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Speed humps

  1. Mar 17, 2005 #1
    Wonder if anyone can help?

    I need to find out the maximum force experienced by a suspension system when it is travelling over a triangular speed bump.

    I am doing a set of experiments on a suspension test rig and I would like to know the extremes of suspension forces.

    The inital speed is 40mph or 17m/s. And the angle of the speed bump is 30 degrees. It is 30cm wide.

    I need to find out how much it decelerates when it hits the speed bump so I can work out the force. I tried using the equation of motion s= ut+1/2at2 but could not figure out values for v and t.

    Can anyone help me out. I am going ever so slightly mad........

    Thank you in advance.....
     
  2. jcsd
  3. Mar 17, 2005 #2
    Your equation is totally unapplicable, it only makes sense in cases with constant acceleration.

    An equation that will be useful is Hooke's law, which says that a spring force is proportional to the displacement of the string from equilibrium:

    [tex]\vec{F}=-k\vec{x} [/tex]

    (negative sign because displacement in corresponds to force going out, etc)

    So what you do, is find the spring constant (say 4000 Newtons / meter) and figure out how far the spring will be compressed.

    The way I would do the analysis is like this, the work done by the spring is:

    [tex]W = \int\vec{F(x)} \cdot \vec{dx} = \frac{k x^2}{2} [/tex]

    And the spring compresses untill its weight (the torque produced by its weight) + the upward force of acceleration over the hump = spring force max.

    So now that you know k, and x, compute the work W. For the purposes of this discussion, that W is the change in the car's kinetic energy.
     
  4. Mar 21, 2005 #3
    spring constant

    Thanks for your reply Crosson. Thanks for pointing out my school boy error.
    I was trying to calcualte the spring constant on the system. The problem is a I don't know the springs extension when it is in equilibrium.
    If I use F = kx the value for F and displacment will be based on the compressed value for the spring.
    In order to find out the uncompressed value for the spring I would have to take the spring off the car.
    Is there any other way? :confused:
     
  5. Mar 21, 2005 #4

    russ_watters

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    Staff: Mentor

    Stand on the back bumper and measure the displacement of the shocks.

    Using your weight and the weight of the car, you now have two points from which to construct a spring displacement equation.
     
  6. Mar 25, 2005 #5
    I gave that a go and I have got a figure for the spring constant. Can anyone tell me what the 'd' stands for in the above equation. I apologise for my stupidity. I was always more artistically inclined.....
     
  7. Mar 25, 2005 #6

    brewnog

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    Science Advisor
    Gold Member

    I'm surprised you questioned the 'd' and not the big s shaped squiggle! The 'd' tells you what to integrate with respect to, perhaps look up integration in a basic calculus book to get an idea, or hang around...
     
  8. Mar 30, 2005 #7
    integration

    I am struggling with the integration equation above.......

    I have looked at a couple of books on calculus but im afraid im lost!

    W = S F(x) . d(x) = kx2
    2

    The k value i have is 3250N/m and the x value is 0.07m

    If anyone could help me I would be greatly appreciative!
     
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