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Coefficient of friction/weight

  1. Oct 15, 2004 #1
    Can anyone help me with this problem? What equations should I use?

    A block weighing 5.3 N requires a force of
    3.2 N to push it along at constant velocity.
    What is the coefficient of friction for the

    A weight W is now placed on the block and
    6.6 N is needed to push them both at constant
    velocity. What is the weight W of the block?
    Answer in units of N.
  2. jcsd
  3. Oct 15, 2004 #2
    Fk=Ukn so Fk/n=Uk where Fk is the force and n is the normal force... draw a free-body diagram and do the equations to solve for it... Help any? Also, the crate is in equilibrium since it moves at constant velocity... Any angles involved?
  4. Oct 15, 2004 #3
    I'm sorry... that doesn't make any sense to me.
  5. Oct 17, 2004 #4
    Since you know the block is moving at constant velocity, the sum of all forces on it must be zero...otherwise it would be accelerating. So what are the forces in the horizontal direction? Wel,, there are only two: the one that you are applying to move it [tex]F_{push}[/tex] and the force of kinetic friction [tex]F_k[/tex]. You can then use the definition of kinetic friction: [tex] F_k = \mu_k N[/tex] here [tex]N[/tex] is the normal force on the box. Of course the normal force is simply the weight of the box in this case. Plug the stuff into [tex]\Sigma F = ma = 0[/tex] and solve for [tex]\mu_k[/tex]. For the second part of the problem you will have to use the value of [tex]\mu_k[/tex] that you just found.
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