Friction, Normal Force

[ahyaa]

After looking for this answer online I was surprised I couldn't find it.. so naturally I came here

Anyways, I was taught in class that friction = coefficient*normal force. My question is what happens if we have an angled normal force (with respect to our axes), would the normal force in the friction equation be one of the components, or the magnitude of the entire normal force?

[imatfaal]

Friction arises from an object moving on a surface - it acts parallel to the surface and opposes motion (it is the parallel component of the contact force of an object on a surface). The Normal force is by definition perpendicular to that surface (it is the perpendicular component of the contact force). The Normal is perpendicular to the surface regardless of the direction of the axes.

 

Your axes are merely computational aids; they cannot affect your answer - the result of the calculation must always be the same regardless of things like co-ordinate systems, axes, which way is positive etc. So your choice of axes - and it sometimes pays to take a lot of care in choosing as it may make a hard problem into a doddle - does not change the rules.

 

What may be confusing you is that the normal force on an object from a slope is often shown as m.g.cos(theta) - this is because the Normal is equal and opposite to the perpendicular component of the objects force on the slope due to gravity (ie m.g times by the cosine of the angle of the slope) . But the friction is still F_fric = mu.N.

 

And just a little aside - "magnitude of the entire normal force" No, it is not the magnitude - it is the force itself. The magnitude is a scalar - you want the vector Force. Remember those little arrows on the top of vectors - I find it useful to put them in like the textbooks, it concentrates the mind and allows you to avoid slip ups.

[studiot]

 

 

Sorry quoting or copy pasting no longer works, no doubt a Christmas present from Microsoft.

 

However you should have been taught that

 

Limiting Friction = coefficient times the Normal Reaction.

 

The Normal Reaction is always at right angles to the frictional force.

 

However the frictional force is variable. That is it is only as large as it needs to be to oppose motion.

 

So it is always exactly equal but opposite to the pulling or pushing force up to the 'Limit' when the object starts to move.

 

At the limit the frictional force equals the coefficient times the normal force.

 

Before this the frictional force is less than this.

[studiot]

I should have added that if the load that provides the sideways push or pull is applied at an angle, the load resolves into two components.

 

One component is parallel the direction of the friction and is the only component that opposes it.

 

The other component is parallel to the direction of the normal reaction and does not affect the friction.

This component adds (subtracts from) to the normal forces acting, but does not appear in the relation friction = coefficient times the normal reaction ( or the force required to maintain equilibrium if not at limiting friction).

 

Also, when the object is moving, there is a different situation in action and equilibrium no longer applies.

[ASG]

the co-efficient of restitution has very strong limitations. variable frictional force can be said to be a no force if it cannot exist on its own.