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Force vs Time graph (Impulse)

  1. Jan 21, 2013 #1
    1. The problem
    I have trouble understanding slope and area on graphs. For a force as a function of time graph the area is equal to the impulse. Why is that? And what is the slope equal to? How do I know?


    2. Relevant equations
    p=mv
    fΔt=mΔv



    3. The attempt at a solution
    So the slope should be equal to f/t?
    What on earth is force divided by time equal to?
    And is the area underneath the graph always equal to y*x (in this case y=force, x=time)? So area = f*t = impulse?
     
  2. jcsd
  3. Jan 21, 2013 #2

    ehild

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    The impulse of a force is defined as I=FΔt for a force F acting for a very short time. And the change of momentum p=mv is equal to the impulse of the force. I=Δp=mvf-mvi.

    See the first picture: a constant force is plotted. Its impulse is FΔt= F(tf-ti), the area under the line F(t) between the initial and final times.

    The second picture shows a force which grows linearly with time. Its impulse is equal to the area under the line, which is the same as the average force multiplied by the elapsed time: I=FavΔt=Fav(tf-ti).
    The picture on the right shows a general force-time plot. Again, the impulse of the force between ti and tf is equal to the area under the curved line.

    If you know the impulse of a force imparted to a body, it is equal to the change of momentum. I=mvf-mvi.


    "f " means the average force during the time period Δt.

    The slope of the f(t) graph is irrelevant. It is not f/t, but the limit of Δf/Δt when Δt get shorter and shorter. (If you study Calculus you will learn that it is the time derivative of f(t).)

    The area is fΔt in case f is constant. Otherwise you slice the whole area and add the small pieces f(t)Δt together. You will learn that the area is equal to the integral of f(t) between ti and tf.

    ehild
     

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