How can I determine the center of mass of a vehicle?

[Altairs]

Homework Statement

I have to find the center of mass of any vehicle.

Homework Equations

These are what I have to make.

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 ?

 

[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.

 

[tiny-tim]

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]

True - although if you tip the car up on two wheels you can calculate the height.

 

[Altairs]

True - although if you tip the car up on two wheels you can calculate the height.

How would I calculate the height ?

 

[Altairs]

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]

… 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]

Right, and what about the tipping part ?

 

[tiny-tim]

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]

Just one useful hint for the tipping part.

 

[Altairs]

Got it. Thanks.

 

[Altairs]

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]

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).

 

[Altairs]

(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]

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)

 

[Altairs]

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]

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]

That is what I am going to use to weigh the whole vehicle.

I don't 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.

 

[tiny-tim]

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.

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!

 

[Altairs]

But the problem is that let's 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 ?

 

[tiny-tim]

But the problem is that let's 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 ?

Yes, that's the whole point … the reading only gives the weight supported by the tyres that are on the platform.

So by moving the car around you can find how the weight is distributed between the tyres.

 

[Altairs]

So by moving the car around you can find how the weight is distributed between the tyres.

I am lost.

 

[tiny-tim]

ok … suppose that when the car is on flat ground, two-thirds of the weight is over the front tyres, and one-third over the back tyres.

If you had two weighing-platforms, close together, you could find this by placing the car so that one pair was on one platform, and the other pair on the other.

Then one platform would show 600kg, say, and the other would show 300kg.

If you only have one platform, you can achieve the same just by putting the front tyres on the platform. Then it will show 600kg. Then turn the car round so that the rear tyres are on the platform. Then it will show 300kg. (Put only one tyre on, it will show half that.)

Finally, you tilt the car by raising the front tyres, and measure, say, 400kg for the rear tyres at an angle of 45º.

 

[Altairs]

If you only have one platform, you can achieve the same just by putting the front tyres on the platform. Then it will show 600kg. Then turn the car round so that the rear tyres are on the platform. Then it will show 300kg. (Put only one tyre on, it will show half that.)

Why turn it around when subtracting reading of front tyres from total can do it?

Finally, you tilt the car by raising the front tyres, and measure, say, 400kg for the rear tyres at an angle of 45º.

Why this ?

What I don't get is that this method will only give the ratios of weight distribution amongst the four tyres. What I requre are the 'absolute' values of reactions.

 

[tiny-tim]

Hi Altairs!

Why turn it around when subtracting reading of front tyres from total can do it?

ah … because then I can add the two weights to give me the total weight of the vehicle.

(I know you want to add the weights of all the components … but I think my way is simpler! )

What I don't get is that this method will only give the ratios of weight distribution amongst the four tyres. What I requre are the 'absolute' values of reactions.

I don't follow you.

Surely the weight on the platform is the reaction?

 

[Altairs]

ah … because then I can add the two weights to give me the total weight of the vehicle.

(I know you want to add the weights of all the components … but I think my way is simpler! )

I will weigh the whole car on the platform.

I don't follow you.

Surely the weight on the platform is the reaction?

But those reactions changes with the position of tyres over the platform. And consequently the COM calculated would change.

 

[tiny-tim]

But those reactions changes with the position of tyres over the platform. And consequently the COM calculated would change.

But the COM does change if the car is tilted, and the reactions change accordingly, and the weights on the platform equal those reactions.

 

[Altairs]

I wasn't talking about the tilting operation. The other one, where you said that simply putting the tyres over the platform would give the reactions. What I am saying is that, those reactions will change with the position of tyres over the platform and those reactions would therefor be dependent on the position of the tyres and wouldn't be 'absolute'.

 

[fluidistic]

I have to find the center of mass of any vehicle.

Just calculate the barycenter of the vehicle. Check out http://en.wikipedia.org/wiki/Barycenter.

 

[tiny-tim]

What I am saying is that, those reactions will change with the position of tyres over the platform and those reactions would therefor be dependent on the position of the tyres and wouldn't be 'absolute'.

Hi Altairs!

I'm wondering whether we're talking about the same thing.

What do you mean by "the position of tyres over the platform"?

I've never used one of these platforms, but I was assuming that it didn't matter which part of the platform the tyre was on.

Am I missing the point you're making?

 

[Altairs]

Okay. I'll explain what I perceived from your posts.

That you said that what I need to do in order to find reactions at the tyres is simply to put the front two tyres and get their reactions and repeat this for the back two tyres. My point here is that the reading on the scales wouldn't be the 'real' or 'absolute' reaction of the tyres as the reading would change with the amount of portion of car over the platform with even only two tyres over it. So, the readings would change if we move the car around.

 

[tiny-tim]

My point here is that the reading on the scales wouldn't be the 'real' or 'absolute' reaction of the tyres as the reading would change with the amount of portion of car over the platform with even only two tyres over it. So, the readings would change if we move the car around.

Hi Altairs!

The reading doesn't change with the amount of portion of car over the platform, it only depends on which tyres are on the platform (and on the angle, of course).

If we put the front two tyres on the edge of the platform, and the rest of the car over solid level ground, the weight shown will be exactly the same as if we move the car forward so that the rear two tyres are almost on the platform.

 

[Altairs]

Then all I need to do is to weigh the front two tyres and divide it by two (assuming that they support equal weight) and do the same for the back two tyres.

 

[tiny-tim]

Then all I need to do is to weigh the front two tyres and divide it by two (assuming that they support equal weight) and do the same for the back two tyres.

Yup!

(except why bother to divide by two … you only need the ratio … )

 

[Altairs]

When I'll take side view and front view than I'll only be considering the two tyres in view then won't I be assuming that the whole weight of the car is supported by those two tyres ?

