# Center of mass of a vehicle

by Altairs
Tags: mass, vehicle
 P: 127 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 ?
 Sci Advisor HW Helper P: 8,953 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.
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P: 26,148
 Quote by Altairs 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 ?
Hi Altairs!

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

 Sci Advisor HW Helper P: 8,953 Center of mass of a vehicle True - although if you tip the car up on two wheels you can calculate the height.
P: 127
 Quote by mgb_phys True - although if you tip the car up on two wheels you can calculate the height.
How would I calculate the height ?
P: 127
 Quote by mgb_phys 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.
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.
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P: 26,148
 Quote by Altairs … 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.
Hint: don't sum … divide and conquer!
 P: 127 Right, and what about the tipping part ?
 Sci Advisor HW Helper Thanks P: 26,148 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.
 P: 127 Just one useful hint for the tipping part.
 P: 127 Got it. Thanks.
 P: 127 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.
 Sci Advisor HW Helper Thanks P: 26,148 Hi Altairs ! hmm … 45º seems a bit dangerous … wouldn't it be better to use a lesser angle? Yes, your method seems fine. 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).
P: 127
 Quote by tiny-tim (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).
(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.
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P: 26,148
 Quote by Altairs But I don't see how can I tip the vehicle once and get x and y solved.
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 = … ?

(and you can get x on the flat, so then you only need y)
 P: 127 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 ?