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Center of mass of a vehicle

by Altairs
Tags: mass, vehicle
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Altairs
#1
May28-08, 02:57 AM
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 ?
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mgb_phys
#2
May28-08, 03:02 AM
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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.
tiny-tim
#3
May28-08, 03:10 AM
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Quote Quote by Altairs View Post
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!

mgb_phys
#4
May28-08, 03:12 AM
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Center of mass of a vehicle

True - although if you tip the car up on two wheels you can calculate the height.
Altairs
#5
May28-08, 06:21 AM
P: 127
Quote Quote by mgb_phys View Post
True - although if you tip the car up on two wheels you can calculate the height.
How would I calculate the height ?
Altairs
#6
May28-08, 06:25 AM
P: 127
Quote Quote by mgb_phys View Post
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.
tiny-tim
#7
May28-08, 06:35 AM
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Quote Quote by Altairs View Post
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!
Altairs
#8
May28-08, 06:43 AM
P: 127
Right, and what about the tipping part ?
tiny-tim
#9
May28-08, 06:49 AM
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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.
Altairs
#10
May28-08, 06:49 AM
P: 127
Just one useful hint for the tipping part.
Altairs
#11
May28-08, 06:52 AM
P: 127
Got it. Thanks.
Altairs
#12
May28-08, 08:48 AM
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.
tiny-tim
#13
May28-08, 10:24 AM
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Hi Altairs !

hmm 45 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).
Altairs
#14
May28-08, 12:09 PM
P: 127
Quote Quote by tiny-tim View Post
(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.
tiny-tim
#15
May28-08, 02:03 PM
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Quote Quote by Altairs View Post
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 45 say, R1 + R2 + R3 + R4 = ?

(and you can get x on the flat, so then you only need y)
Altairs
#16
May29-08, 09:03 AM
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 ?
tiny-tim
#17
May29-08, 12:09 PM
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Quote Quote by Altairs View Post
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 ?
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?
Altairs
#18
May29-08, 01:52 PM
P: 127
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.


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