 

[tiny-tim]

Hi Altairs!

When I'll take side view and front view than I'll only be considering the two tyres in view then won't I be assuming that the whole weight of the car is supported by those two tyres ?

(I assume that by "view" you mean "on the platform")

Nooo … the total weight of the car is still supported by all four tyres.

You are measuring how much is supported by those two (and the rest of the total weight of course is supported by the other two, and you take the ratio).

(oh … but why take "side view" at all, if the c.o.m. is in the middle?)

 

[Altairs]

No with side view I mean the real side view. With the doors in view. By taking moment along one tyre and as I would know the reaction and the total mass of the whole car won't I be able to find the centre of mass in one axis ?

Doing the same for the front view (the windscreen) one I can find the COM in another axis ?

Is this feasible ?

 

[tiny-tim]

No with side view I mean the real side view. With the doors in view. By taking moment along one tyre and as I would know the reaction and the total mass of the whole car won't I be able to find the centre of mass in one axis ?

Doing the same for the front view (the windscreen) one I can find the COM in another axis ?

Is this feasible ?

Yes, that's right … if you know the total weight, then you can find the x-coordinate of the c.o.m. by weighing the front tyres on the flat, the y-coordinate of the c.o.m. by weighing one set of side tyres on the flat, and the z-coordinate of the c.o.m. by weighing any pair of tyres on a slope.

 

[Altairs]

Right. Let's say that in the side view the tyres are T1 and T2. And in front view the tyres are T1 and T4. For calculating the coordinate in the side view I'll only need one of T1 and T2 and in the same way for calculating the coordinate in the front view I only need one of T1 and T4. I hope that I am right here.

 

[tiny-tim]

Right. Let's say that in the side view the tyres are T1 and T2. And in front view the tyres are T1 and T4. For calculating the coordinate in the side view I'll only need one of T1 and T2 and in the same way for calculating the coordinate in the front view I only need one of T1 and T4. I hope that I am right here.

I'm confused … are you saying that you only need to put one tyre on the platform each time?

i] You can't put one tyre on, and then just double the weight;

ii] it's physically much easier to put two tyres on anyway, isn't it?

 

[Altairs]

I will be weighing two tyres at once.

My confusion still exists. I'll explain.

If there are two tyres over the platform (let's say front two) then changing the position of the car over the platform in such a way that rear two tyres are still on the ground should change the reading on the meter because the more the portion of the car is over the platform the more the reading will be. Is this the case ? You have already said both yes and no in previous posts.

 

[Altairs]

If there is the whole car on the platform and we lift it up at an agle such that it is just enough to put the rear two tyres in air and then take the reading. Then repeat this by tipping it on the back tyres then will the two readings equal the weight ?

 

[tiny-tim]

If there are two tyres over the platform (let's say front two) then changing the position of the car over the platform in such a way that rear two tyres are still on the ground should change the reading on the meter because the more the portion of the car is over the platform the more the reading will be. Is this the case ? You have already said both yes and no in previous posts.

It's definitely no … the reading depends only on which tyres are on the platform (and the angle of the car), not on the "overlap".

I didn't intend to say "yes" … where do you think I did?

If there is the whole car on the platform and we lift it up at an agle such that it is just enough to put the rear two tyres in air and then take the reading. Then repeat this by tipping it on the back tyres then will the two readings equal the weight ?

Do you mean lifting the rear tyres just an inch off the ground, or do you mean "balancing the car on its front legs"?

 

[Altairs]

Do you mean lifting the rear tyres just an inch off the ground, or do you mean "balancing the car on its front legs"?

Just an inch. BTW what's the difference?

 

[Altairs]

It's definitely no … the reading depends only on which tyres are on the platform (and the angle of the car), not on the "overlap".

It means that while the front two tyres are over the platform the reading will remain constant and it will shoot to the maximum value (weight) only when the previous two tyres are also over the platform. And in this duration (i.e. just before the rear two tyre are on the platform and just after the front two tyres are over platform) the reading won't change ?

 

[tiny-tim]

Just an inch. BTW what's the difference?

Well, if you lift the rear tyres an inch off the ground, then that means that you are now supporting the same weight that the ground was supporting … so the platform reading (from the front tyres) will be the same.

But if the car is balanced (so that if you let it go, it would stay there), then all the car's weight is supported by the front tyres, and so the platform reading will show the whole weight.

 

[Altairs]

Here it is.

But the problem is that let's 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 ?

Yes you can … if you put one or two tyres on the scale, and the others on the ground, then the scale will measure 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!

 

[tiny-tim]

It means that while the front two tyres are over the platform the reading will remain constant and it will shoot to the maximum value (weight) only when the previous two tyres are also over the platform.

Yes, that's right … drive the car slowly forward,and the platform reading will stay the same right until the rear tyres reach the platform.

 

[Altairs]

Well, if you lift the rear tyres an inch off the ground, then that means that you are now supporting the same weight that the ground was supporting … so the platform reading (from the front tyres) will be the same.

But if the car is balanced (so that if you let it go, it would stay there), then all the car's weight is supported by the front tyres, and so the platform reading will show the whole weight.

The idea is that the whole car will be over the platform. Then the car will be lifted from it back by men or something so that they will be standing on the ground. My point was that there is no difference in this and simply putting front two tyres over and rear two tyres off the platform but my friends wouldn't agree. Second point is that will this procedure (the tipping one) when performed for both the front and the rear tyres equal the weight ?

 

[tiny-tim]

Here it is.

But the problem is that let's 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 ?

Ah … now I see …

No, when you said 1/4 th of the car, I thought you meant that only one tyre was on the platform