# Paradox in the two block problem [closed]

### UPDATE (regarding duplicate) :

This question is not a duplicate of another question. Sure, the situation in both the questions is same and, yes, both questions ultimately provide a methodology to solve the problem and finding the correct value of friction, but the moderator should realize that here I was NOT asking for a method to solve for friction. The problem was that I did not even realize that I will have to solve for friction.

I had solved a lot of such problems way back in school and I had got into a habit of assuming static friction to be the equal and opposite to whatever force is applied against friction (upto a maximum limit of friction). This was a blunder I made, and this is what made it appear like a "paradox". Moreover, as it turns out, I asked the same problem to a few friends of mine and many of them made the same mistake.

So, essentially the "another" problem is just asking for a general methodology to solve such problems, while this problem is like a puzzle which presents the user a methodology of solving the problem by considering different systems and the contradictions that arise due to them. I believe, a user who knows the general methodology presented in the other question is susceptible to the confusion/paradox that this problem presents.

### ORIGINAL QUESTION:

This is a familiar problem with the setting as given below: $\mu$ is the coefficient of friction.

Now, with the given applied force of $F = 10N$, and taking $g = 10m/s^2$, we know that the maximum friction force between $m_1$ and $m_2$ can be $30N$. Since, $F = 10N$ is less than the maximum friction force $30N$, friction force will be $f = 10N$ and the 2 blocks will, therefore, move together.

The acceleration of the combined system will therefore be:

$$a = F/(m_1+m_2)$$ $$a = 10/(5+3) = \frac{10}{8} m/s^2$$

Now, if we only consider the $m_1$ block and create its free-body diagram, we see that in the horizontal direction, there is only one force : force applied by $m_2$ block due to friction. And this friction is $f = 10N$. Now, if we calculate the acceleration of $m_1$ block, we find that:

$$a = f/m_1 = \frac{10}{5} = 2 m/s^2$$ This is paradoxical. How come the value of acceleration comes out to be different when we only consider the second block in our problem?

• It's not a paradox, you changed the system (from 8 kg to 5 kg). – LDC3 Jun 1 '15 at 4:19
• @LDC3, that isn't the error, the error is that OP assumed the friction force to be $10N$. The actual method of computing the friction force is given by the answers. – Cicero Jun 1 '15 at 4:21
• @Cicero It depends on the point of view; either the system was changed, or he has an invalid assumption. – LDC3 Jun 1 '15 at 4:37
• An easy way to see why the friction force is not 10N: If it were, the upper block would have a net force of zero N, and it wouldn't move at all! – Sanchises Jun 1 '15 at 8:53
• @sanchises Thanks for pointing this out. However, this was also part of the paradox/confusion. – shivams Jun 1 '15 at 9:01

You are making a major flaw here. The friction between the 2 blocks is not going to be $10N$. It is going to be something, but we will have to calculate it.

Assume the friction force to be $f$ such that the acceleration for both the blocks is same.

Now, equations for $m_2$ and $m_1$ separately are:

$$F-f = m_2*a = 3a$$ $$f = m_1*a = 5a$$

Solving both the equations, you will get:

$$f = \frac{50}{8}N$$

This friction force is now compatible. Using this friction force, the acceleration will now come out to be $$a = \frac{10}{8}m/s^2$$, which is same as the original answer.

This is an interesting paradox! I suggest resolving it in the following way:

First, note that the coefficient of friction between the two blocks allows friction "up to 30N" to occur - but if the two blocks are not moving relative to each other, the friction will be "whatever it is".

Second, look at the force balance on the top block. We have the force of the rope, 10 N, and we have whatever force of friction between the blocks is. The difference between these two is the force available to accelerate the top block. Assuming the acceleration of the top block is $a$, then we find for the force of friction

$$F_f = F_{pull} - m_2 \cdot a\tag1$$

This is the force available for accelerating the second block, so we find for the acceleration of the second block:

$$a = \frac{F_f}{m_1}\tag2$$

We can now eliminate $F_f$ from the equation to solve for $a$:

$$a = \frac{F_{pull} - m_2 \cdot a}{m_1}$$

Which we can rearrange to give

$$a = \frac{F_{pull}}{m_1 + m_2}\tag3$$

This is of course the "usual" result - the acceleration we would expect when the two blocks move together under the influence of $F_{pull}$. But we can now substitute this value of $a$ into (1) and get the actual force of friction:

$$F_f = F_{pull}\left(1 - \frac{m_2}{m_1+m_2}\right)$$

Done in this way, there is no more paradox... Good question!

• Wow. Thanks for the nice explanation. I don't how I made that blunder of assuming the friction to be 10N. Force of habit, I guess. However, I've accepted @mahesh's answer since he answered before. – shivams Jun 1 '15 at 4:30
• @shivams no need to explain why you selected a particular answer. I am glad my explanation helped at least a little in resolving this paradox in your mind. I thought it was a good question. – Floris Jun 1 '15 at 4:35
• @Floris This formula works only when $F_{pull}$ is less than a threshold force beyond which the upper block slips. In this case finding this threshold force is quite simple. The maximum acceleration of the block of mass $m_1$ is $2 m/s^2$. So, the maximum value of $F_{pull}$ for which the two blocks move without slipping is $16N$. But this way of attacking the problem when there are a lot of blocks (say 5) is very cumbersome. Is there an alternate way to solve the problem or perhaps a very organised (algorithmic) way of solving it? – user161940 Mar 30 '17 at 17:52
• @Floris What I don't understand is why does applying force on the upper block causes same acceleration in the lower block......Do these blocks act as a system or something – Hydrous Caperilla Apr 9 '18 at 10:50
• @HydrousCaperilla if the force needed to accelerate the bottom block is less than the force of static friction between the block then they won’t move relative to each other - so yes, in that case they will move as a system. But you don’t know that until you do the math carefully - it is an assumption that can be tested after you do the work. – Floris Apr 9 '18 at 11:25

In the second calculation, the $10N$ force is not acting on the m1 block. The only force on the block is the friction force from m2. The frictional force will be exactly the amount required to keep m1 and m2 moving together. The frictional force will be given by: $$F = m_1 a = (5 kg)(1.25 m/s^2) = 6.25 N$$

Checking with just the top block: $$a_2 = (F_{pull} - F_{friction})/m_2 = (10 - 6.25)/3 = 1.25 m/s^2$$ which is the same as the original answer.