2 blocks being pulled across the floor

[mujadeo]

Homework Statement

A block of mass m1 = 2 kg and a block of mass m2 = 3 kg are tied together and are pulled from rest across the floor by a force of P = 22 N. The coefficient of friction of the blocks with the floor is µ = 0.26.
m1 is on left, pulling force on right

Homework Equations

What is the tension in the string between the blocks?

The Attempt at a Solution

I've already got the acceleration of the 2 blocks (first part of prob).
This is [22N/(2kg+5kg)] - [5*9.8*.26] = 1.85m/s/s = acceleration of whole system.

I'm stuck on the tension though...but here's what i did.
F=ma
F=2*1.85 = 3.7N

Then to get the opposing (frictional) force i went
F= µn
F= .26 (2*9.8) = 3.7 = 5.1N

Then i find the difference to get net force: 5.1 - 3.7 = 1.4N

this is wrong though? please help, and thanks!

 

[Doc Al]

I've already got the acceleration of the 2 blocks (first part of prob).
This is [22N/(2kg+5kg)] - [5*9.8*.26] = 1.85m/s/s = acceleration of whole system.

OK.

I'm stuck on the tension though...but here's what i did.
F=ma
F=2*1.85 = 3.7N

This is the net force on m1.

Then to get the opposing (frictional) force i went
F= µn
F= .26 (2*9.8) = 3.7 = 5.1N

OK.

Then i find the difference to get net force: 5.1 - 3.7 = 1.4N

That's not the net force!

What other force acts on m1?

Realize that F = ma is the net force. Identify (on a diagram) all individual forces acting on each block. The sum of those forces on each block must equal the net force on that block from F = ma.

 

[mujadeo]

all forces acting on m1 are weight, normal, tension and friction
If F=ma is the net force on m1, doesn't that = the pulling tension in the string?

 

[Doc Al]

all forces acting on m1 are weight, normal, tension and friction

Right. The weight and normal force cancel, but the others do not.

If F=ma is the net force on m1, doesn't that = the pulling tension in the string?

No. The tension is just one of the forces acting on the block. How can it equal the net force?