Weight Distribution Question
 

Which wheel in the attached drawing bears a greater percentage of the weight of the box? I'm thinking it's A, but I want some support from my friends on this one.

Thanks,

Eric.

I would have to say wheel A because the weight is tipped back onto that wheel (A)

Hope I helped

Brian

A is the correct answer. More of the box is closer to, outside of, wheel A.

The answer cannot be determined from the information given because you have not specified the location of the center of mass.

If you assume that the CG is in the center of the box, and that the box is at rest, wheel A will support more of the weight. You can do a quick free-body diagram to proove it to yourself. (See attached)

Notice that the forces balance, which is probably what your friends were thinking. But when you include the torques in the equation, the ground has to push on wheel a more than wheel b to keep the box from tipping.

Eventually you get the equations:

Fa = W/(1 + Ra/Rb)

and

Fb = W/(1 + Rb/Ra)

and it is unnecessary to make any assumptions about the location of the CG, other than that it lies between the wheels.

Substitute the distances Ra and Rb into the above equations, and out pops the force distribution on the wheels.

If we further assume that the wheels are a distance "L" apart,
and the CG is a distance "H" above the line of the axles and centered between them (with the body horizontal), and the body makes an angle A with horizontal, then we have for Ra and Rb :

Ra = ((L/2) - H tanA) cosA, and
Rb = ((L/2) + H tanA) cosA.

(Check behind me - I did this quickly on scrap paper.)

Now you only need to know how wide the robot is, how high above the axles the CG is (good luck, but try to estimate), and what angle the body makes with the floor, and you can calculate the weight distribution. If the CG is not centered, the geometry is a little more complicated, but not that bad.

Thank you all for your help. The CG was not particularly relevent because we are simply trying to shift more weight from the 'B' wheel to the 'A' wheel based on, um, field tests :)

Thanks again,

Eric.

I know what you mean - a lot of times all the calculations in the world can't do as much a a quick whack with the hammer.