ladders

Here is the question:

A uniform ladder XY. Length l, Weight W. X on rough ground Y on smooth wall. coefficient of friction = 0.2. Mass 9W at top of ladder. Horizontal force P applied to ladder at X towards the wall. Ladder in equilibrium at angle theta where tan theta = root 3.
a) find in terms of W the reaction at the wall (done).
b) Find, in terms of W, the range of possible values of P for which the ladder remains in equilibrium (stuck).

I have the worked solution but some of it I don't get.

Here is the answer
R = W + 9W (get it)
for the ladder to be in limiting equilibrium F = mu R = 2W (get it)

If P + F > S (S is reaction at wall and R at ground) the ladder will slide towards and up the wall. (I sort of get this except that if the base of the ladder moves towards the wall, won't friction oppose motion and change direction in which case what?)

If P < S-F the ladder will slide away from the wall. (get it)

BUT the next line is
Therefore S-F <=P <= S+F (Why? if you rearrange the thing in bold, above, then P> S-F so it can't be bigger or smaller)

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I've tried to make sense of this but I'm flummoxed.

Thanks

It's similar to the other question you asked.
F is the limiting friction, P is a "variable" horizontal force.
P must oppose S and if there was no horizontal friction P=S is the only way for equilibrium to occur
Adding ground friction gives you a buffer in both directions. When P<S and the ladder wants to slide away from the wall or when P>S and the ladder wants to move towards the wall.
So P can be a bit bigger or smaller than S, and the system will remain in equilibrium, until it exceeds the limit and equilibrium is broken. The limit on each side is F.

I'm not ignoring you I'm just trying to work out what it means.

I get this line : P must oppose S and if there was no horizontal friction P=S is the only way for equilibrium to occur
It's after that that I'm working on.

Just take one of the cases. If P<S, then the ladder would slide away from the wall if there was no friction. With friction present, P=S-F is the limiting value. It will remain in equilibrium if P>=S-F but will slide away when P<S-F. The least value P can be is S-F as friction will be on limiting value and you need to provide extra horizontal force.

The larger F can be (rougher surface) the smaller force P needs to be.

Similarly for the other case of movement towards the wall.

The question is really about a person P pushing a ladder at the bottom when another person (9W) has climbed to the top. In this case, ground friction alone is generally not enough to maintain equilibrium.

Let me try

Without friction if P < S it slide away from the wall. This I get.
But there is friction and it opposes possible motion so it's in the same direction as P. So P + F = S in limiting equilibrium.
The question has said that mu = 0.2. So in my mind, the Reaction at the ground can't change so Friction can't change so if P got bigger the bottom ladder would move towards the wall wouldn't it?

Limiting friction can't change. It's the max value friction can take. So
0 <= friction <= F
Where you're treating friction a scalar so you handle the sign (direction) elsewhere.

When
S-F < P <= S+F
The person is using more force than necessary to hold the ladder in place. They would do the minimum when P=S-F and friction would be at its limiting value, and the ladder would be on the verge of slipping. When
necessary friction for equilibrium = |S-P|
exceeds the limiting value F, movement will occur.

Another scenario would be when P=S. There would be zero net horizontal force as P-S=0, so no net horizontal acceleration, so the force exerted by friction would be zero, even when the surface is rough so mu>0. Mu, multiplied by R, gives the limiting friction value, the actual friction force generated is often less in which case things remain in equilibrium. Friction is only equal to its limiting value, when analysing the extreme equilibrium cases, where the system is on the equilibrium boundary.

Thanks for putting up with me. These are now my notes:

When a mass is on the point of moving, Friction is at its maximum and Fmax = uR
R cannot change unless the scenario changes.
A body can be in equilibrium without being in limiting equilibrium in which case friction < uR.
In the problem I am doing friction can be smaller than Fmax but if it is F <S so P has to compensate for the difference
As a minimum P = S-F

BUT P can bigger that that without moving the ladder - Yep, I know I can push things harder and harder without me moving them.

Are all my thoughts OK up to now?

Now this is the bit I can't get - the right hand side of the inequality. There must be a point where P is big enough but is Fmax now working opposite to P so P <= S+Fmax? If it's not that reason then I don't get it.

Basically ok.

When you push harder than S, friction now acts against the push as you're trying to push the ladder into the wall and the net force (ignoring friction) is towards the wall. Friction will stop this until limiting friction is reached.

The coefficient of friction has a simple interpretation which is sometimes ignored/ forgotten/... If you have a mass on a plane,
mu = tan(theta)
Where theta is the maximum angle the plane makes with the horizontal before equilibrium breaks. For angles less than this value, the mass is stationary, i.e. the required friction to hold the mass in place is less than the limiting value. For angles greater than it, equilibrium breaks and the mass slides. So when
mu = 1, limiting equilibrium is at 45 degrees
mu = Sqrt(3), limiting equilibrium is at 60 degrees
Mu = 1/sqrt(3), limiting equilibrium is at 30 degrees

See Angle of Friction in
https://en.m.wikipedia.org/wiki/Friction

Thank you for the "basically OK". I can live with that.

Thanks again. I love mechanics but I'm pretty confused a lot of the time.

I don't know if this will help, but for me, the "canonical example" of friction is a heavy object on a flat rough surface.

In the case where the object isn't moving, friction is always going to be equal and opposite to the sum of the other horizontal forces on the object (if the object isn't moving, then the sum of all horizontal forces including friction has to be zero).

So, for example, if you imagine two people having a "push of war" on opposite sides of the object, you can see that the frictional force will having to keep changing magnitude and/or direction depending on who is pushing harder at any particular time.