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A box of mass 8.0kg is being dragged along the ground by a force of 30N.

  1. May 14, 2008 #1
    [SOLVED] A box of mass 8.0kg is being dragged along the ground by a force of 30N.

    1. The problem statement, all variables and given/known data
    A box of mass [tex]8.0kg[/tex] is being dragged along the ground by a force of [tex]30N[/tex].
    (a) If the friction force is [tex]26N[/tex] what is the resulting acceleration?
    (b) If the frictional force is one-quarter of the normal reaction force, what is the acceleration?

    2. Relevant equations
    F = m.a

    3. The attempt at a solution

    (a) [tex]m = 8.0kg \ , \ F = 30N[/tex]

    [tex]F = 30 - 26 = 4N[/tex]

    [tex]F = m.a[/tex]

    [tex]a = \frac{F}{m} = \frac{4}{8} = 0.5ms^{-2}[/tex]

    (b) [tex]m = 8.0kg , \ , F = 30 \times \frac{3}{4} = 22.5N[/tex]

    [tex]a = \frac{F}{m} = \frac{22.5}{8.0} = 2.8ms^{-2}[/tex]

    Are any answers correct?
     
  2. jcsd
  3. May 14, 2008 #2

    Hootenanny

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    Correct :approve:
    Careful here, notice what the question says:
     
  4. May 14, 2008 #3
    Thnx Hootenanny,

    Normal Reaction Force [tex]= 4N[/tex]

    [tex]m = 8.0kg \ , \ F = 4 \times \frac{3}{4} = 3N[/tex]

    [tex]F = m.a[/tex]

    [tex]a = \frac{F}{m} = \frac{3}{8} = 0.4 ms^{-2}[/tex]

    I don't thinks its correct, can you please explain?

    By Normal reaction force, it means the reaction without the friction?
     
  5. May 14, 2008 #4

    Hootenanny

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    No, the normal reaction force is the force which acts on the box and is normal to the surface. It is the force which prevents the box from entering the surface.
     
  6. May 14, 2008 #5
    I understood what normal reaction force means, but tell now I don't know how to calculate it. Can you please show me an example or explain?

    Thanks...
     
  7. May 14, 2008 #6

    Hootenanny

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    Consider your example, what are all the forcing acting in the vertical plane?
     
  8. May 14, 2008 #7
    Forces acting on the vertical plane are 30N and 26N
     
  9. May 14, 2008 #8

    Hootenanny

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    No, they are the forces acting on the horizontal plane. What are the forces acting in the vertical plane, i.e. in the 'up and down' direction?
     
  10. May 14, 2008 #9
    [tex]F = m.a[/tex]

    [tex]m = 8.0kg \ , \ a = g(gravity) = 9.81ms^{-2}[/tex]

    [tex]F = 8.0 \times 9.81[/tex]

    [tex]F = 78.5N[/tex]

    So, the normal reaction force is [tex]78.5N[/tex] ?
     
  11. May 14, 2008 #10

    Hootenanny

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    Correct :approve:

    Personally I would leave it as 78.48 N to prevent rounding errors later on.
     
  12. May 14, 2008 #11
    Thnx Hootenanny, So answer (b) would be:

    [tex]F = m.a[/tex]

    [tex]F = 30 - \left(\frac{1}{4} \times 78.48\right) = 10.38N \ , \ m = 8.0kg[/tex]

    [tex]a = \frac{F}{m} = \frac{10.38}{8.0} = 1.3ms^{-2}[/tex]

    correct?
     
  13. May 14, 2008 #12

    Hootenanny

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    Correct indeed :approve:
     
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