How Do You Solve Tension and Acceleration in a Two-Pulley System?

[astronomerc]

Homework Statement

Two blocks are connected by a rope that passes around two pulleys as indicated in the figure. (Attached)

a. Determine the tension in the rope.
b. Determine the acceleration of each block (Hint: The two blocks do not have the same acceleration).

The Attempt at a Solution

I call m1 the left mass and m2 the right.
I think the Tension is the same for the whole rope.
My m1 free body has normal up (+y) weight down, kinetic friction left, Tension right (+x)
My m2 free body is 2T up (-y) and weight down.
I think a1=2a2

For m1x Fx=max
T-f=m1a
T=2ma2+um1g

For m2:
m2g-2T=m2a2
T=(m2g-m2a2)/2

I put my two equations together to eliminate T
4m1a2+2um1g=m2g-m2a2

I solved for a2
a2=(m2g-2um1g)/(4m1+m2)

I plugged what I had for a2 into the original to solve for u:

I ended up with m2g-2um1g-m2g=2um1g
everything cancels to -u=u

Can someone please point out my flaw, be it in my algebra or my logic?

 

[Doc Al]

I solved for a2
a2=(m2g-2um1g)/(4m1+m2)

Everything up to here looks fine.

I plugged what I had for a2 into the original to solve for u:

I ended up with m2g-2um1g-m2g=2um1g

But I don't understand what you are trying to do here. I presume u is given, just like the two masses.

You found a2, now find a1 and T.

 

[astronomerc]

u is not given. The problem in it's entirety is posted... I am struggling to find u, but think I can do the rest from there.

I thought about assuming u was zero, but that doesn't seem right either.

 

[Doc Al]

u is not given. The problem in it's entirety is posted... I am struggling to find u, but think I can do the rest from there.

I thought about assuming u was zero, but that doesn't seem right either.

It sounds like you are just assuming that there's friction. I see no mention of it in the problem statement that you gave.

In general, unless the problem explicitly mentions friction, assume that all surfaces are frictionless. (If not, there's not enough information to solve the problem.)