Where is the Center of Gravity of a Car Located?

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SUMMARY

The center of gravity (CG) of a car can be calculated using the forces on the front and back axles. In this discussion, the front axle experiences a force of 2100 lbs, while the back axle has a force of 1800 lbs. By applying the torque equilibrium principle, the CG is determined to be 60 inches from the front axle, which is consistent with the calculated forces. The initial method of converting weights to mass and distances to meters was incorrect for this calculation.

PREREQUISITES
  • Understanding of torque and equilibrium principles
  • Basic knowledge of forces and weight distribution
  • Familiarity with unit conversions (lbs to Newtons)
  • Concept of center of mass in physics
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Find the center of gravity of a car with respect to the front axle. There is a distance of 130 inches between the front and back axles. There is a force of 1800lbs on the back axle and 2100lbs on the front axle.

I changed 130 inches to meters and I got 3.30m. I changed both 1800lbs and 2100lbs to Newtons and I got 8010N and 9345N respectively. I divided each by 9.8 to get their masses.

I then divided 3.30m by the sum of 817.3g + 953.6g. I got .0019m. So that means that the center of gravity is .0019m away from the front axle or .075inches.

Did I do this correctly? My answer does not seem to make much conceptual sense...
 
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No, you didn't do this correctly. The center of mass is defined as:
[tex](m_1 x_1 + m_2 x_2 + ...)/M_{total}[/tex]

But you can also think of the center of mass as that point about which the net torque due the object's weight would be zero. (No need to do any unit conversions.) You have one torque pulling clockwise, the other counterclockwise: set them equal to each other.
 
That makes sense.

I set the CG x inches away from the front axle, which means that it has to be 130-x inches away from the back axle.

1800(130-x)=2100x
234000=3900x
x=60

So the CG must be 60" away from the front axle, which makes sense.

Thanks
 

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