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When is reaction force perpendicular and when is it angled?

I'm revising for CCEA Mechanics 1 for AS Maths. I'm having difficulty understanding when the reaction force of an object is perpendicular to the wall or angled when an object is in equilibrium.

For example, a ladder always has its reaction force perpendicular to the wall unless it is resting on the top corner of a wall. I know that one.

But consider this:

The reaction force at P occurs perpendicular to the wall, but what information tells us this?

Yet in this example:

The reaction force at A occurs at an angle. What information tells us this?

Basically, how do I distinguish whether the reaction force is an angle or perpendicular?
reactionPer.jpg33.8KB
Reply 1
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Zacken
22
Original post by TiernanW
I'm revising for CCEA Mechanics 1 for AS Maths. I'm having difficulty understanding when the reaction force of an object is perpendicular to the wall or angled when an object is in equilibrium.


Reaction is always perpendicular to the surface causing the reaction.


The reaction force at P occurs perpendicular to the wall, but what information tells us this?


The particle is held against the wall and hence the wall causes the reaction, so the reaction is perpendicular to the wall.

Yet in this example:
The reaction force at A occurs at an angle. What information tells us this?


The rod is hinged to the wall and hence the hinge causes the reaction, so the reaction is perpendicular to the hinge. But the whole point of a hinge is that it can be hinged at any angle, so the reaction is angled.

Basically, how do I distinguish whether the reaction force is an angle or perpendicular?


There are two main cases at A-Level:

1. Held against or leaning against or just contact with no intermediary device: perpendicular to it. Whether it's the ground or wall or whatever.

2. Hinged to something: reaction is angled.
Original post by Zacken
Reaction is always perpendicular to the surface causing the reaction.


This isn't true. The reaction is always normal for an interaction between smooth surfaces, but may not be if the surfaces are rough i.e. if they can generate friction.
Reply 3
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Zacken
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Original post by atsruser
This isn't true. The reaction is always normal for an interaction between smooth surfaces, but may not be if the surfaces are rough i.e. if they can generate friction.


Ah, yes. Good shout, thanks.
Reply 4
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TiernanW
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Original post by atsruser
This isn't true. The reaction is always normal for an interaction between smooth surfaces, but may not be if the surfaces are rough i.e. if they can generate friction.


So if a ball was being pushed into a wall by a force not perpendicular to the wall, but there was enough friction to stop it moving the reaction would occur at an unknown angle?
Original post by TiernanW
So if a ball was being pushed into a wall by a force not perpendicular to the wall, but there was enough friction to stop it moving the reaction would occur at an unknown angle?


Yes. In fact that's a pretty good scenario.

As another example, consider a rock on a rough flat horizontal surface. With no external force applied to the rock except gravity, there is a normal reaction force from the ground on the rock. However, if you now try to move the rock, by pushing it sideways, then there is now a non-perpendicular reaction force on the rock, being the resultant of the frictional force opposing motion, and the normal reaction.
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TiernanW
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Original post by atsruser
Yes. In fact that's a pretty good scenario.

As another example, consider a rock on a rough flat horizontal surface. With no external force applied to the rock except gravity, there is a normal reaction force from the ground on the rock. However, if you now try to move the rock, by pushing it sideways, then there is now a non-perpendicular reaction force on the rock, being the resultant of the frictional force opposing motion, and the normal reaction.


So does this mean that Friction is part of the reaction force? Would saying there is a normal perpendicular reaction on the wall and a frictional force at a right angle to it, be the same as one reaction force at non-perpendicular angle?

Sorry just making sure I clarify it fully.
Original post by TiernanW
So does this mean that Friction is part of the reaction force?


Yes. Usually you consider the resultant of the normal force and friction to be the reaction. This resultant will not be perpendicular to the surface.

Would saying there is a normal perpendicular reaction on the wall and a frictional force at a right angle to it, be the same as one reaction force at non-perpendicular angle?


Yes. Usually there is not much we can say about the precise angle though, unless the body is on the point of slipping i.e. we have reached the value of limiting friction. When this happens, the reaction is often called the "total reaction", and with normal reaction NN, coefficient of friction μ\mu, it has components:

R⊥=NR∥=μNR_{\perp}= N \\ R_{\parallel} = \mu N

and it makes an angle θ\theta with the normal to the surface, where:

tan⁡θ=R∥R⊥=μNN=μ\tan \theta = \frac{R_{\parallel}}{R_{\perp}} = \frac{\mu N}{N} = \mu

Then θ\theta is often called the "angle of friction". (Exercise: I place a mass on a rough surface, coefficient of friction μ\mu, making angle θ\theta with the horizontal. At what value of θ\theta does the mass start to slip?)
Reply 8
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TiernanW
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Original post by atsruser
Yes. Usually you consider the resultant of the normal force and friction to be the reaction. This resultant will not be perpendicular to the surface.



Yes. Usually there is not much we can say about the precise angle though, unless the body is on the point of slipping i.e. we have reached the value of limiting friction. When this happens, the reaction is often called the "total reaction", and with normal reaction NN, coefficient of friction μ\mu, it has components:

R⊥=NR∥=μNR_{\perp}= N \\ R_{\parallel} = \mu N

and it makes an angle θ\theta with the normal to the surface, where:

tan⁡θ=R∥R⊥=μNN=μ\tan \theta = \frac{R_{\parallel}}{R_{\perp}} = \frac{\mu N}{N} = \mu

Then θ\theta is often called the "angle of friction". (Exercise: I place a mass on a rough surface, coefficient of friction μ\mu, making angle θ\theta with the horizontal. At what value of θ\theta does the mass start to slip?)


For the exercise, friction points up the slope. As it is about to slip the acceleration is still equal to 0, so mgSinθ\theta = μ\mumgCosθ\theta, so it slips at an angle of tanθtan \theta = μ\mu
Original post by TiernanW
For the exercise, friction points up the slope. As it is about to slip the acceleration is still equal to 0, so mgSinθ\theta = μ\mumgCosθ\theta, so it slips at an angle of tanθtan \theta = μ\mu


Right! I didn't actually expect you to do the exercise there and then though. Worth a rep.
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TiernanW
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Original post by atsruser
Right! I didn't actually expect you to do the exercise there and then though. Worth a rep.


Well who would I have to check it? Haha I'm on study leave. Thanks for the rep!
Reply 11
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TiernanW
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Original post by atsruser
Right! I didn't actually expect you to do the exercise there and then though. Worth a rep.


Sorry one last thing. xD AB is a uniform rod and AC is a light in-extensible string, and BC is a wall.


In this example do we not include T in the moments because it attaches to C? I know you never include forces that act on the point you're taking moments from because its distance is 0, but curious because T acts on the rod AB and so does 3g.
MC.jpg11.4KB

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