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The Normal Force and Inclined Planes

  1. Jan 30, 2008 #1
    1. The problem statement, all variables and given/known data
    A woman pushes her 150kg motorcycle up a slope of 5 degrees with a constant speed of 2 m/s. She achieves this by exerting a force on the bike of 450N parallel to the slope. What is the magnitude of the frictional force acting on the bike?

    2. Relevant equations
    a= gsin(angle)

    3. The attempt at a solution

    a= (150 x 9.8) x sin5
    F= 150 x 128
    F= 19216.84

    but the answer is ment to be 322 N, im not realy sure where I am going wrong, any help would be greatly appriciated
  2. jcsd
  3. Jan 30, 2008 #2


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    The motorcycle is pushed at a constant speed up the slope. This means that the forces (components) acting along the direction of the slope cancel each other out.
  4. Jan 30, 2008 #3
    ΣFx = 450N – Ff – 150kgx9.8ms-1xSin5
    Ff = 321.8810582 = 322N

    Or you could use the formula:
    Ff = μFn

    but you arent given μ...
    Last edited: Jan 30, 2008
  5. Jan 31, 2008 #4


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    Spot on with your first way of solving for Ff. It seems that you might be a bit unsure about getting to the solution due to the -1 in front of the sin?

    We know that the sum of the forces along the slope need to be zero (no acceleration along the slope)

    [tex]\Sigma F_x = 0[/tex]
    [tex]0 = 450 - Ff - mg\sin(5^o)[/tex]
    [tex]Ff = 450 - mg\sin(5^o)[/tex]
    Last edited: Jan 31, 2008
  6. Jan 31, 2008 #5
    Ahh yes, sorry.. I made a mistake.

    While calculating the Fn, I mistaken the g (in Fn=mg) to be a velocity... what i was supposed to write was:

    150kg x 9.8ms^-2 x Sin5

    Sorry for the confusion.
    Last edited: Jan 31, 2008
  7. Jan 31, 2008 #6


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    Do you understand the solution? It seems you got it from somewhere else?
  8. Jan 31, 2008 #7
    Oh yes, I understand it... I study Engineering Technology and last year we studied all these types of questions.
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