AP Physics Free Response: Friction between two blocks

[be547]

Homework Statement

Block A of mass 2M hangs from a cord that passes over a pulley and is connected to block B of mass 3M that is free to move on a frictionless horizontal surface. The pulley is a disk with frictionless bearings, having a radius R and moment of inertia 3MR^2. Block C if 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 t each other

vertical tension in rope on right of pulley = 2Mg-2Ma

horizontal tension on rope on left of pulley = 2Mg-5Ma

Q: if a (linear accel) = 2m/s^2 determine the coefficient of kinetic friction between blocks B and C

Homework Equations

static friction is less than or equal to coefficient of static multiplied by Fnormal

kinetic F = kinetic coefficient multiplied by Fnormal

The Attempt at a Solution

i played around with a number of things but really have no clue what I am doing.
i wrote down the idea that the horizontal tension which was found to be 2mg-5ma must be more than the maximum static friction. then I isolated blockc (4M) and found the normal force to be 4MG ( i think) but i cannot figure out where the kinetic frictional force is acting since the both blocks are moving. AP exam on monday. its killing me that i can't get this. PLEASE HELP!

 

[Doc Al]

The kinetic friction acts between blocks B and C.

Hint: Identify the forces on block B, the pulley, and block A. Write separate force (or torque) equations for each body, using Newton's 2nd law. You'll get three equations, which you can solve together to find the coefficient of friction.

 

[be547]

ok. so i must be missing something fundamental. yesterday i finally found an answer of .1 for the coefficient of kinetic f. but now, isolating block b with friction pulling to the left and horizontal tension as a force on the right i am getting some odd answers that cannot fit a coefficient. ie 4, -2/3

what i did yesterday was i used nsl on block b
T.hor-frict.kin= 3Ma
T.hor = 2Mg-5Ma (from nsl for rotation)
so...
2Mg-5Ma-mk4Mg= 3Ma
cross out Ms
2g-5a-3a=mk4g
sub in g=10 and a=2
20-10-6=mk40
divide by 40
mk=0.1

this seems to be a reasonable answer but i don't understand (if it is right) why it worked with the normal force from block c.
is this even correct?

 

[Doc Al]

ok. so i must be missing something fundamental. yesterday i finally found an answer of .1 for the coefficient of kinetic f. but now, isolating block b with friction pulling to the left and horizontal tension as a force on the right i am getting some odd answers that cannot fit a coefficient. ie 4, -2/3

If you want to understand what went wrong, post your work for that solution.

what i did yesterday was i used nsl on block b
T.hor-frict.kin= 3Ma

OK. That's for block B.

T.hor = 2Mg-5Ma (from nsl for rotation)

This is the combined equations for block A and the pulley.

so...
2Mg-5Ma-mk4Mg= 3Ma
cross out Ms
2g-5a-3a=mk4g
sub in g=10 and a=2
20-10-6=mk40
divide by 40
mk=0.1

Sounds good.

this seems to be a reasonable answer but i don't understand (if it is right) why it worked with the normal force from block c.

Not sure I understand your question. The friction between C and B equals μN. N is just the weight of block C.