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Homework Help: Torque problem

  1. Jun 28, 2010 #1
    1. The problem statement, all variables and given/known data
    The following camping van with a total mass of 800 kg is connected at point P to the back of a car. The car accelerates with constant acceleration a during 4 seconds from rest until it achieves 60 km/h. Assuming there is no resistance (from air and ground), calculate the normal reaction in the wheels and the components of the contact force acting at point P.

    2. Relevant equations

    3. The attempt at a solution

    I'm really confused in this one, so any help to start would be nice.

    Attached Files:

  2. jcsd
  3. Jun 28, 2010 #2


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    Let's start simple then.

    So by Newton's 2nd law, how much force is the car producing and in what direction is it (use x and y for the horizontal and vertical directions)

    On the wheel, what direction is the normal reaction?

    At the contact force, there will be a reaction in two directions, what directions will those be ? (vertical and/or horizontal?)
  4. Jun 28, 2010 #3
    The force in the x direction that makes the van move:

    [tex]v = \frac{60 \times 10^3}{3600} m/s \approx 16.667 m/s [/tex]

    [tex]a = \frac{16.667}{4} = 4.16675 m/s^2 [/tex]

    [tex]F_x = 800 \times a [/tex]

    I suppose the normal reaction acts in the y direction and is located under the wheels. The F force acting at point P is made of two components Fx and Fy acting in x direction and y direction, respectively. Fx is the force previously calculated.

    The motion equation in the x direction is:
    [tex] F_x = m a [/tex] (already solved)

    and in the y direction:
    [tex] F_y + R_n - W = 0[/tex]

    Now the torque equation is missing. But I don't know where to put the axis of rotation and calculate the corresponding torques.
  5. Jun 28, 2010 #4


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    You can put it anywhere you want, remember if it is in equilibrium, the sum of moments about any point is zero.

    So you can take moments about the contact point , or the normal reaction of the wheels or the weight. Though I'd suggest about the contact point P.
  6. Jun 28, 2010 #5
    Something like this?

    [tex] 1.2 \times W - 1.2 \times R_n = 0 [/tex]

    [tex] 1.2 \times 800 \times 9.8 - 1.2 \times R_n = 0 [/tex]
  7. Jun 28, 2010 #6


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    Yes that should work out correctly.
  8. Jun 29, 2010 #7


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    You have three forces on the van, W FP and FC.

    The question asks for FP, but not for the whole of FC. so it's quickest to take moments about C

    that immediately gives you the direction of FP, doesn't it? :wink:
  9. Jun 29, 2010 #8
    Wow!!Take the axis at point P is very easy to calculate. Thanks .
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