Solving Friction Problem: Acceleration of Lower Block

[crazy_nuttie]

Homework Statement

In the diagram shown below, the lower block is acted on by a force, F, which has a magnitude of 64.1 N. The coefficient of kinetic friction between the lower block and the surface is 0.346. The coefficient of kinetic friction between the lower block and the upper block is also 0.346. What is the acceleration of the lower block, if the mass of the lower block is 4.04 kg and the mass of the upper block is 1.86 kg?

http://img266.imageshack.us/img266/7450/knpic0828.png

Homework Equations

ΣF y = ma y

ΣF x = ma x

The Attempt at a Solution

So i set up 2 freebody diagrams, one for each object and I made F net equations. For the top block:

ma = T - μk*mg (1)

The friction comes from the friction between the top block and the bottom block.

and for the bottom it would be:

ma = Fa - Fk - T (2)

In this case the friction comes from the friction between the bottom block and the ground.


So in the second equation I solved for T, subbed it into equation 1 and solved for a, however I am not getting the correct answer. Any ideas?

 

[Donaldos]

What happened to the friction between the two blocks in your second equation?

 

[crazy_nuttie]

How would I write the friction between the two blocks?

ma = Fa - Fk - T - Fk (?)

So would I get,

ma = F - μk* m1g - μk * m2g - T

 

[Donaldos]

Yes, I think that's correct.

$$m_2 a= F-(m_1+m_2)g\mu_k-T$$

 

[crazy_nuttie]

Hey I still can't get it with that formula.. Can someone help?

 

[Donaldos]

Did you find the expression for $$a$$?