This is going to be a long post-like question so make sure you have enough time before reading this. And this I got after thinking continuously for about 4 hours on the same question and I want to verify that if my ideology about this question is correct. First, what the question actually is?


Please refer to the diagram in image

Explanation of the Image (You can skip this)

The image of the question is just below, please refer to it for better visualization. There are two blocks A and B. Block A has a mass of $m$ while block two has a mass of $2m$. They are placed on top of each other. The surface between them (Block A and B) has a coefficient of friction $\mu$ and that between Block B and ground is $2\mu$. In front of these two blocks is a wall that has a pulley attached to it (the wall). Now the block A is attached to a string which then runs over the pulley which is attached to the wall. And from that pulley, the string runs over another pulley which is attached to Block B and this string finally terminates getting attached back to the wall ( refer image for better visualization ). Now a force $\overleftarrow{F}$ is applied on the Block A.

enter image description here

Actual Question

What would be the minimum value of $F$ so that the blocks start to move?

My Solution

One Note Before Starting This

Whenever I say a force being "paid by a system", I mean either it gets canceled by another force on the system or it gets exerted on the system. Also, the source of the force mentioned above in the main title of the question refers to the main force $\overleftarrow{F}$ acting on the system.

Now the solution

The most shocking thing I learned while solving this question was that the friction between block A and B will be paid two times all differently by the same force $\overleftarrow{F}$ acting in disguise on the two blocks. This was shocking as I first thought that the friction will be the usual force paid by the two blocks in some ration and not two times. But what I came to figure out was that it will be paid by each block separately and the amount of force of friction paid will be the same for two blocks and not in some ratio and those two will actually be the action-reaction pairs. One way we can thing about this is that the first block let's say starts moving in the left direction then it will exert a force of friction (which would be the reaction pair of the friction on block A) on the second block. Now to overcome this force the second block will pay the same amount of force in opposite direction. Hence the two pairs of force which are action-reaction pairs will be paid differently by each block. Now, these two pairs will be equal to $\mu N$ where $N$ is the Normal force by Block A on B. And $N = mg$. Hence

Friction b/w Block A and B $ = \mu N \\ \rightarrow \mu mg$

Friction b/w Block B and ground $= 2\mu N \\ \rightarrow 2\mu(3m)g \\ \rightarrow 6\mu mg$

For Block A

Now on block A, there is a force $\overleftarrow{F}$ and a friction $\mu mg$ which would be paid completely by this block so the tension $T$ in the string would be

$$ T = F - \mu mg $$ Let's say this equation one.

For Block B

Now there is $2T$ on block B and friction from the ground and surface above and remember that this block will pay its reaction pair separately. Hence

$$ 6\mu mg + \mu mg = 2T $$

Let's say this equation two.

Analyzing equation 1 and 2 we get

$$7\mu mg = 2F - 2\mu mg \\ F = \frac{9\mu mg}{2}$$

Final Question

Now here I want to ask that was this way of thinking correct or was it just a coincidence that I reached this solution? And thank you for spending your precious time on this question to help me out. I appreciate your efforts.

Image with all forces labelled

enter image description here

  • $\begingroup$ Your talk of forces being "paid" is confusing. Your equations are correct, but I am having a hard time following your concepts. Typically we take the force diagram you have provided and then set up the equations you did using Newton's second law. There isn't a need to get confused even further by talking about forces being paid for. Are you wanting someone to check your work, or are you wanting a conceptual understanding of how forces work? $\endgroup$ Oct 2, 2020 at 18:58
  • $\begingroup$ Your diagram is a bit confusing to me. It seems like you're treating the string as if it's inextensible and the pulleys are fixed in place. But if that is the case, then no amount of force should be able to move block A. $\endgroup$ Oct 2, 2020 at 18:59
  • $\begingroup$ @BrainStrokePatient No, the force diagram is correct as long as the pulleys are massless. $\endgroup$ Oct 2, 2020 at 19:00
  • $\begingroup$ @BioPhysicist Here paying of forces just means that the force is being applied on the body and if any force resists it then it will get cancelled or if there is no restriction then it will cause the body to accelerate $\endgroup$
    – Ritanshu
    Oct 2, 2020 at 19:02
  • 1
    $\begingroup$ So by "paid for" you just mean that a force is acting on an object? $\endgroup$ Oct 2, 2020 at 19:02

3 Answers 3


I suppose that when the system was assembled, some small tension $T$ was applied on the string to attach it to the wall, so that it can be straight. That is before $F$ comes to play.

When some $F$ is applied, while $F - T < \mu mg$ nothing moves. And the tension in the string doesn't change because the string is not affected by $F$.

As soon as $F - T = \mu mg$ block A can move. Let's say it moves a small displacement $\Delta L$.

