I'm doing my IB HL Math Exploration, where I use a bit of physics. Say I have a car with mass m, center of gravity at height h, and width of 2b, and I am turning a corner, how much centrifugal acceleration is needed for the inner wheels of the car to leave the ground?

I'm trying to solve this question using this website. However, as the IB Physics course does not cover torque :(, I am unable to understand the working out.

I understand the working until this line:

Taking moments about the centre of gravity :

What does it mean by taking moments?

$R_2b = Fh$ and so $F = \frac{R_2b}{h} = \frac{mgb}{h}$

How are these equations derived?

The car in my exploration is not necessarily travelling in a circular path, but in paths including those modeled by the function $f(x) = x^2$. However, I do have the acceleration of the car. Therefore, can I simply use Newton's Second Law $F = ma$ and the derived $F=\frac{mgb}{h}$, and equate $a = \frac{gb}{h}$?

Thank you for your help!


1 Answer 1


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Look at this figure and take the sum of the torques about point B.

  • $\begingroup$ How would I find the torques of the three forces shown in the diagram? Do I have to resolve them into vectors that are perpendicular to the like from B to the center of gravity? Do I need to consider the horizontal frictional force not shown as well? $\endgroup$ Jul 25, 2019 at 9:56
  • $\begingroup$ For a static equilibrium $\sum _{\tau }=\dfrac {mv^{2}}{r}\cdot h-mg\cdot b+F\cdot 2b=0$ $\endgroup$
    – Eli
    Jul 25, 2019 at 10:44

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