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Is there any difference in terms of fore/aft weight distribution between the two wheels in the two pictures?

In other words: Is the location of the center of gravity of the 100kg weight the only factor that matters for fore / aft weight distribution between the two wheels?

Assumptions made: Differences in the center of gravity of the structure that holds the weight are neglected. Both structures have the same mass. Update (30th of March 2023): it's a static scenario (no acceleration or deceleration), thanks to JohnAlexiou for pointing this out.

I have an intuitive sense what the correct answer is but I'd like to be sure. It would be great if an answer could make a reference to the physical principles/laws that apply here.

Update (30th of March 2023): I just want to say that I find it incredibly cool how strangers on the internet come together to solve real world problems (this is a question I had with implications for a bicycle I'm designing) while engaging with their passion (bicycle design and physics in this case). The answers by Eli and Bob D come to the same conclusion namely that there is no difference in fore / aft center of gravity between the two drawings. I have not marked an answer as the correct one because I think it's best for the reader to assign their own confidence rating to the answers that were given by reading the comments under Eli's answer. So thanks again to all who replied!

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    $\begingroup$ Is this static, or dynamic? Is the vehicle accelerating? or moving at a constant speed? $\endgroup$
    – John Alexiou
    Commented Mar 28, 2023 at 17:33
  • $\begingroup$ @JohnAlexiou it's static, no acceleration or deceleration. Makes sense as a clarification. $\endgroup$
    – aehhhhmm
    Commented Mar 28, 2023 at 20:49

2 Answers 2

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Lets look at the equations

Static equilibrium

to obtain the unknows forces $~F_1~,F_2~$ you take the sum of the forces $~\sum F_i=0~$ and the sum of the torques about point A $~\sum_A\tau_i=0~$. you get two equations for the unknows $~F_1~,F_2~$

I)

$$\sum F_i=F_1+F_2-m\,g=0\\ \sum_A\tau_i=F_1\,a-F_2\,b-m\,g\,c=0$$

solution

$$F_1=m\,g\frac{b+c}{a+b}\\ F_2=m\,g\frac{a-c}{a+b}$$

II)

$$\sum F_i=F_1+F_2-m\,g=0\\ \sum_A\tau_i=F_1\,d-F_2\,e=0$$ with $~d=a-c~,e=b+c~$

you obtain $$\sum_A\tau_i=F_1\,a-F_2\,b\,-\underbrace{c(F_1+F_2)}_{m\,g\,c}=0$$

the equation of case I.

thus there is no difference in terms of fore/aft weight distribution

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    $\begingroup$ In your last equation you are missing the $b$ coefficient for $F_2$. Otherwise, nice job. $\endgroup$
    – Bob D
    Commented Mar 28, 2023 at 15:31
  • $\begingroup$ @Eli Thanks for taking the time! I'm trying to understand your equations. What does "τ" stand for? And "A" is the wheelbase (distance between the two axels)? $\endgroup$
    – aehhhhmm
    Commented Mar 28, 2023 at 15:36
  • $\begingroup$ @BobD your remark doesn't effect the outcome that "there is no difference in terms of fore/aft weight distribution", right? $\endgroup$
    – aehhhhmm
    Commented Mar 28, 2023 at 15:39
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    $\begingroup$ @aehhhhmm No, Eli is correct except that the b coefficient appears to be inadvertently omitted to make the second term a torque and not just a force. There is no difference in the weight distribution. My answer supports the conclusion from a different perspective. $\endgroup$
    – Bob D
    Commented Mar 28, 2023 at 16:25
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    $\begingroup$ @aehhhhmm see additional information about the static equilibrium $\endgroup$
    – Eli
    Commented Mar 28, 2023 at 16:51
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The line of action of the force of gravity on the 100 kg weight in the first diagram acts through the same point on the structure as the physical connection of the vertical support of the weight in the second diagram, as shown in the diagrams below. So the weight distribution on the two wheels should be the same for the two diagrams as long as the sum of the torques about A remains zero for both diagrams.

That is the case because the decrease in the counter clockwise moment about A due to the reactions $F_1$ and $F_2$ when A is moved to the right from diagram 2 to diagram 1 equals the increase in the counter clockwise moment of $mgc$ about A due to the weight in diagram 1.

Hope this helps.

enter image description here

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  • $\begingroup$ @John D, could it be that the torque arrow in your drawing points in the wrong direction? $\endgroup$
    – aehhhhmm
    Commented Mar 30, 2023 at 9:41
  • $\begingroup$ @aehhhhmm The torque arrow was not intended to show the torque produced by the weight (which would be counter clockwise) but rather the internal structural reaction to the torque produced by the weight, like a cantilever support. But I can see how that might confuse someone, so I'm considering revising my answer. Thanks. $\endgroup$
    – Bob D
    Commented Mar 30, 2023 at 15:11

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