Consider the following situation. Two blocks A and B connected by an inextensible string rest on a rough horizontal surface. The mass of A is 2kg and that of B is 3kg. The coefficient of static friction between block A and the surface is 0.1 and the coefficient of static friction between block B and the surface is 0.2

Two external forces of magnitudes 1N and 8N are now applied on A and B. The situation is shown in the figure below:

enter image description here

My goal is to find the tension in the string connecting the blocks and the friction acting on each block after the forces are applied.

First, I calculated the values of limiting friction corresponding to A and B, using (friction coefficient)*(normal contact force). For A it turns out to be 2N and for B 6N.

Now, considering the two blocks plus the string as a single system, the total external force is 8-1=7N rightwards and the maximum possible static friction is 6+2=8N leftwards. From this I concluded that the system cannot accelerate under the application of the given forces. Moreover, the sum of the static friction forces acting on A and B must be exactly 7N in order to balance out the external force.

To find the tension in the string, we need to find the individual magnitudes of the two friction forces. But I can't think of any reasonable way to obtain this using Newton's laws. I would like to know the right way to approach this problem.

The correct answer:

  1. Friction on block B is 6N leftwards.
  2. Friction on block A is 1N leftwards.
  3. Tension in the string is 2N, the system is in equilibrium.

From the solution, it seems that the friction on block B has been taken as the limiting(or maximum) value, 6N.

How did they arrive at this conclusion? Is there a way to explain this just using Newton's laws and motion constraints, or is some additional reasoning and extra conceptual insight required? Are these the ONLY possible values of friction forces? It seems like the forces could be anything as long as they add up to 7N and do not exceed their respective limiting values(although my gut feeling tells me that only one solution is possible for a practical situation such as this). The tension would then be decided accordingly, depending on the values of the friction forces (assuming that the breaking stress of the string is large enough, of course). In other words, Newton's laws predict an INFINITE number of solutions. What would be the most accurate and way of arriving at the answer to the question?

Final question: Is my way of concluding that the system does not accelerate correct? Or is there a more solid argument?

  • $\begingroup$ Hi Kalyan. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$
    – Qmechanic
    Commented Nov 22, 2016 at 15:17
  • $\begingroup$ Oh I'm sorry for that. I thought that tag generally wouldn't get you answers. I'll add it back again. $\endgroup$
    – Newton
    Commented Nov 22, 2016 at 15:19
  • $\begingroup$ You are correct. With no further information this problem does not have a unique solution. $\endgroup$
    – Farcher
    Commented Nov 22, 2016 at 15:29
  • $\begingroup$ @Farcher But how is that physically possible? There must be some definite interaction between the ground and the block right? What are the factors upon which the different solutions depend? For example are there any initial conditions or does the manner in which the forces are applied affect the result in any way? $\endgroup$
    – Newton
    Commented Nov 22, 2016 at 15:38
  • $\begingroup$ You have to ask yourself, how do I decide on the division of the 7 N frictional force between the two blocks? $\endgroup$
    – Farcher
    Commented Nov 22, 2016 at 15:59

2 Answers 2


Such problems of indeterminate forces (eg Block on a block problem, with friction and Paradox in the two block problem) can be dealt with in 2 stages :
stage 1 : look at the system as a whole to determine whether it accelerates, and if so by how much
stage 2 : apply $F=ma$ to each part of the system individually.

Regarding the 2 blocks and string together as the system, the net applied force on it is 7N to the right. The maximum friction force is 8N (left or right). So the system does not accelerate. The net force on each block must be zero.

Applying the equilibrium conditions for each block we have
block A : $T-1 \le 2$ and $1-T \le 2$ giving $-1 \le T \le 3$
block B : $8-T \le 6$ and $T-8 \le 6$ giving $2 \le T \le 14$.

There is also a constraint for the string : $T \ge 0$.

Combining the inequalities, the only feasible solutions are in the interval
$2 \le T \le 3$.

I agree with Pirx : a unique solution does not exist. The answer given is not the only solution. If you think the IITJEE exam question proves otherwise, please upload the question and official solution.

  • $\begingroup$ You are right. The paper in 1997 unfortunately got leaked and thus had to be replaced. So there is no official answer key as such. The solution mentioned in my question was from a reference book and was not the official one. After reading your answer I believe that we can at best predict the range in which the forces lie, the exact values being indeterminate. $\endgroup$
    – Newton
    Commented Nov 23, 2016 at 18:16
  • $\begingroup$ How did you determine them to be systems......I suck at identifying them? $\endgroup$ Commented Apr 9, 2018 at 10:45
  • $\begingroup$ @HydrousCaperilla That is quite a difficult question to answer in a comment. There does not seem to be a question on this site asking "What is a system?" You could post a question, using an example of your choice. $\endgroup$ Commented Apr 10, 2018 at 5:33
  • $\begingroup$ @HydrousCaperilla Briefly, what a system consists of depends on your purpose, the question you are trying to answer. The inputs are environmental conditions such as applied forces and their responses. In this case the external forces are the 1N and 8N forces and the friction forces. Their response is the (potential) motion of the blocks. These external factors define the boundary between the system (the blocks and string) and its environment. $\endgroup$ Commented Apr 10, 2018 at 5:40
  • $\begingroup$ @HydrousCaperilla In this example I first consider the 2 blocks and string together to be the system, to find the condition for the whole to move as one. Then I consider each block in turn to be a system, to find out the conditions for them to move separately. So the system changes according to what I am trying to find out. $\endgroup$ Commented Apr 10, 2018 at 5:45

The answer will indeed depend on the initial conditions. Here is how I think about this problem to arrive at the solution suggested to you: Very roughly, if you consider a situation with the rope slack at the beginning, then block B will start moving until the rope is tight. While the block is moving, the friction force on it will be equal to its weight times the kinetic friction coefficient for the moving surface (which in general will be lower than the limiting value for static friction). At some point the rope will tighten, and exert a force on Block B, retarding its motion and ultimately stopping it. The assumption seems to be that, at the point when block B stops, the friction force returns to the limiting value of static friction. Sounds reasonable to me, but reality might be more complex than this.

  • $\begingroup$ Wow, I never would have thought about it in that direction. Thanks for the response! $\endgroup$
    – Newton
    Commented Nov 22, 2016 at 16:04
  • $\begingroup$ And why does it seem "reasonable" that the friction returns back to the limiting value? Just looking for some intuition. $\endgroup$
    – Newton
    Commented Nov 22, 2016 at 16:12
  • $\begingroup$ All I meant with that last comment is that I consider this a plausible scenario. I could, however, imagine that performing the corresponding experiment may produce different outcome(s), depending on the properties of the surfaces in question, as well as the rope. $\endgroup$
    – Pirx
    Commented Nov 22, 2016 at 16:20

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