Friction and Tension

  • #1

Homework Statement

A block of mass M = 2m sits on a table and is attached to two other blocks, each having mass m, by strings passing over pulleys as shown (if you need picture let me know, one mass is hanging on the left and the other is hanging on the right). The coefficient of static friction between the large block and the table is 0.55 and the coefficient of kinetic friction between the large block and the table is 0.3. a) If M = 6 kg, what is the tension in the left string? b) What is the friction force on M? c) The string on the right is now cut. Describe in as much detail as you can the subsequent motion of all of the blocks. d) After the string is cut, what is the friction force on the large block?

Homework Equations

Fnet = ma
Fsmax = musN
Fk = mukN

The Attempt at a Solution

I believe a and b are connected by principle. Now I'm under the belief that static friction is the friction that must be overcome in order to cause movement, and that kinetic friction is the retained friction opposing the force during movement. Yet for part A since the two masses hanging over the pulley are in direct opposition of each other it seems to me that the friction (in this case static) would be offset in both directions and therefore zero. So the force in T of the left string is = Fg for the hanging mass. Then for part b the force of friction on mass M, right now is 0. Yet he gives us static friction so it seems to me that either it's just there (cause in part c and d we'd be using kinetic friction I believe) or I'm missing something in my thought pattern.

Any help would be greatly appreciated as always! Thank you!

Answers and Replies

  • #2
Well, for starters nothing is going to move before the string is cut!
So the tension in the string is just the force of gravity on m.
  • #3
Indeed friction is something that needs to be overcome, so its budget is given by the familiar μ times the normal weight.

As to your pulleys if they are horizontally located and opposing and equal then it should all be static.

After cutting the string, then the thing that needs determining is whether the unbalanced force is sufficient to overcome the static friction that would hold it at rest, and if it is, then you can figure the system acceleration based only on the kinetic μ, because it is less than the static and once in motion that's all you're dealing with.
  • #4
I'm a little confused with part c in determining if the big mass M will move against the static friction. Mainly since I'm not entirely sure how to determine if there is some increase in Ft that'll cause it to move. Can you make sure I'm doing my math right for it?

For M, Fsmax = .55(mg) = .55(60) = 33
We still have Ft = -Fgofm = 10(3) = 30 (*we're supposed to use g=10 for his class)

So I'd guess since Ft < Fsmax then there's no movement in M? Or am I forgetting something (or just doing it wrong entirely)?
  • #5
Correct! You have 30 N force driving the motion and 33 N holding it back, so no movement.

The wording suggests that movement is expected so you must look very, very carefully for any error in typing the question. In particular, is the hanging mass half the mass on the table? Is the coefficient of static friction greater than one half? These two things alone guarantee no movement.

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