# Determine the kinetic coefficient of friction

#### EP

Block A of mass 2m hangs from a cord that passes over a pulley and is connected to block B of mass 3m. Block B is free to move on a frictionless horizontal surface. The pulley is a disk with frictionless bearings, having radius R and moment of inertia 3MR^2. Block C of mass 4m is on top of block B. The surface between blocks B and C is not frictionless. shortly after the system is released from rest, block A moves with a downward acceleration a, and the two blocks on the table move relative to each other.

a.) In terms of the given quantities and g, find the tensions, T1 and T2 for each section of string.

For this I have the Sum of Fy= 2m(g-a)=T1 for Fx= T2= 7ma which the anwer given is different. The t1 is correct! But is this a typo or am I wrong?

b.) If a=2m/s^2, determine the kinetic coefficient of friction between blocks B and C.

I was thinking the force of friction would be equal to the tension in the x direction but when I work that out I dont get the right answer. Is that the right concept?

c.) What is the acceleration of block C at this time?

I dont have any work for this part yet but I'll put it on here just in case.

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#### andrevdh

Homework Helper
a) I agree with you on T1. But for T2 block B is moving wrt block C, therefore B will experience a kinetic frictional force, $f_K$ coming from C. C will experiences a similar force on it (action-reaction pair), but in the opposite direction. You need to clarify the direction of $f_K$ for yourself as far as B and C is concerned.

#### EP

Well the Fk would be going in the opposite direction of the system. The equation for the Fx would then be Ft2-Fk=Fx. But then theres no coeffecient in the answer given which is 2mg-5ma. I'm lost on what that frictional force is going to be equal too. I just cant seem to get that answer.

#### andrevdh

Homework Helper
EP said:
The equation for the Fx would then be Ft2-Fk=Fx.
That's right. What you have now is the resultant force on block C (block B has only Fk in the x direction and it is not accelerating at a since it is slipping).

#### EP

When you work it out do you get the answer given? I have Coeffecient of friction in my answer. I ended up getting Ft=7ma+u4mg which is not the answer given.

#### andrevdh

Homework Helper
EP said:
The equation for the Fx would then be Ft2-Fk=Fx
You have the the resultant force, Fx, for block B (sorry,I changed B and C inadvertently in my previous posts!) here. Which according to the symbols in the drawing is
$$T_2 - f\prime = F_x$$
now since this is the resultant force for B
$$F_x=m_B a$$
which will therefore give
$$T_2=3ma + f\prime$$
and since
$$T_1=2m(g-a)$$
one can get $f\prime$ in terms of m,g and a by setting $T_2=T_1$

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#### Feldoh

You have the the resultant force, Fx, for block B (sorry,I changed B and C inadvertently in my previous posts!) here. Which according to the symbols in the drawing is
$$T_2 - f\prime = F_x$$
now since this is the resultant force for B
$$F_x=m_B a$$
which will therefore give
$$T_2=3ma + f\prime$$
and since
$$T_1=2m(g-a)$$
one can get $f\prime$ in terms of m,g and a by setting $T_2=T_1$
Sorry to bring up an old thread but I have this same exact problem and stuck on part b. I don't think you can set T1 = T2 because the pulley isn't massless, wouldn't you need to torque equations to solve this?

#### stewartcs

Sorry to bring up an old thread but I have this same exact problem and stuck on part b. I don't think you can set T1 = T2 because the pulley isn't massless, wouldn't you need to torque equations to solve this?
As long as the string is assumed to be massless (or not accelerating), which is normally the case in text book problems, the tension throughout the string is the same.

#### Feldoh

So it doesn't matter that the pulley isn't massless?

#### stewartcs

So it doesn't matter that the pulley isn't massless?
It does matter if a pulley has mass or not, but I didn't see where the problem states that it has a mass. So it is assumed to be massless unless otherwise stated.

#### stewartcs

Well after re-reading the original post it appears that it does have a mass associated with it, so yes you have to consider the torque.

#### YellowTaxi

mass of the pulley adds more inertia to the whole system.
Inertia always slows the effects caused by the applied forces.
But all the forces here are the same whether the pulley has mass or none at all.
(hope that makes sense)

Remember that all the forces stay the same as if the pulley were weightless.
It's just the acceleration that's affected by the mass of the pulley

They've given you the moment of inertia which is the circular equivalent of mass
It's a hint, And they obviously want you to factor that in.

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