Now, the tension in the string is $T' = T + E\frac{\Delta L} {L}$, where E is the modulus of elasticity and $L$ the total length.

But the meaning of a string is being rigid, so we can translate this to an infinite $E$.

So, even for a very small displacement, $T'$ in the string becames big enough to move block B, what means: $2T' = (2\mu)3mg + \mu mg => T' = \frac{7}{2}\mu mg$.

Back to block A, when it starts to move $T -> T'$, so: $F - T' = \mu mg => F = \frac{9}{2}\mu mg$

  • $\begingroup$ Thanks, This is what I wanted to know that what is logically going on there. This helped me a lot, Best answer $\endgroup$
    – Ritanshu
    Oct 3, 2020 at 4:38

Now here I want to ask that was this way of thinking correct or was it just a coincidence that I reached this solution?

Your thinking is seems OK and solution correct, but it could have reached more quickly with applicable free body diagrams and realizing that a constraint is that both blocks must move at the same time. I will respond to your response to @BioPhysicist asking you what your really asking, that is:

I am asking that both the pairs of friction force which are action-reaction pairs on body A and B are opposing the force 𝐹 and is that actually what is happening?

What is really happening here is that linking the movement of both masses with the pulley system gives you the constraint that in order for either mass to move, both must necessarily move. With that knowledge you can start by treating both blocks as a combined system.

  1. Draw a free body diagram for the combined masses as a system showing all external forces acting on the combined masses. Those forces would be $F$ and $f_{2}=(2u)(3mg)$ acting to the left and $3T$ acting to the right. That gives you one equation and two unknowns $F$ and $T$. Note that for the combined system, $f_1$ is an internal force that is not included.

  2. Draw a free body diagram of block A alone. For it you have $F$ acting to the left and $T+f_{2}$ where $f_{2}=umg$ acting to the right. Solve that for $T$ and plug into the first equation gets you the answer.

Hope this helps.

  • $\begingroup$ I wanted to solve this question without any law of Newton and more intuitively $\endgroup$
    – Ritanshu
    Oct 2, 2020 at 19:32

I am asking that both the pairs of friction force which are action-reaction pairs on body A and B are opposing the force F and is that actually what is happening?

Friction opposes relative motion$^*$ between surfaces, not other forces (at least not explicitly).

Pulling on block A to the left will tend to cause block A to slide to the left across block B. Therefore, static friction will act to the right on block A. By Newton's third law, this static friction force must also act on block B to the right, but you can look at it for the same reason. If block A would slide to the left relative to block B, then block B would be sliding to the right relative to block A. Hence, friction opposes this sliding.

The same is true between block B and the ground. Block A moving to the left would cause block B to move to the right, thus sliding to the right relative to the ground. Static friction between block B and the ground opposes this and acts to the left.

Of course force $F$ sets all of this up, but the friction forces themselves are not acting because of force $F$ directly. They are just opposing relative motion. An easy way this "force opposition" line of thinking can confuse you is to say, "Well why isn't friction on block A opposing the tension force? How can friction know which force to oppose?" And the answer to this is that friction isn't opposing forces, its opposing relative motion.

$^*$Of course with static friction it is impending relative motion rather than actual relative motion, but for sake of brevity I will say relative motion to refer to that case as well.

  • $\begingroup$ I know that friction doesn't oppose a force and it opposes relative motion but that is just an indirect way of saying that it opposes the net force on the block. The answer to your question on How does the friction know which direction to oppose is that it always opposes in the direction of net force.Please understand that your argument here is just like the argument that which force is applied first of the action-reaction pairs. Everyone knows that none comes first but one causes other, similarly friction opposes relative motion which instatly lead to result that if opposes main force $\endgroup$
    – Ritanshu
    Oct 2, 2020 at 19:25
  • $\begingroup$ @user921307 No, "friction opposing net force" is also false. If static friction is present, then the net force is always $0$ (at least for your blocks on horizontal surfaces here). In the case on kinetic friction, the net force does not have to be in the direction of motion, so there it fails too. $\endgroup$ Oct 2, 2020 at 19:37
  • $\begingroup$ Ok, I understand that but I think you also understand that for this context this statement would be applicable $\endgroup$
    – Ritanshu
    Oct 2, 2020 at 19:44
  • $\begingroup$ @user921307 ok, yeah that is fine for static friction, but it doesn't work for kinetic friction. And I would argue the "net force" line of thinking is the indirect way because friction is an interaction between the two surfaces, so motion between them is what is causing the interaction. But I think we both understand everything at this point. Thanks for the discussion $\endgroup$ Oct 2, 2020 at 19:44
  • $\begingroup$ Thanks for sharing your perspective $\endgroup$
    – Ritanshu
    Oct 2, 2020 at 19:46

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