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Homework Help: Work problem: Pulling wagon and finding angle

  1. Mar 17, 2009 #1
    1. The problem statement, all variables and given/known data
    In many neighborhoods, you might see parents pulling youngsters in a four-wheeled wagon. The child and the wagon have a combined mass of 50kg and the adult does 2.2 x 10^3 J of work pulling the two 60m at a constant speed. The coefficient of friction for the surfaces in contact is 0.26

    1) Determine the magnitude of the force applied by the parent

    2) Determine the angle at which the parent is applying this force.

    3. The attempt at a solution

    m = 50kg
    W = 2.2 x 10^3 J
    d = 60m
    mu(kinetic) = 0.26

    W = Fa cos(angle) d
    Fa = W / cos (angle) d

    Do I need to find the angle to solve this?


    Constant speed so no acceleration, force applied = kinetic friction
    Fk = (Fn)(mu kinetic)
    (Fn)(mu kinetic) = W / cos(angle)d
    No vertical movement so Fn = weight = ma
    cos(angle) = W / (m)(a)(mu kinetic)(d)

    cos(angle) = (2.2 x 10^3 N/m) / (50kg)(9.8N/kg)(0.26)(60m)
    (angle) = 73 degrees

    Go back to 1):

    Fa = W / cos (angle) d
    Fa = (2.2 x 10^3 N/m) / cos(73)(60m)
    Fa = 125.41N

    Is this correct? Or does that value of work already have the angle included in its calculation..?
  2. jcsd
  3. Mar 18, 2009 #2


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    No vertical movement so Fn = weight = ma
    This is not true. If F is the pulling force on the wagon, F*sin(theta) will act in the opposite direction to mg. So net Fn will be mg - Fsin(theta).
  4. Mar 28, 2009 #3
    Ok this is what I have so far:
    W = Fcos(angle)d
    Fa = W / cos(angle)d
    Constant speed so Fa = Fk
    Fk = W / cos(angle)d
    Fk = Fn(mu kinetic) so
    Fn(mu kinetic) = W / cos(angle)d
    cos(angle) = W / Fn(mu kinetic)
    But Fn = mg[sin(angle)]. Now I am stuck with the arithmetic: (assuming I did everything else right?)
    cos(angle) sin(angle) = W / mg (mu kinetic)
    Do they turn into tan somehow?
  5. Mar 29, 2009 #4


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    Fn = mg[sin(angle)].
    This is not correct.
    Fn = mg - Fa*sin(theta).
  6. Mar 29, 2009 #5
    Fn(mu kinetic) = W / cos(theta)d
    Fn = mg - Fa*sin(theta) so,
    [mg - Fa*sin(theta)] (mu kinetic) = W / cos*(theta) d
    [mg - Fa*sin(theta)][cos*(theta)] = W / (mu kinetic) d
    Not sure what to do now :/
    Can I factor our (theta)? Does Fa = ma?
  7. Mar 29, 2009 #6


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    Find Fa*cos(theta)
    Rewrite {[mg - Fa*sin(theta)][cos*(theta)] = W / (mu kinetic) d}
    as cos(theta) =.....
    Square both the sides and write [1-sin^2(theta)] =...
    Then solve the quadratic to find the sin(theta) and then Fa.
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