Friction with two stacked blocks and friction

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Homework Statement



Here is the problem:

Block A in the figure weighs 1.05 , and block B weighs 3.90 . The coefficient of kinetic friction between all surfaces is 0.32. Find the magnitude of the horizontal force necessary to drag block B to the left at constant speed if and are connected by a light, flexible cord passing around a fixed, frictionless pulley.See the attachment for a picture.

https://www.physicsforums.com/attachment.php?attachmentid=44607&stc=1&d=1330645523


Homework Equations



ƩFx=Max, Friction=uk*n

The Attempt at a Solution



My FBD for the upper block has N going up MG going down Fk going left and T going right (positive)

My FBD for lower has N going up MG going down (I included both weights) Tension and then both frictions going to the right, and the force required going to the left (which i made positive for this one)

I am not sure what i am missing in solving this. Help me see if I have the right set up please. For the bottem block I have:

Sum of forces in x = M*Ax

Pull - Tension - .32(4.95) -.32(1.05)= M*Ax

For the top block I have:

T-.32(1.05)=M*Ax

Am I correct for these equations, especially in the bottem block where I have subtracted both tensions, I am assuming this is correct because the table exerts friction but so would the top block as it is sliding across the top block.
 

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M*Ax is mass * accel in the x direction from Newton 2

Sum of forces in x = mass time accel in x
 
Can you possibly point me in the direction of what i am missing... i have 3 equations with 2 unknowns right now.
 
EEintraining said:
Can you possibly point me in the direction of what i am missing... i have 3 equations with 2 unknowns right now.

You meant two equation with three unknowns, didn't you?

What do you know about the tension along a single rope ?

Do you know the acceleration if the blocks if they move along a straight line with constant speed?

ehild
 
Ehild thanks for your help yeah I did mean 2 eq and 3 unknown. After looking at this question that was exactly what I was was missing I must have read it too quickly constant speed tells me exactly what the acceleration is.