Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Problem on Friction Coefficient

  1. Oct 10, 2004 #1
    A force F is applied on a block of mass m in such a way that it stays in contact with another block of mass M over a frictionless surface. What is the required force for the two blocks to stay in contact? The static coefficient friction between the two block is [tex]\mu_{s}[/tex].

    NOTE: mass m is not in contact with the ground, and M > m.

    My solution is

    Acceleration on mass m is

    Acceleration on mass M is

    Since, m > M, the reaction force on m is

    [tex]\mu_{s}m(\frac{F}{m}-\frac{F}{M}) \geq mg[/tex]

    So the required force is
    [tex]F \geq \frac{Mm}{\mu_{s}(M-m)}g[/tex]

    Is this correct?
    Last edited: Oct 10, 2004
  2. jcsd
  3. Oct 10, 2004 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Do a sanity check on your answer. What if m = M? Would the force be infinite? What if the coefficient of friction were zero? F is infinite again? :yuck:

    I assume the problem is to find the maximum force that you can push on the top mass without having it slide off the bottom one. What are the forces on the top mass? And on the two masses taken together? Apply Newton's 2nd law to each case.
  4. Oct 10, 2004 #3
    Thank you for your help Doc Al. :blushing:

    Okay, I think I might've got it.

    Second attempt.

    By Newton's 3rd law, there must be an equal opposite reaction force.

    Using, F = (M+m)a, a = F/(M+m).

    So the force on the top mass = mF/(M+m).

    Action force = Reaction force. Therefore,

    Reaction force = Force on the top mass + Force on the bottom mass.
    Force on the bottom mass = Reaction force on the top mass.

    F = mF/(M+m) + F(b)

    F(b) = F - mF/(M+m).

    [tex]\mu(F-\frac{mF}{M+m}) \geq mg[/tex]

    [tex]F \geq \frac{(M+m)m}{M\mu}g[/tex]

    How does it look?

    After thinking about it, I think if coefficient is 0, the force would be infinite, for the friction force would equal 0.
    Last edited: Oct 10, 2004
  5. Oct 10, 2004 #4

    Doc Al

    User Avatar

    Staff: Mentor

    I really don't know what you're talking about with all the action-reaction stuff. But you are getting closer. :smile:

    Try this:
    (1) Consider the top block. Apply Newton's 2nd law.
    (2) Consider the system of both blocks together. Apply Newton's 2nd law.

    Combine the equations to solve for the applied Force. (Remember, you are solving for the maximum force.)
  6. Oct 11, 2004 #5
    Okay, I've been thinking about the problem, and your hints. I still don't get it.

    I can't see any connection between the top mass and the whole system, and I don't see anything wrong with my second solution...

    This is what I meant by the reaction force:

    Since there is an applied force on the system, whomever is applying the force must experience an equal and opposite reaction force. The reaction force must come from the two blocks. So the reaction force produced by the bottom mass M must also be the reaction force on mass m.

    What's wrong with that?
  7. Oct 11, 2004 #6

    Doc Al

    User Avatar

    Staff: Mentor

    The connection is simple: both have the same acceleration. Part of your error is confusing the horizontal force between the two blocks (what you, for some reason, call the reaction force) with the normal force between the two blocks. (If you wish, I can go over it line by line. Let me know.)

    There are severals ways to solve this problem, all equivalent.

    True, if something pushes on the blocks, the blocks must push back equally. (But that doesn't help.) More usefully, the two blocks do exert equal and opposite forces on each other. And in the horizontal direction, the force they exert on each other is just the friction (which, at it's maximum, is [itex]\mu N = \mu mg[/itex]).

    So let's solve this problem using the suggestions I gave earlier:
    (1) Consider the top block. Apply Newton's 2nd law. This gives us:
    [tex]F - \mu mg = ma[/tex]
    (2) Consider the system of both blocks together. Apply Newton's 2nd law. This gives us:
    [tex]F = (M + m)a[/tex]

    These equations describe the horizontal forces, which is where the action is. (Note that I have assumed that the force F is applied horizontally.) I'll leave it to you to solve for the maximum force, F.
  8. Oct 11, 2004 #7
    Thank you very much! :smile:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook