# Centripetal force on a car turning

1. May 28, 2004

when a car is cornering on a banked road, what force provides the centripetal force? in my book it says that it is the horizontal component of the normal reaction, and the vertical component of the normal balances the weight.
But i thought the normal reaction was equal to the component of the weight that is perpendicular to the slope. hence the vertical component of the normal reaction could never balance the whole weight.

2. May 28, 2004

### remcook

The normal force is indeed normal to the surface. It basically does whatever it takes to stop the car from falling through the road. Since it is the only force with a vertical component and the car is in equilibrium vertically (it isn't falling through the road), the vertical part of the normal force must be equal to the weight of the car. Since there is also a horizontal component, the actual magnitude of the normal force is larger than the magnitude of the weight. This last case is when you have a horizontal road.

When the centrifugal force is higher (you speed up for intance), the weight of the car is still the same. So: so is the normal force. The excess horizontal (centrifugal) force will cause your car to move outwards. This is what you would expect, right?

Hope I explained it right...

Edit- Friction has a vertical component too with a banked road...

Last edited: May 28, 2004
3. May 28, 2004

yeah, thats ok i guess, but in every example i have seen, the normal force is smaller in magnitude than the weight. As i said, it is always the same magnitude as the component of the wieght that is perpendicular to the slope. (a number of these example are when a box is being pulled up a slope etc, does this change the problem?)

4. May 28, 2004

### remcook

remember that the box is neither falling through the floor nor flying into the sky.

so all vertical components must add up to zero.

If you drag a box up there is also friction, which has a vertical (downwards) component.

So...the vertical component of the normal force should...

another thought: if you aren't pulling the box up, and there is no friction, what causes the horizontal movement? (sliding down the slope) think about it!

Where have you seen "every example" and could you give one?

5. May 29, 2004

all the examples are pretty much in the same vein. they tell you to take axes along and perpendicular to the slope, and this shows that the normal force must be equal to the compnent of mg perpendicular to the slope. friction and mg down slope are at right angles so have no component.
also, when the car is cornering on a banked road, what is the component of mg down the slope doing? i thought that may be centripetal?

6. May 29, 2004

### Staff: Mentor

Right! (In the case of no friction.)
You are thinking of a case in which the car (or box or object) is on an incline and the net force is zero perpendicular to the surface. This is not the case here! The car is accelerating towards the center.

The net force is zero in the vertical direction:
$$Ncos\theta - mg = 0$$
But in the horizontal direction the net force is not zero:
$$Nsin\theta = \frac{mv^2}{r}$$

7. May 29, 2004

ok. but when the object is sat on the slope, going up or down the slope, the net force perpendicular the slope will be zero wont it? am i right in thinking that this only changes when the object starts moving 'along' the bank and not up or down it? thanks

8. May 29, 2004

also, does the car/object not fly off the track if the net force perpendicular to the slope is not zero? weight has a component down the slope, so what balances this? thanks again.

Last edited: May 29, 2004
9. May 30, 2004

### Staff: Mentor

The weight of the car is balanced by the vertical component of the normal force.

Remember, a car going in a circle is not in equilibrium: it is accelerating towards the center of the circle. So forces in the horizontal direction do not balance. But vertical forces do.

10. May 30, 2004

### turin

It may be a poor approach to orient your axes the same as you would for the inclined plane. In the inclined plane case, it is usually good to orient the axes according to the slope. I the case of an acceleration, it is usually a good idea to orient the axes according to the acceleration. If you orient them thusly, then they should not line up with the slope in this case.