block, 3 kg block and 7 kg block? Take the value of g=10m/s^2.

enter image description here

what is the acceleration of each of the block ?

Source of question : HC VERMA ,CHAPTER 6,Question 23


maximum value of static friction between m2 and m3 =15 N
maximum value of static friction between m1 and m2 =4 N
applied force = 10 N

considering the surface between m2 and m3
applied force on m2 =10N < 15 N ; no relative motion between m2 and m3.

considering the upper surface m2 and lower surface of m1
applied force on m2 =10N > 4 N ; there is relative motion between m2 and m1.

acceleration of m1 = force/2 = (maximum value of static friction)/2 = 4/2 = 2 m/s^2

acceleration of m3 = force/7 = (value of static friction)/7 = 10/7 m/s^2.

since there is no relative acceleration between the surface of m2 and m3,acceleration of m2 = that of m3= 10/7 m/s^2.

  • $\begingroup$ This is quite a wall of text, it's hard to tell what's really being asked and how all the other information relates to it. $\endgroup$
    – David Z
    Jul 9, 2016 at 18:44
  • $\begingroup$ The question as it now stands violates the homework policy, which requires that you ask about a specific physics concept and that you show some effort to work through the problem. You have removed your working and also the 2 conceptual questions which you asked. The problem with the original question was not that there was too much text but that it was badly presented, making it difficult to read. I think it was ok (just about) after I edited it for you. $\endgroup$ Jul 10, 2016 at 11:08
  • $\begingroup$ This is NOT a homework problem, i can across this question while i was revising for a course in college. $\endgroup$ Jul 10, 2016 at 13:52

1 Answer 1


First off, you should draw free body diagram for each block.

enter image description here

Then, you should recognize if relative motion between blocks occurs or they move simultaneously. For this purpose, you can consider to the maximum friction force between the blocks. If second block ($m_2$) moves with respect to the other blocks, then friction forces $f_1$ and $f_2$ must reach to their maximum amount ($\left(f_1\right)_{\textrm{max}}=\mu_1N_1$ and $\left(f_2\right)_{\textrm{max}}=\mu_2N_2$). If so, then the block $m_2$ cannot move at all because $F\lt \left(f_1\right)_{\textrm{max}}+\left(f_2\right)_{\textrm{max}}$. If block $m_2$ doesn't move, then the other blocks mustn't move but as you see in the diagrams, they ($m_1$ and $m_3$) will move because there is a nonzero force acting on each of them. So, friction forces have not reached to their maximum amount and thus relative motion won't occur.

It may that you say relative motion will occur only between $m_1$ and $m_2$. If so, we will have two blocks: $m_1$ and $m=m_2+m_3$

enter image description here

If relative motion occur between $m_1$ and $m$ then we have $f_1=\left(f_1\right)_{\textrm{max}}=4\mathrm N$ and then $a_1= \frac{f_1}{m_1}=2\mathrm {m/s^2}$

On the other hand, we have: $F-f_1=ma$ thus $a=\frac{F-f_1}{m_2+m_3}=0.6\mathrm {m/s^2}$

As you see, $a\lt a_1$ and it is impossible.

It is obvious that relative motion won't occur only between $m_2$ and $m_3$ too. Because $F=10\mathrm N\lt \left(f_2\right)_{\textrm{max}}=15\mathrm N$

I suppose that you can continue yourself and find the acceleration.

  • $\begingroup$ maximum friction between m2 an m3 is 15 (>10), m1 and m2 =4(<10). So shouldn't there be relative motion between m2 and m1, and now realtive motion between m2 and m3? $\endgroup$ Jul 10, 2016 at 4:10
  • $\begingroup$ @user1559814 I'll update the answer. Please wait! $\endgroup$
    – lucas
    Jul 10, 2016 at 4:29
  • $\begingroup$ i have edited the question with my answer, can you please point out where i have made conceptual mistake $\endgroup$ Jul 10, 2016 at 13:51
  • $\begingroup$ @user1559814 As I said in my answer,when relative motion doesn't occur between $m_2$ and $m_3$ they can be substituted by a single block $m$. Your second and important mistake is that you don't consider to possibility of motion of $m_1$. You have obtained $a_1=2$ and $a_2=10/7$ (although this is wrong but there is no problem with what I want to say). This is impossible. Because $m_2$ pulls $m_1$ follow itself and so, $m_1$ cannot overtake the $m_2$. $\endgroup$
    – lucas
    Jul 10, 2016 at 15:54

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