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The jumping motorcycle

  1. May 6, 2009 #1
    Hey guys,

    New to the forum but full of so many questions, of which I assume will be a cakewalk for many of you. I'm no physics expert as you will soon discover.

    Looking foward to your input and assistance.

    The first question is how do I calculate the force associated with a falling mass? For example, if you take a 10 pound brick and drop it 10 feet on a scale, what would be the maximum reading that I would see on the scale? Wind resistence and such aside.

    The next question is similar but a little more complicated.

    If I know the angle and height of a jump, the speed of an appoaching motorcycle, and the total weight of the bike with rider, how would I calculate the amount of force that the bike/rider has as it returns to earth? In other words, I assume that the falling mass will hit with a force greater than the total weight of bike and rider, with respects to the curve that it takes as it flys through the air.

    Let's say I'm just trying to calculate how strong the landing ramp needs to be. Something like that. Putting the matter of the suspension and such aside.

    Hope that makes sense.

    Thoughts?
     
  2. jcsd
  3. May 12, 2009 #2
    The downward force on falling objects is always constant (if you put air resistance aside). The downward force would always be the mass of the object times the gravitational acceleration of the earth (roughly 9.8 m/s2). The curve with which the bike travels doesn't make any difference.
     
  4. May 12, 2009 #3
    This question is asked over & over again in many different guises. The question is not "how much force."

    Nerd is right, as far as he/she goes: mass times gravitational acceleration equals the force. But that just tells you the weight of the object (in your case, mcycle + rider). It doesnt tell you anything about the momentum (which depends on both mass and speed). And when you hit the ramp, it is the change in momentum that you are concerned with.

    What you want to know is the "impact load" on your landing-ramp. That question is probably better asked in the engineering section of the forum. The answer you're going to get there is, "well that depends..." Because, well, it depends on how long it takes for you and your bike to decelerate from your landing speed to the final speed. This time is a few fractions of a second, and the shorter it is, the more impact load your ramp has to withstand. The time it takes depends alot on the suspension - since a long springy suspension will stretch out the time and make for a softer landing. That's why you bike has springs to begin with. Eventually you will find that the only way to really know is to do some experiments. There may also be some empirical correlations (that's a fancy name for basing your estimate on other people's experiments, or thumb rules).

    Good luck.
     
  5. May 12, 2009 #4
    Ah...good stuff and thanks for the info.

    So...just to make sure that I have this correct, for as elementary as it may seem, if we have a ball that is 10 kilograms in weight (the mass) and we drop it any distance (assuming is can reach its maximum velocity) the force of the impact is 98 kgf?
     
  6. May 12, 2009 #5

    Thanks!

    So what if we simplify this by saying that we are rolling a bowling ball at a given speed, at a ramp that is 2 feet long and one feet high?

    Can we calculate how much force the ball will hit the ground (landing) with?

    And then what happen if the ramp is curved instead of being flat?
     
  7. May 13, 2009 #6
    The unit for force is Newton (N). But the 98N is just the weight on the ball (which is always there, whether it's bouncing, flying or rolling, etc). If a ball is lying still on a surface, the weight gets canceled out by the force that the surface exerts on the object. But if it bounces, the force exerted by the surface is bigger than the weight (the force is the reason it is accelerated upwards again), or equal to the weight if we assume it doesn't bounce and there is no suspension etc (so this statement doesn't make a lot of sense in the first place).


    It makes no sense to simplify it down any further and to work with forces here.
    I'd recommend to read gmax's post again. That's pretty much all there is to it. You might want to read up on momentum and Newton's laws of motion.
     
  8. May 13, 2009 #7
    The motorcycle + rider mass and weight is m and mg respectively. If the motorcycle + driver's downward velocity times mass is mv, and it hits a horizontal ramp, then the impulse (force times time) is integral [F*dt]= mv. Using 200 kg and 2 meters/sec for m and v, and dt = 0.1 seconds, then Fmax = about 4000 Newtons, or about 2 g's.
     
  9. May 14, 2009 #8
    OK, and using 200 kg and 2 meters/sec for m and v, and dt = 0.2 seconds, then Fmax = about 2000 Newtons; using the same and 0.05 seconds you get 8000 N, etc.

    That's why I said above "it depends." Unless you quantify the time to decelerate to zero, you don't have a number you can use to design your ramp.
     
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