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Newton's Laws / Friction

  1. Aug 14, 2009 #1
    1. The problem statement, all variables and given/known data

    A 4.1-kg box is pushed along a horizontal floor by a force of magnitude 21 N at an angle θ = 35° with the horizontal. If the coefficient of kinetic friction between the block and the floor is 0.20, calculate the acceleration of the box. Be careful when calculating the normal force.

    2. Relevant equations

    [tex]\Sigma[/tex]F = ma
    Ffr= [tex]\mu[/tex]k Fn
    Fn=mg

    3. The attempt at a solution

    Fcos35°- Ffr = ma
    21Ncos35° - [tex]\mu[/tex]k mg = ma
    21Ncos35° - 0.20 (4.1 kg) (9.8 m/s2) = (4.1 kg) a
    (17.2021N - 8.036N) / 4.1 kg = a
    2.235 m/s2 = a


    Is this correct or on the right track?
     

    Attached Files:

  2. jcsd
  3. Aug 14, 2009 #2

    PhanthomJay

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    You ignored the warning! The normal force is not the objects weight. Sum forces in the y direction to calculate it.
     
  4. Aug 14, 2009 #3
    Right. I thought that the Fn=mg was the tricky part. I never know when they equal to each other or not. Any tips?

    Second try:

    Fn - mg - F sin35° = 0
    Fn = mg + F sin35°
    Fn = (4.1kg)(9.80m/s2) + 21N sin35°
    Fn = 40.18N + 12.045N
    Fn = 52.22N

    F cos 35° - Ffr = ma
    F cos 35°- [tex]/mu[tex] k= ma
    21N cos 35° - (0.20)(52.22N) = (4.1kg) a
    (17.2021N -10.444N) / 4.1kg = a
    1.648 m/s2 =a
     
  5. Aug 14, 2009 #4

    PhanthomJay

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    I didn't check your math, but your method is now correct. Always draw free body diagrams of the object, and identify the forces acting on it. Break up those forces into their x and y components where necessary. Then apply Newton 1 or 2, as applicable, in the x and y directions, separately. This is what you have done. There is no net force in the y direction, since the block remains in contact with the table and does not accelerate vertically; in the x direction, there is a net force, and hence, an acceleration in that direction.
     
  6. Aug 14, 2009 #5
    Great! Thanks again.:smile:
     
  7. Aug 14, 2009 #6
    I can't see the attachment yet, but consider the following.
    If a force has a component pushing -up- on the mass, then the normal will be smaller. The force is supporting some of its weight, so the normal force doesn't have to be as strong in order to obtain equilibrium in that direction.
    If the force has a component pushing -down- on the mass, then the normal will be larger. The force pushes the mass harder into the floor, meaning the reaction force will be greater.

    Make sure your analysis makes sense in light of the above. I haven't seen the diagram, so I don't know if "35° with the horizontal" means it's being pushed up or down.
     
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