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Homework Help: Two blocks held against each other

  1. Oct 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Two blocks m (16kg) and M (88kg) are as shown in the figure. If the co-efficient of friction b/w the blocks is 0.38 but the surface beneath the block is frictionless, what is the minimum force required to hold m against M?

    How would you go about solving this kind of a problem? which concept am i required to use?

    2. Relevant equations

    F(friction)= c.o.f * F(normal)

    3. The attempt at a solution

    So i tried solving it (its too lengthy to write exactly what i did) but i ended up with a final answer of F = 2.68e3 N. Can someone double check that? thanks

    Attached Files:

    Last edited: Oct 9, 2009
  2. jcsd
  3. Oct 9, 2009 #2


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    Homework Helper

    The solution need not be lengthy.
    If F is the applied force, what is the common acceleration of M and m?
    What is normal reaction on m by M?
    What is the frictional force between m and M? What is its direction?
    Which force pulls m down ward direction?
  4. Oct 12, 2009 #3
    How are you supposed to get the system's acceleration though?

    Fa = Force Applied
    Ff = Force Friction
    Fg = Force Gravity
    Fn = Force Normal

    [tex]\Sigma[/tex]Fx = MA

    Fa = 104a
    a = Fa/104?

    Doesn't really help unless I'm missing something.

    Force causing the friction (gravity) is:

    Fg = 16(9.81)
    Fg = 157N

    Therefore, the block has 157 Newton's being pulled down.

    [tex]\Sigma[/tex]Fy = MA

    The block is held still, therefore, no acceleration
    [tex]\Sigma[/tex]Fy = 0
    Ff - Fg = 0

    [tex]\mu[/tex]Fn - 157 = 0
    Fn = 157/(.38)
    Fn = 413N

    This would be, in theory, the force that the big 88 Kg block exerts on the small block... But it still doesn't help me find the dang acceleration

    Edit: Oh nevermind... ha ha... Yeah it does. I can then use the system of the 2nd block to find the acceleration of the full system.
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