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Static/kinetic friction

  1. Aug 8, 2005 #1
    The coefficient of static friction is 0.60 between the two blocks in figure. The coefficient of kinetic friction between the lower block and the floor is 0.20. Force causes both blocks to cross a distance of 5.0 m, starting from rest.
    What is the least amount of time in which this motion can be completed without the top block sliding on the lower block?

    this is wht i think as long at the two blocks are moving together they have the same accleration.

    bottom box:
    x: Ff= ukFn= 0.20*7kg*9.81m/s/s=7*a

    top box:
    x: Ft-ff= Ft- usFn= 0.60*4*9.81m/s/s=4*a

    and then cant you just use the kinematics equation xf=xi+vt+.5at^2
    to find the time?

    Attached Files:

  2. jcsd
  3. Aug 8, 2005 #2
    Hmm--I prefer to keep the static friction coefficients as [itex] \mu _{S,1} [/itex] and [itex] \mu _{S,2} [/itex] until the end, where
    [tex] \left\{ \begin{gathered}
    \mu _{S,1} = 0.60 \hfill \\
    \mu _{S,2} = 0.20 \hfill \\
    \end{gathered} \right\} [/tex]
    and [tex] F_{net,1} [/tex] is the net force for the top block, [tex] F_{net,2} [/tex] is the net force for the bottom block. Here's my solution:
    [tex] \left\{ \begin{gathered}
    F_{net,1} = a \cdot 4.0kg = F_T - 39N\mu _{S,1} \hfill \\
    F_{net,2} = a \cdot 3.0kg = 39N\mu _{S,1} - 69N\mu _{S,2} \hfill \\
    \end{gathered} \right\} \Rightarrow [/tex]
    [tex] a = \frac{{F_T - 39N\mu _{S,1} }}{{4.0kg}} \Rightarrow [/tex]

    [tex] 3.0kg\frac{{F_T - 39N\mu _{S,1} }}{{4.0kg}} = 39N\mu _{S,1} - 69N\mu _{S,2} \Rightarrow [/tex]
    [tex] \frac{{3.0}}{{4.0}}\left( {F_T - 23N} \right) = 9.6N \Rightarrow F_T = 36N [/tex]
    [tex] \therefore a = \frac{{36N - 23N}}{{4.0kg}} = 3.3\frac{m}{{s^2 }} [/tex]

    *Because [itex] v_0 = 0 \frac{m}{s} [/itex], and I like to set [itex] x_0 = 0 m [/itex],
    [tex] \therefore v = \left( {3.3\frac{m}{{s^2 }}} \right)t \Rightarrow x = 5.0m = \left( {3.3\frac{m}{{s^2 }}} \right)\frac{{t^2 }}{2} \Rightarrow t = 1.7s [/tex]

    Is this the correct answer?
    Last edited: Aug 8, 2005
  4. Aug 8, 2005 #3
    thanks! ooo i see where i messsed up. on the bottom box for the mass, i made it 7 instead of just 3kg.

    thanks alot!
  5. Aug 8, 2005 #4
    No problem, welcome to PF:smile:
    Drawing the free-body diagrams can help tho*

    -When you draw it for the bottom mass,
    you can draw two weight vectors, and one large normal force opposing
    the stacked masses that relates to the friction of the bottom block against the floor...the second mass has its own free-body diagram/properties, its own 3.0kg, unique to the system (rather than just combining into 7.0kg)
    Last edited: Aug 9, 2005
  6. Aug 9, 2005 #5

    Doc Al

    User Avatar

    Staff: Mentor

    The easy way to solve for the acceleration is to just consider forces on the bottom block. No matter how hard the rope is pulled, the only force transmitted to the bottom block is that from the static friction between the blocks. Once that force reaches its maximum value ([itex]\mu_2 m_2 g[/itex]) the bottom block will have the maximum acceleration. You can compute the net force on the bottom block ([itex]F_{net} = \mu_2 m_2 g - \mu_1 (m_1 + m_2) g[/itex]) and thus calculate that maximum acceleration. The rest is kinematics.
  7. Aug 9, 2005 #6
    Hmm--I guess I need a better book than Barron's Review! (I never took Physics C, I'm studying from a review book)-->your method is much faster, I agree'''
    (So far I took CalcII in school, and have a bunch of Phys B knowledge w/o the math-->so I figured a review book would be faster, though not necessarily better than a real textbook!)
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