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I have a 4-wheeled vehicle that is 10 x 10 cm. Each wheel has its own motor. The wheel diameter is 3.1 cm. The total weight of the vehicle is 500 grams. The friction coefficient of the rubber wheels to concrete is 0.6-0.85 (found this online, from the research I did I think it is needed for calculating torque). How much torque is needed on the wheel before the motor becomes too powerful and just cause itself to slip.

If someone share the steps and formulas needed to solve the problem it would be of great help.

So far I only calculated the weight on each wheel (I'm not even sure if this is correct)

$$\begin{align} W &= \text{wheel}_1 + \text{wheel}_2 + \text{wheel}_3 + \text{wheel}_4 \\ 500\text{ g} &= 4\text{wheel} \\ 125\text{ g} &= \text{wheel} \\ \end{align}$$

so 125 g per wheel (assuming weight is distributed equally).

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  • $\begingroup$ why not add more motors and tires to increase area ? $\endgroup$
    – Yrra Uy
    Sep 13 '18 at 14:23
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Before start to solve the problem, i like to use your assumption about load distribution. And i assume you found the static coefficient of friction. Notice there is a difference between static and dynamic coefficient of friction. The dynamic coefficient is slightly smaller in amplitude than static coefficient.

Look at the free body diagram of one of the wheels: enter image description here

To make the wheel move, you need to apply enough torque to cancel out the force $F_t$: $$F_t = \mu F_N$$ $$F_N = mg$$ Here $\mu$ is the static coefficient of friction, $F_N$ normal or reaction force on the wheel, $g$ is constant, about $9.81 \frac {m}{s^2}$ and $m$ is the mass of the vehicle in this case $m$ is just a fraction of it nl. one fourth.

Now we can calculate the torque:

$$\tau = F_t \frac{d}{2} =\frac{d}{2} mg \mu $$

Here $d$ represent the diameter. Notice this is enough to bring the vehicle in movement, but if you want to accelerate the vehicle then this amount of torque is not enough.

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  • $\begingroup$ Thank you for your response. Give me a second as I try to solve it $\endgroup$
    – Jack
    Sep 13 '18 at 13:04
  • $\begingroup$ @Jack You did point it out in your question, it's 3.1 cm. Is that what you mean ? $\endgroup$ Sep 13 '18 at 13:07
  • $\begingroup$ Yes, I apologize, i have edited my question. I came up with an answer of 1.05 N.m . Did i do it correctly? $\endgroup$
    – Jack
    Sep 13 '18 at 13:18
  • $\begingroup$ @Jack hmm, sounds not right to me, look , $\tau = 0.5*0.031*9.81*0.85$ $\endgroup$ Sep 13 '18 at 13:28
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    $\begingroup$ oops forgot to multiply the diameter/2 sorry. it seems you did not multiply its mass in kg. τ=0.5∗0.031∗9.81∗0.85*0.125 = 0.01615584375 N.m . Quite low than i expected. $\endgroup$
    – Jack
    Sep 13 '18 at 13:33

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