# Friction problem involving a pulley

## Homework Statement

I am posting the snapshot of the problem

## Homework Equations

Moment of Inertia equations

## The Attempt at a Solution

Now, the pulley has the moment of inertia $I = 1.5MR^2$. This is strange, since the coefficient of $MR^2$ is less than or equal to 1. So, this means that $M$ is not the mass of the pulley or $R$ is not the radius of the pulley. Do you think my suspicion is correct ? #### Attachments

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haruspex
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Gold Member
M is not the mass of the pulley
Correct. M is given as the mass of the suspended mass at A. This is being used as the unit of mass throughout.

Right. So $R$ must be the radius of the pulley. Now let $T_2$ be the tension in the vertical part of the string and $T_1$ be the tension in the horizontal part of the string. Then we can write the following equations if $a$ is the magnitude of the downward acceleration of block A.
$$\tau_{net} = I\alpha = 1.5MR^2 (a/R)$$
$$\tau_{net} = (T_1 - T_2) R$$
$$Mg - T_2 = Ma$$
$$T_1 = 5Ma$$
so solving, we get $T_2 = M(g-a)$. Now let $f$ be the kinetic friction between blocks B and C. Also, let $a_2$ be the acceleration of block C in ground frame. So we can write the equations
$$f = 3M a_2$$
$$T_1 - f= 2M a$$
And this leads us to conclude that
$$5Ma - 3M a_2 = 2M a$$
$$\to a_2 = a$$
But this does not seems right. I think the acceleration of block B and C are not the same. And I am getting it the same. Whats happening here ?

haruspex
Homework Helper
Gold Member
$$T_1 = 5Ma$$?
No.

There is no friction between block B and the ground. And friction between blocks B and C can be considered the internal forces. So I considered the blocks B and C as a system and then applied the Newton's second law to them. Is that a problem ?

haruspex
Homework Helper
Gold Member
There is no friction between block B and the ground. And friction between blocks B and C can be considered the internal forces. So I considered the blocks B and C as a system and then applied the Newton's second law to them. Is that a problem ?
Yes, because you are told the two blocks move relative to each other, so do not have the same acceleration. Your equation would be ok if the acceleration in it referred to their common mass centre.

But can't we take both blocks B and C as a system ?. So there is just one external force in the horizontal direction on this system, which is $T_1$.

haruspex
Homework Helper
Gold Member
But can't we take both blocks B and C as a system ?. So there is just one external force in the horizontal direction on this system, which is $T_1$.
Yes, but as I wrote that would mean the acceleration in question is that of the common mass centre, not of either of the individual blocks.

Oh, ok I got what you are saying. So let me set up separate equations for blocks B and C. For block B, I would have
$$T_1 - f = 2Ma$$
and equation for block C would be
$$f = 3M a_2$$
and we also get the third equation which is from the torque equation.
$$T_1 - T_2 = 1.5 Ma$$
so with the help of above equations, we can now say that
$$T_1 = 2Ma + 3Ma_2$$
$$T_2 = Mg - Ma$$
and after plugging this in the torque equation, we get
$$(2Ma + 3Ma_2) - (Mg-Ma) = 1.5Ma$$
This simplifies to
$$a_2 = \frac{g}{3} - \frac{a}{2}$$
Now, if $\mu_k$ is the cofficient of kinetic friction between blocks B and C, then we have $f = \mu_k (3Mg)$. So we have the equation
$$f = \mu_k (3Mg) = 3M a_2$$
after plugging for $a_2$ , we get
$$\mu_k = \frac{1}{3} - \frac{a}{2g}$$
And, we are given in the problem that $a = 2\;m/s^2$, so after plugging for that, we get $\mu_k = 0.23$. Does this sound right ? I think something went wrong since we get that $a_2 = 2.27\; m/s^2$. This means that block C's acceleration in ground frame is greater than the acceleration of block B. This can not happen since block C is sliding to the left relative to the block B.

haruspex
Homework Helper
Gold Member
the third equation which is from the torque equation.
Check the signs on that one.

• issacnewton
Ah, yes, so we have
$$T_1 -T_2 = -1.5 Ma$$
Then the equation for $a_2$ becomes
$$a_2 = \frac{g}{3} - \frac{3a}{2}$$
This will give us $a_2 = 0.27 \; m/s^2$, which now make sense. So coefficient of kinetic friction would now become
$$\mu_k = \frac{1}{3} - \frac{3a}{2g}$$
So, with $a= 2\; m/s^2$, we get $\mu_k = 0.027$, which matches with the final answer given.
Thanks haruspex. These torques always cause me confusion about the signs.