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Dynamics - the physics of tobogganing and forces at an angle

  1. Jan 5, 2013 #1
    1. The problem statement, all variables and given/known data

    A child is tobogganing down a hillside. The child and the toboggan together have a mass of 50.0kg. The slope is at an angle of 30.0° to the horizontal.
    Assume that the positive y-direction is pointing in the direction of the normal force. Assume that the positive x-direction is down the incline.

    Find the acceleration of the child

    a) in the case where there is no friction

    b) if the coefficient of friction is 0.15

    3. The attempt at a solution

    a) Fn + FgI = mgcosθ
    = (50)(9.8)cos 30°
    = 424N
    Fnet = FgII = mgsinθ
    = (50)(9.8)sin 30°
    = 245 N (+ x-direction)
    Fnet = ma
    a = Fnet/m
    = 245/50
    = +4.9m/s/s
    Therefor the child's acceleration in the case where there is no friction is 4.9 m/s/s.

    b) FgI = 424 (- y-direction)
    FgII = 245 (+ x-direction)

    Fnet = FgII (+ x-direction) + Fk (- x-direction)

    Fn = FgI = 424N (down)

    Fk = μkFn
    = (0.15)(424)
    = 63.6N (- x-direction)

    Fnet = FgII + Fk
    = (+245)+(-63.6)
    = +181.4N
    Fnet =ma
    a = Fnet/m
    = 181.4/50
    = +3.63m/s/s

    Therefore the child's acceleration is +3.63 m/s/s if the coefficient of friction is 0.15.
     
  2. jcsd
  3. Jan 5, 2013 #2

    Dick

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    Science Advisor
    Homework Helper

    I think your answers are right. I'm having a hard time puzzling through your designations for the various forces though. Could you explain them?
     
  4. Jan 6, 2013 #3
    yeah I wasn't sure how to write them on the computer, basically I am following my book.
    Fg - force of gravity
    FgI - Fg(cosθ)
    FgII = Fg(sinθ)
    Fn - normal force
    Fnet - net force
    Fk - force of kinetic friction
    m - mass
    a - acceleration
    g - gravity
    μk - coefficient of kinetic friction

    please let me know if I missed any or if you still have any questions. Thanks
     
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