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Homework Help: Forces again (review)

  1. Oct 24, 2007 #1
    A 25.5 kg block is connected to an empty 1.00 kg bucket by a cord running over a frictionless pulley (Fig. 4-57). The coefficient of static friction between the table and the block is 0.400 and the coefficient of kinetic friction between the table and the block is 0.320. Sand is gradually added to the bucket until the system just begins to move.

    I am having the most trouble finding out how to utilize the coefficient of static AND kinetic friction. Do I need to use both? I have my FBDs all drawn up, but I keep on ending up with 2 equations and 3 unknowns.

    Take to right as + and down as +...from the FBD of the Block I get [tex]\sum F=m_Ba[/tex] implies [tex]-Fr+T=m_Ba[/tex] where [tex]Fr=\mu*N[/tex] (but I have yet to decide which mu to use?? I would assume since there is an acceleration, I need to use mu of kinetic friction)

    [tex]N_B=m_Bg[/tex] B=Block. b=bucket

    From FBD of bucket I get [tex]-T+m_bg=m_ba[/tex]

    Find the mass of the sand added to the bucket:

    Find the acceleration of the system:

    A hint would be dope!
  2. jcsd
  3. Oct 24, 2007 #2
    i have a similar problem like this

    i was confused on both givens, since your last sentence states that it gradually moves ... use kinetic friction.
  4. Oct 24, 2007 #3
    Yeah....I keep coming up with too many unknowns...

    I get from FBD_B: [tex]-m_Bg*\mu_k+T=m_Ba[/tex]
    so [tex] T=m_B(a+g\mu)[/tex] and from FBD_b [tex]T=m_b(g-a)[/tex]

    So I have T, a, and the mass added to the bucket that are unknown...

    I was thinking of maybe solving for mu_k AND mu_static......That might provide me with a third equation. Since mu_static-mu_k=0.4-0.32
    That might work.....

    EDIT: this is an fing mess....any ideas?
    Last edited: Oct 24, 2007
  5. Oct 24, 2007 #4
    I am still jammed on this....do you see what I am missing?
  6. Oct 24, 2007 #5
    First calculate force of static friction (force required to make the block move):
    [tex]F_{f} = \mu_{s} m g = 0.4(25.5)(9.8) = 99.96 N[/tex]
    Now use this to find the mass of sand:
    [tex]F_{f} = F_{w} = m g[/tex] ==> [tex]m = F_{f}/g = 99.96/9.8 = 10.2 kg[/tex]
    but of course we must subtract the mass of the bucket, so the mass of sand is 10.2kg -1kg = 9.2 kg.
    Now to find the acceleration let's first calculate the force of kinetic friction:
    [tex]F_{f} = \mu_{k} m g = 0.32(25.5)(9.8) = 79.968 N[/tex]
    Thus the net force on the block once in motion is the force due to the weight of the bucket minus the force of kinetic frictions: F = 99.96 - 79.968 = 19.992 N. Using F = ma (or a = F/m) we have a = 19.992/25.5 = 0.784 m/s^2
    Last edited: Oct 24, 2007
  7. Oct 24, 2007 #6
    Clearly it has been a while since I have used these terms. So N*mu_s is what is required to move the system? and N*mu_k is what is needed to keep it moving. Ah. I vaguely remember a graph in which static friction increases until it maxes out and then it drops to a constant. Thia makes much more sense now.

  8. Oct 24, 2007 #7
    yep, you've got it now :)
  9. Oct 24, 2007 #8
    ah i have learned something tonight, thanks too!
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