• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Center of mass of a vehicle

  • Thread starter Altairs
  • Start date
127
0
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 ?
 

mgb_phys

Science Advisor
Homework Helper
7,744
11
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

Science Advisor
Homework Helper
25,790
242
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! :smile:

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

mgb_phys

Science Advisor
Homework Helper
7,744
11
True - although if you tip the car up on two wheels you can calculate the height.
 
127
0
True - although if you tip the car up on two wheels you can calculate the height.
How would I calculate the height ?
 
127
0
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

Science Advisor
Homework Helper
25,790
242
… 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! :smile:
 
127
0
Right, and what about the tipping part ?
 

tiny-tim

Science Advisor
Homework Helper
25,790
242
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:
 
127
0
Just one useful hint for the tipping part.
 
127
0
Got it. Thanks.
 
127
0
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

Science Advisor
Homework Helper
25,790
242
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).
 
127
0
(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

Science Advisor
Homework Helper
25,790
242
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 = … ? :smile:

(and you can get x on the flat, so then you only need y)
 
127
0
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

Science Advisor
Homework Helper
25,790
242
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? :smile:
 
127
0
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.
 

tiny-tim

Science Advisor
Homework Helper
25,790
242
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! :smile:
 
127
0
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:

tiny-tim

Science Advisor
Homework Helper
25,790
242
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 ?
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. :smile:
 

tiny-tim

Science Advisor
Homework Helper
25,790
242
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. :smile:
 
127
0
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.
 
Last edited:

tiny-tim

Science Advisor
Homework Helper
25,790
242
Hi Altairs! :smile:
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! :wink:)
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. :confused:

Surely the weight on the platform is the reaction? :smile:
 

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top