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

Force Problem with coefficient of static friction

  1. Apr 3, 2008 #1
    [SOLVED] Force Problem

    A 4.0kg block is put on top of a 5.0 kg block. The coefficient of static friction between the two blocks is u = 0.23 . The assembly of blocks is placed on a horizontal, frictionless table. What is the magnitude of the maximum horizontal force F that can be applied to the lower block so that the blocks will move together without any slippage?

    The only relevant equations I can think of are
    F = ma and
    F = uN

    I assumed that if the blocks are moving together than the fictional force between the blocks is greater than or equal to the forward force on the bottom block. I got that the maximum static frictional force is F = (.23)(4.0 * 9.8) = 9.016N but I don't know where to go from there, any ideas? Thanks
  2. jcsd
  3. Apr 3, 2008 #2

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    That's a good start. You also have to take acceleration into account. The acceleration is clearly nonzero, as the fricitonal force is the only force acting on the top block. So you need to use Newton's second law to write down 2 equations, one for the horizontal motion of each block. Use the fact that the two blocks move together to conclude that they have a common acceleration.
  4. Apr 3, 2008 #3
    (gone - my apologies)
    Last edited: Apr 3, 2008
  5. Apr 3, 2008 #4
    Ok so the acceleration of both blocks is equal. Also the only force acting on the top block is the static frictional force. Therefore:

    The force of the bigger block is

    F = umg = ma

    F = 9.016 = ma

    a = 2.254m/s2

    Then since the acceleration is equal I plug the acceleration into the bottom equation of F = ma.

    F = (4 + 5) * 2.254 = 20.286N.

    The answer key says that it is 20.25N though...any ideas where I went wrong?
  6. Apr 3, 2008 #5

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Mr. Vodka,

    1. Your solution is an absolute mess. You're using 2 different symbols to represent the same acceleration, and you're using the same symbol (F) to represent 2 different forces.

    2. We forbid homework helpers from posting complete solutions around here anyway. It's up to the student to do that.

    Please keep that in mind when posting in the future.


    That's wrong. There are two forces acting on the bigger block in the horizontal direction. One is friction, and the other is the applied force.
  7. Apr 3, 2008 #6
    Ok now I think I'm getting a little closer and yet still further away at the same time.

    So the equation for the top block is:

    Fs = Frictional Force
    F = applied force
    m1 = mass of top block
    m2 = mass of bottom block

    Fs - F = ma

    while the equation for the bottom block is:

    F = ma

    then I substituted in the variables and solved for acceleration in each one.

    a = (Fs - F) / m1

    a = F / (m1 + m2)

    then set them equal to each other and cross multiply

    (m1 + m2)(Fs - F) = (m1)(F)

    (4 + 5)(9.016 - F) = 4F

    81.144 - 9F = 4F

    13F = 81.144

    F = 6.24

    but I'm still wrong....
  8. Apr 3, 2008 #7

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This is wrong. It says that the acceleration points in the opposite direction of the applied force F. Does that make sense to you?
  9. Apr 3, 2008 #8
    Shouldn't both blocks be accelerating in the positive x axis?

    Or are you saying that the equation is:

    F - Fs = ma ??

    shouldn't the frictional force be positive since it is pointing towards the acceleration, similiar to how a car works?
  10. Apr 3, 2008 #9
    I get 20.286N as well Nanuven.

    if you reduce the precision on a = 2.254m/s^2 to a = 2.25m/s^2 see what happens.
  11. Apr 3, 2008 #10

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes, they should.


    No, certainly not. The frictional force on on the top block points in the direction of acceleration. It points in the opposite direction for the bottom block.

    Think about it for a minute. Imagine you've got the bottom block all by itself, and you pull on it. Now imagine that you put the smaller block on top, and pull it again. Does it make sense to you that the frictional force exerted on the bottom block by the top block is actually going to help you?
  12. Apr 3, 2008 #11
    Ok, I got it. Thanks everyone!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook