# Normal/Tangential Force

1. Jun 18, 2012

### luis20

Hi. I have two questions

All contact forces are normal to the surface right? If there was no friction, there wouldn't be any tangential force?

Thanks

2. Jun 18, 2012

### jfy4

The normal force is indeed always normal to whatever surface you are talking about. The frictional force is tangential to that surface, and is proportional to the Normal force in elementary analysis.

3. Jun 19, 2012

### haruspex

Not quite. The maximum static friction force is proportional normal force. The actual tangential frictional force is anything up to that.

4. Jun 19, 2012

### luis20

Thanks for the replies

What I want to know is if all tangential forces are due to friction.

In a fast collision, the tangential force can be neglected?

5. Jun 19, 2012

### nasu

Not at all. Any kind of force may have a component tangent to the trajectory.
See for example motion on inclined plane. The gravity has a tangential component.

For what purpose?

The effect of tangential forces on the final momentum distribution may become important for collisions that are not "head-on". For example, throwing a ball towards a wall at an angle. If the ball-wall friction is large the tangential component of the momentum will decrease and the angle of "reflection" will be smaller than the angle of incidence.

6. Jun 19, 2012

### luis20

What I meant was all tangential forces on a surface are due to friction. The tangential component of gravity will only affect the surface if there is friction. So I think my statement may be correct.

The thing you said about the ball is what I wanted to read. My teachers neglect the tangential component when we talk about collisions. Maybe that's because they are engineers.

Edit: But if you see collision articles, they talk only about the force which is normal to the contact surface! (tangential neglected) Collision is considered fast so the tangential force might be very little?

7. Jun 19, 2012

### nasu

Well, what kind of collisions are discussed by your teachers? Are they head-on collisions?
I mean, is there a tangential component of the momentum of the colliding object(s) or not?

8. Jun 19, 2012

### luis20

There is a tangential component. It's like your example of the ball.

Collisions which produce torque, where the normal force is not in the direction of the center of mass, those have tangential force for sure right?

9. Jun 19, 2012

### haruspex

It's not a question of whether the normal force passes through the centre of mass. The test is whether the relative motion of the points of contact immediately before collision is normal to the contact plane. Thus, a ball striking a flat surface at an angle will experience a tangential frictional force on impact unless it happened to be spinning in exactly the right way so that it made rolling contact instantaneously.
Also, I don't think one can argue that the contact is too brief for the tangential force to matter. The maximum tangential force is proportional, at each instant, to the actual normal force, just as in normal static arrangements, and the tangential and normal forces operate over the same brief time period.
If a ball radius R rebounds from a surface with speed V at θ to the normal, no sliding, I would think that the ball is now spinning at rate V sin(θ) / R.

10. Jun 20, 2012

### jbriggs444

This would be the case if the ball were to smoothly roll off the wall. That model is inaccurate in two ways.

1. The ball may slide across the wall instead of rolling without slippage. This would tend to apply in, for instance, the case of a steel ball hitting a steel wall at a glancing angle.

In the limit your spin rate omega(final) = omega(initial).

2. Rotational rebound.

In my youth we used to play with "super balls". These were hard rubber balls that rebounded pretty efficiently (90 to 95% restitution). They were also sticky enough that they would catch immediately and roll without slipping. If you threw one of these at the floor and got a good spin on it you could watch it go into a back and forth bouncing pattern with the direction of rotation reversing at each bounce.

In the limit your spin rate omega(final) = 2 V sin(θ) / R - omega(initial)

The spin rate resulting from any actual collision could be expected to lie somewhere in this range.

11. Jun 20, 2012

### haruspex

I did specify "if no sliding"
True - I did forget about that.