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Frictional Force

  1. Oct 19, 2006 #1
    Question: A force of [tex]29.0 N[/tex] is applied to a [tex]3.00kg[/tex] block which is initially at rest on a rough horizontal surface. A [tex]2.00kg[/tex] block is initially at rest on the [tex]3.00kg[/tex] block. The coefficient of friction between the blocks is [tex]\mu_s=0.35[/tex] and the kinetic friction between the [tex]3.00kg[/tex] block and the table is [tex]\mu_k=0.47[/tex].
    a) Find the acceleration of the [tex]3kg[/tex] block.
    b) Find the acceleration of the [tex]2kg[/tex] block.
     
  2. jcsd
  3. Oct 19, 2006 #2
    I think, One thing you could do is that, find the force of friction and then find then thr net force, after that since you have the net force and the mass, you can find acceleration.
     
  4. Oct 20, 2006 #3
    Would it be something like this:
    Let the applied force of 29N be F.
    a) For the 3.0kg block, the frictional force between it and the table = [tex]\mu_k(3.0kg+2.0kg)g=0.470(5g)N[/tex]
    The frictional force between the blocks also act upon the 3.0kg block = [tex]0.38(2g)N[/tex]
    Writing, [tex]F_{net}[/tex] for the 3.0kg block, we have [tex]29N-0.38(2g)-0.470(5g)=3a[/tex]
    Solving for [tex]a[/tex], we have [tex]a=-0.49m/s^2[/tex]
    b) For the 2.0 kg block, the only force affecting its motion is the frictional force between it and the 3.0 kg block. That frictional force = [tex]0.38(2g)N[/tex]. So [tex]F_{net}=0.38(2g)N=2a[/tex], solving for [tex]a[/tex], we have [tex]a=3.7m/s^2[/tex]

    Please, correct me if I am wrong, it would help me greatly. Thanks!
     
  5. Oct 20, 2006 #4

    OlderDan

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    There are some errors here. Does it make sense to you that the 3kg block has negative acceleration? Does it make sense that the 2kg has a positive acceleration while the 3kg block has a negative aceleration?

    I'm not sure if the .35 you state in the problem or the 0.38 you use in the computation is the correct number. Be sure to check this. Either way, you cannot assume that the actual force of friction between the blocks is the maximum possible force. If the blocks do not slip, the force can be much less than the force you calculated.
     
  6. Oct 20, 2006 #5
    Alright thanks alot OlderDan for replying.

    I rethought of the problem and may someone please tell me if my logic now is correct.

    The static friction between the blocks is 0.35, the kinetic friction between the 3.00kg block and the surface is 0.47, we can find kinetic friction first.

    [tex]F_{kf}=\mu_k(m_1+m_2)g=0.47(5.00kg)(9.8m/s^2)=23.03N[/tex]
    So the horizontal applied force is [tex]29N-23.03N=5.97N[/tex]

    Now we calculate the static friction between the blocks, if it is greater than [tex]5.97N[/tex], then the [tex]2.00kg[/tex] block on top does not move.

    [tex]F_{sf}=\mu_s(m_2)g=0.35(2.00kg)(9.8m/s^2)=6.86N[/tex]. Because [tex]6.86N>5.97N[/tex], the [tex]2.00kg[/tex] block does not move.

    So we can treat the two blocks as one massive block with mass of [tex]5.00kg[/tex]. Then [tex]5.97N=(5.00kg)a[/tex] and [tex]a=1.194m/s^2[/tex]

    Part b) wouldn't the [tex]2.00kg[/tex] and the [tex]3.00kg[/tex] mass then have the same acceleration of [tex]1.194m/s^2[/tex]?
     
  7. Oct 20, 2006 #6

    OlderDan

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    This is the correct conclusion, but the comparison of forces that led you to it is not quite right. The 6.86N is the maximum possible force on the 2kg block, but the 5.97N force is the net force on the combined 3kg+2kg system assuming the 2kg block does not slide. What you should be comparing to verify that the 2kg block does not slide is the maximum possible force to the force required to give the 2kg block the acceleration you computed. The net force on the system could have been greater than 6.86N and the 2kg block still not slide.

    I know what you meant, but it is not correct to say the 2kg block does not move. What you mean is that it does not move relative to the 3kg block or does not slide. To an observer watching the blocks react to the applied force, the 2kg block will certainly move.
     
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