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Find coefficient of friction

  1. Oct 2, 2004 #1
    The question is:

    "Two blocks are accelerated across a horizontal frictionless surface as shown. Frictional forces keep the two blocks from sliding relative to each other, and the two move with the same acceleration. What is the magnitude of the horizontal component (frictional force) of the force exerted on the small block by the large block?"

    The mass of the small block on top is: M ,the mass of the bottom block is: 2M , and the force, P is: 3 N.

    I drew the FBD's, block M has three forces: n1 (up), Mg (down), and f (right). Block 2M has 5 forces??? P (right), 2Mg (down), n1 (down), n2 (up), and f (left).

    I can't seem to get it, there appears to be too many unknowns to me.

    I appreciate any help. Thanks!
     
  2. jcsd
  3. Oct 2, 2004 #2
    No picture for us to see?

    [itex]\sum \vec F=m\vec a[/itex]

    You know the acceleration. Without a picture its hard to determin if your assessments are correct. The force of friction will act in the same direction as the motion of the lower block. You know the masses of the blocke. you can form two equation--one f=ma for each block and you have two unknowns. You should be able to get an answer.

    Of those 5 forces can't you relate them so that they are all in terms of M and f? Give it a try and see. Remember the top block's M is the same as the bottom block M--there's just two of them on the bottom though. The frictional force on the top block is the same as on the bottom block too.

    Anyway, hope this helped. Show what you've done if you need more help. From what I can see your assessment of the forces is correct but without a picture I can't be 100% sure.
     
  4. Oct 2, 2004 #3
    Sorry about no picture. It is a surface with two boxes stacked on each other with the smaller one on top. The force P pulls the bottom box to the right.

    These are my equations but then I don't know where to go.
    \sum Fy (M) = n1 - mg so n1=mg
    \sum Fx (M) = f1 = ma so \mu mg = ma

    \sum Fy (2M) = n2 - n1 - 2Mg = 0 so n2=3Mg
    \sum Fx (2M) = P - f2 = 2Ma so P-\mu(3Mg)=2Ma
     
  5. Oct 3, 2004 #4
    OHH!!! I think I finally got it!!!!

    I took both of the \sum Fx equations and solved them for ma then set them equal to each other and then solved for f.

    f=ma and (P-f)/2=ma

    f=(P-f)/2
    P=f/3

    I was confused by the question, I thought that it wanted me to solve for \mu, but it was really asking about f.

    Right?
     
  6. Oct 3, 2004 #5
    That last equation should be P = 3f so f = P/3.
     
  7. Oct 3, 2004 #6
    Oops, you are exactly right. The excitement from finally conquering the problem caused my fingers to malfunction!!! Thanks for the help!!!
     
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