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

In summary: You can tip the car on one pair of tyres to get the x and y coordinates. :)In summary, Homework Equations state that in order to find the center of mass of a vehicle, you must find the individual masses of the major components, and then equate that with the total mass of the vehicle to find the center.
  • #36
Hi Altairs! :smile:
Altairs said:
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). :smile:

(oh … but why take "side view" at all, if the c.o.m. is in the middle?)
 
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  • #37
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 ?
 
  • #38
Altairs said:
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. :smile:
 
  • #39
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.
 
  • #40
Altairs said:
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? :confused:

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? :smile:
 
  • #41
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.
 
  • #42
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 ?
 
  • #43
Altairs said:
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?
Altairs said:
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"? :confused:
 
  • #44
tiny-tim said:
Do you mean lifting the rear tyres just an inch off the ground, or do you mean "balancing the car on its front legs"? :confused:

Just an inch. BTW what's the difference?
 
  • #45
tiny-tim said:
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 ?
 
  • #46
Altairs said:
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. :smile:
 
  • #47
Here it is.

Altairs said:
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 said:
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! :smile:
 
  • #48
Altairs said:
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. :smile:
 
  • #49
tiny-tim said:
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. :smile:

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 ?
 
  • #50
Altairs said:
Here it is.
Altairs said:
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
 
  • #51
tiny-tim said:
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

Oh...right..
 
  • #52
Altairs said:
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.

If the men are lifting it by holding onto the rear axle, then your friends are wrong, and you are right … the distribution of the forces (or reactions) depends only on the positions of the points of support on the car.

(But if the men are holding it by the tow-bar under the rear bumper, then the reading will change, because the c.o.m. is further from the tow-bar.)
Second point is that will this procedure (the tipping one) when performed for both the front and the rear tyres equal the weight ?

Sorry … I'm not following this at all … why would you get men to tip the car when you can get the same reading by leaving the rear tyres on the ground?
 
  • #53
when we tilt the car and find the reaction thn v'l hav a frictional force acting on the tyre,in contact with the ground apart from the surface reaction so will we consider it as limiting friction or how do we calculate it?
 
  • #54
Welcome to PF!

Whatevr said:
when we tilt the car and find the reaction thn v'l hav a frictional force acting on the tyre,in contact with the ground apart from the surface reaction so will we consider it as limiting friction or how do we calculate it?

Hi Whatevr! Welcome to PF! :smile:

The weighing platform will only measure the vertical reaction, and the friction force is entirely horizontal, so the friction force will make no difference. :smile:
 
  • #55
but when u find the z distance u take moments abt a point nd therefore u hav 2 kno all the forces how else wud u claculate the distances
 
  • #56
Whatevr said:
but when u find the z distance u take moments abt a point nd therefore u hav 2 kno all the forces how else wud u claculate the distances

Hi w'v'r! :smile:

ah, you've made a good point … if the higher pair of tyres is supported on a sloping surface, then the friction will spoil the equations.

So it is important that the higher pair of tyres (or its axle) be supported on a raised horizontal surface. :smile:
 
  • #57
no i don't get it:( that won't b possible i thnk or the frictional component wud stil exist ??
friction will exist if we raise the car frm one end
 
  • #58
Whatevr said:
no i don't get it:( that won't b possible i thnk or the frictional component wud stil exist ??
friction will exist if we raise the car frm one end

If all the wheels are on horizontal surfaces, why would the car want to move anywhere (even with the brakes off)?

There'd be no gravitational advantage in moving. :smile:
 
  • #59
no but thn how'd we calculate the z distance i.e. the height part?
 
  • #60
Whatevr said:
no but thn how'd we calculate the z distance i.e. the height part?

If we can do away with the friction part than it is quite simple.
 
  • #61
ummmmmm that's the prob that friction exists and we hav 2 b as exact as possible so v hav 2 consider friction
 
  • #62
I still don't see any point in taking friction. Secondly, I don't have the wildest idea that how will we calculate friction in our case( I am even confused that while the car is in horizontal position with no external horizontal force acting over the platform the friction is acting or not !). Its definitely not limiting. Thirdly, we have to give a 'standard procedure'. With friction involved it won't remain standard as we would need to know the different values of [tex]\mu[/tex] for different contacting surfaces (tyres and platforms).
 
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  • #63
tiny-tim said:
if the higher pair of tyres is supported on a sloping surface, then the friction will spoil the equations.

So it is important that the higher pair of tyres (or its axle) be supported on a raised horizontal surface. :smile:

This situation can also be simulated by hooking the front end of the car (from bumper or something) and then lifting.
 
  • #64
Altairs said:
This situation can also be simulated by hooking the front end of the car (from bumper or something) and then lifting.

Hi Altairs! :smile:

Yes … provided you ensure that the rope is vertical once the car is raised!

Though there is a practical problem … I don't think there's anything at the front of a car that's strong enough not to be damaged. :cry:
 
  • #65
tiny-tim said:
Though there is a practical problem … I don't think there's anything at the front of a car that's strong enough not to be damaged. :cry:

Something can be attached as an extension to the front side and then the it can be hooked.
 
  • #66
i don't get how ur trying 2 remove friction...
 
  • #67
How are we supposed to calculate the friction present :confused: ?
 
  • #68
Altairs said:
How are we supposed to calculate the friction present :confused: ?

By fixing it so that the friction is zero, which means supporting the wheels on horizontal surfaces, or the towbar on a vertical rope. :smile:
 
  • #69
tiny-tim said:
By fixing it so that the friction is zero, which means supporting the wheels on horizontal surfaces, or the towbar on a vertical rope.

That is exactly what I thought (post 62).

[tex]4F_{s} = 0 [/tex] (fixed and no external forces present horizontally)
 
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  • #70
Altairs said:
That is exactly what I thought (post 62).

[tex]4F_{s} = 0 [/tex] (fixed and no external forces present horizontally)

Hi Altairs! :smile:

(btw, don't send a Visitor Message if it's important, since there's no notification of Visitor Messages, and I might not see it for days. Send PM if you think I've missed a post.)

Yes … ignore friction …

… your post #62, replying to Whatevr, was completely right (which is why I didn't comment on it).
 

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