Solving for Acceleration with Tension and Friction

[m3rc]

Homework Statement

Consider the following diagram:

Assume the strings and pulleys have negligible masses and the coefficient of kinetic friction between the 2.0 kg block and the table (\mu_{k}) is .12. What is the acceleration of the 2.0 kg block?

Homework Equations

Net force = T_{2} - (T_{1} + \mu_{k}F_{g}), where T_{2} is the tension from the 3 kg block, T_{1} is the tension from the 1 kg block, and F_{g} is the force of gravity on the 2 kg block.

a= Net force/mass

The Attempt at a Solution

This seems like a really straightforward problem, but none of the multiple choice answers match my result so I'm confused as to what exactly I'm missing.

I drew a free-body diagram for the 2 kg block with the normal force pointing up, the force of gravity pointing down, T_{1} + \mu_{k}*F_{g} pointing left (presumably the block will accelerate to the right, so the friction force points in the opposite direction), and T_{2} pointing right.

T_{1} = 1 kg * 9.8 m/s^2 = 9.8 N.

\mu_{k}*F_{g} = .12 * 2 * 9.8 m/s^2 = 2.4 N

T_{2} = 3 kg * 9.8m/s^2 = 29.4 N.

Net force = 29.4 N - (9.8+2.4) = 17.2 N

a = 17.2 N/ 2 kg = 8.6 m/s^2.

All of the options for answers are below 4.6 m/s^2.

I'm not looking for the answer; I'd like to just find out conceptually what I'm missing so I can solve it for myself.

What am I missing?

Thanks in advance,
m3rc

 

[Mike Pemulis]

Hi, m3rc! Good question, and thanks for explaining your work so clearly.

Where you're going wrong is that you assume that the two hanging blocks exert a force equal to their weight on the middle block. That would be true if the system was stationary, but it is not true if the system is accelerating, as in this case. The clearest way to see this is to imagine all three blocks were falling freely -- in that case there would be no tension in either string.

What you need to do is draw a free-body diagram for each block, not just the middle one. This will give you three equations of motion, with three unknowns: the tension in each string, and the acceleration. (We know that the accelerations of the three blocks are all equal because they are tied together by the string). Solve the system of equations to obtain the acceleration and the two tension forces.

 

[m3rc]

That did it, thanks so much for your help, Mike!

I drew free-body diagrams for each of the blocks, which led to the equations:

1 kg: a = (T₁-9.8 N)/1 kg, where 9.8 N is the force of gravity on the block
2 kg: a = (T₂-T₁+\mu_k(19.8 N))/2 kg, where 19.8 N is the force of gravity on the block.
3 kg: a = (T₂ - 29.4 N)/3 kg, where 29.4 N is the force of gravity on the block.

I then solved for T₂ in terms of T₁ by setting the second equation equal to the third because the accelerations are equal. I plugged my answer back into the second equation and solved for T₁ by setting it equal to the first equation. I then plugged both answers back into the second equation to come up with the correct answer, 2.8 m/s^2.

Thanks again, Mike!