Solving for Acceleration with Tension and Friction

Click For Summary
SUMMARY

The discussion focuses on calculating the acceleration of a 2.0 kg block in a system involving tension and friction. The coefficient of kinetic friction (\mu_{k}) is 0.12, and the forces acting on the block include tensions from two other blocks and gravitational forces. The initial miscalculation stemmed from assuming the tensions equaled the weights of the hanging blocks without accounting for acceleration. After drawing free-body diagrams for each block and setting up the correct equations of motion, the correct acceleration was determined to be 2.8 m/s².

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of free-body diagrams
  • Familiarity with tension forces in pulley systems
  • Basic concepts of kinetic friction
NEXT STEPS
  • Study the derivation of equations of motion for systems with multiple blocks and pulleys
  • Learn about the effects of acceleration on tension in pulley systems
  • Explore advanced problems involving friction and tension in dynamics
  • Review the principles of free-body diagram analysis in physics
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of tension and friction in dynamic systems.

m3rc
Messages
2
Reaction score
0

Homework Statement



Consider the following diagram:

5970334255_f6653f74cb_m.jpg


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

Homework Equations



Net force = T[itex]_{2}[/itex] - (T[itex]_{1}[/itex] + [itex]\mu[/itex][itex]_{k}[/itex]F[itex]_{g}[/itex]), where T[itex]_{2}[/itex] is the tension from the 3 kg block, T[itex]_{1}[/itex] is the tension from the 1 kg block, and F[itex]_{g}[/itex] 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[itex]_{1}[/itex] + [itex]\mu[/itex][itex]_{k}[/itex]*F[itex]_{g}[/itex] pointing left (presumably the block will accelerate to the right, so the friction force points in the opposite direction), and T[itex]_{2}[/itex] pointing right.

T[itex]_{1}[/itex] = 1 kg * 9.8 m/s^2 = 9.8 N.

[itex]\mu[/itex][itex]_{k}[/itex]*F[itex]_{g}[/itex] = .12 * 2 * 9.8 m/s^2 = 2.4 N

T[itex]_{2}[/itex] = 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
 
Last edited:
Physics news on Phys.org
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.
 
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+[itex]\mu[/itex]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!
 

Similar threads

Replies
2
Views
966
  • · Replies 8 ·
Replies
8
Views
1K
Replies
6
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
15
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
16
Views
3K
Replies
46
Views
7K
  • · Replies 33 ·
2
Replies
33
Views
2K
Replies
6
Views
3K