# Acceleration on a curve for a car

## Homework Statement

As we know, if a car accelerate on curve at certain speed, it will tilt outside the circle track. So i sketch this 2 picture to show how (for me) it works.

http://img193.imageshack.us/img193/2323/66057246.jpg [Broken]
http://img194.imageshack.us/img194/9272/96495741.jpg [Broken]

I just wondering, how can the car tilt or flip outside the circle track if the side force (centripetal force) is pointing inside the circle?

## Homework Equations

My suggestion:
Total Moment about tyre (2); ∑M2=0

## The Attempt at a Solution

So, Fs*h + G*b/2 - FN1*b = 0

I assume that FN1=0 since the car is about to tilt about tyre (2). So here come the problem.

Fs = - G*b / 2h.

The minus sign shows that the side force, Fs, should point outside the circle.

Last edited by a moderator:

Thanks LowlyPion.

So, what I understand from the link u gave me, the friction force should be act towards the centre of the circle and the side force towards outside?

In other words, friction = centripetal and side force = centrifugal ("effective") force?

Hi there,

You might want to forget about centrifugal force in your equations, since it is an imaginary force. The equation of a car winding around a curve, at a constant speed, would looke more like this:
$$\sum F = ma_{cent}$$
$$\text{friction} = m \frac{v^2}{R}$$

Cheers

Thanks fatra2,

The main purpose for me doing this calculation is to get maximum tangential acceleration of the car, since the car is not travelling on the curve with constant speed.

For that, I need to calculate first the maximum side force Fs (the speed and radius is not given). So I use this method to determine the maximum side force without taking the speed and the radius into account.

After I get the side force Fs, I add it to F1 to get the resultant force.

Fres = √(〖F_1〗^2+〖Fs max)〗^2 )

And then, the resultant force should be equal or less than the friction force in x and y-axis (in the picture above, the yellow circle indicate that firction).

Fres ≤ F_μH
(m.a)^2+〖Fs max)〗^2 ≤〖F_μH〗^2

F_μH and m are known. Fs max is also known from the previous calculation. Hence, I get the maximum acceleration,

a_max = √(〖F_μH〗^2-〖Fs max〗^2 )/m

Is my method correct?

Hi there,

If the speed of the car is not constant, the problem gets more complicated. The basic equation stays the same:
$$\sum \vec{F} = m \frac{d^2\vec{r}(x,y)}{dt^2}$$

For which the friction will act in both direction (x,y). Since the car accelerates, the motor develop traction on the wheels, which in turn develop backward friction on the road, which makes the car go forward.

I guess you don't want to have the car spinning and skidding on the road, so you can assume only static friction. But the value $$f=\mu N$$ gives the maximal scalar value of the friction. If the case is where you are far from spinning and skidding, then the value can take a lower value than that.

I am sorry to complicate your porblem, but if you want to have a correct solution, you should not overlook anything. Cheers