Center of mass of a vehicle

  1. 1. The problem statement, all variables and given/known data
    I have to find the center of mass of any vehicle.

    2. Relevant equations
    These are what I have to make.

    3. The attempt at a solution

    Here is what I have thought. I have two solutions :-

    1. To find the mass of all the major components and their centers and then equate with the total mass of the vehicle to find the center. My question. Do I need to find the reactions at the tyres ? I don't think so because after all their sum will be equal to the total weight.

    2. I was thinking that if I have all the four reactions at the tyres computed directly then is there any way I can find the COM ?
  2. jcsd
  3. mgb_phys

    mgb_phys 8,809
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    Yes, just think about it for a minute. Picture a flat board with four legs near the corners, now move a weight around on the board.
    What would the 'weight' on each leg be if the heavy weight was; in the centre, over one pair of wheels, behind a pair of wheels.
  4. tiny-tim

    tiny-tim 26,016
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    Hi Altairs! :smile:

    That will give you the horizontal position of the c.o.m., but not how high it is!
  5. mgb_phys

    mgb_phys 8,809
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    True - although if you tip the car up on two wheels you can calculate the height.
  6. How would I calculate the height ?
  7. But what's confusing me here is that if I take the reactions of the wheels into account and when I'll take their summation then won't they cancel out ? Because the sum of reactions will be equal to the weight of the car.
  8. tiny-tim

    tiny-tim 26,016
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    Hint: don't sum … divide and conquer! :smile:
  9. Right, and what about the tipping part ?
  10. tiny-tim

    tiny-tim 26,016
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    You still divide, and the ratio is still the ratio between the horizontal distances to the c.o.m.

    But when the car is tilted, the c.o.m. gets nearer one pair of tyres (horizontally), and so the ratio of the horizontal distances changes. :smile:
  11. Just one useful hint for the tipping part.
  12. Got it. Thanks.
  13. Here is my rough solution. Please suggest improvements, errors and flaws. Especially if you can tell how to shorten this procedure.

    Everything is in x-y plane only.

    1.First by making FBD and by calculating the individual masses of the components I'll find the reactions at tyres (R1-4).

    2. I'll tip the car at front two tyres and will get an equation in x and y. Something like.

    W = Weight Of Vehicle
    X = Distance in x-axis from the tyres to the C.O.M (when car is horizontal).
    Y = Distance in y-axis from the tyres to the C.O.M (when car is horizontal).
    L = Distance between tyres 1 and 3 or tyres 2 and 4.
    Angle = 45.

    Taking moments about the front two tyres (Tyres 1 and 2).

    Wcos45 * X = Wsin45 * Y + (R3 + R4)cos45 * L

    3. Repeating the same for other two tyres while tipping the car on back tyres (Tyres 3 and 4) will give me another equation in x and y.

    4. Solving simultaneously should give the solution.
  14. tiny-tim

    tiny-tim 26,016
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    Hi Altairs ! :smile:

    hmm … 45º seems a bit dangerous … wouldn't it be better to use a lesser angle?

    Yes, your method seems fine. :smile:

    Though you haven't spelt out how you're going to find W.

    And I think you need to make measurements on the flat also (and you'll only need to tip the car on one pair of tyres).

    (How many tyres can you "weigh" separately and simultaneously? You seem to be planning on separate measurements for two tyres on the same axle, but not for all four, which would be easier, if possible).
  15. (above) Sorry, didn't quite get your point.

    45 was just an example.

    For W :-

    1. I'll sum the individual weights of the engine...etc.
    2. I can say that the car was weighed on that big weighing area where they weigh trucks etc.

    Yes, the four reactions will be calculated on flat. But I don't see how can I tip the vehicle once and get x and y solved.
  16. tiny-tim

    tiny-tim 26,016
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    Because once you know W, you get no extra information by weighing the car twice at the same angle.

    That's because, at 45º say, R1 + R2 + R3 + R4 = … ? :smile:

    (and you can get x on the flat, so then you only need y)
  17. Okay. Now I have the whole procedure. I realized that I can find x and z by keeping the car flat and y by tipping it.

    Now, the worst problem.

    How to find the Rs. I have to give a general procedure for any kind of vehicle. SO, calculating the individual weights of components and then finding the individual Rs isn't an option. Any suggestion ?
  18. tiny-tim

    tiny-tim 26,016
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    uh? I thought you were going to use a standard vehicle-weighing platform?

    They're like very big bathroom scales, aren't they?

    Isn't it obvious how you measure the reaction on one or two tyres? :smile:
  19. That is what I am going to use to weigh the whole vehicle.

    I dont want to use any assumption like equal reactions at all of the four tyres. However, taking equal reactions at the pair of tyres will make things a lot simpler.

    I think I need just one reaction. Others, if needed, can be calculated from that one.

    If I cannot (which I know I cannot ! ) measure the reaction at one tyre by simply putting that one over the scales than I don't know what to do.
  20. tiny-tim

    tiny-tim 26,016
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    Yes you can … if you put one or two tyres on the scale, and the others on the ground, then the scale will meaure the weight that the scale is supporting.

    By Newton's third law, that's equal and opposite to the (vertical component of the) reaction at that point! :smile:
  21. But the problem is that lets say at one time only 1/4 th of the car is on the weighing platform with the front wheels.

    Next time, half of the car is over the platform but still only the front two tires are over the platform. Won't the readings change ?
    Last edited: May 30, 2008
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