Tags:
1. Feb 2, 2015

### minimario

1. The problem statement, all variables and given/known data

When a ladder is learning against a wall with friction both on the wall and against the ground, how do you determine the direction of the 2 friction forces?

2. Relevant equations

3. The attempt at a solution

I found a pic online, why is this true?

2. Feb 2, 2015

### Simon Bridge

Look at the diagram - which way is the ladder going to slide?
What does the friction do to the rate the ladder slides?
Which direction does the friction have to point to do that?

3. Feb 3, 2015

### DEvens

This is actually a harder question than it might appear. The problem is, you have to make assumptions about the nature of the way the ladder is placed on the wall.

Suppose you start the ladder close to vertical and resting on the ground. Then you ever so gently and slowly lean it over until it touches the wall. At this point, there will be very little vertical force at the contact point with the wall. The contact point with the ground has to provide all the friction required to prevent sliding.

Now, let a painter climb up the ladder. He will bend the ladder slightly so the upper end moves down the wall slightly. So there will be frictional force involved, and it will of course point in the direction opposite to the motion. But as to how big this force is, it depends on how big the guy is and how springy the ladder is. And the coefficient of static and dynamic friction of the wall and the ladder. And probably a couple other things I'm not thinking of just now. But it will resist the bowing of the ladder to some degree.

Now let the guy get off the ladder again. The bowing will spring back. And the upper end of the ladder will move up the wall. And there will be friction force in the other direction. This friction will resist the "springing back" of the ladder to some degree.

It's an interesting question to think about. How much "bow" can the friction with the wall sustain? Some interesting geometry in there to resolve the forces.

4. Feb 3, 2015

### haruspex

I agree with all that, but I'd like to go into a couple of details.
As the painter starts to climb the ladder, static friction will initially stop the top end sliding. Assuming it does eventually slide, it will only slide to the point where kinetic friction is sufficient to maintain balance (I'm ignoring momentum). Similarly on the descent, so it is highly likely that after the painter has stepped off the ladder the friction from the wall acts downwards. It will be static friction, but the force's magnitude will correspond to the kinetic coefficient.

5. Feb 3, 2015

### Simon Bridge

This is why the context is important.
There are no physics problems where you do not need to , make some assumptions.

In the above, there is no mention of additional loads on the ladder, so assuming no painter is reasonable. None of the other points make any difference to the question actually stated... re: direction of the friction force. This one only makes a momentary difference... these problems usually concern either a static equilibrium or the ladder is sliding.

All that has been demonstrated is that it is possible to complcate the problem by adding extraneous influences. This is always possible. But if it is not in the problem statement, you need a reason to include it.

It does not matter how little friction is present at the contact points... the context is for working it out: which way you draw the arrows? It may well be that a valid approximation can be made taking one friction force to be zero... however, the supplied diagram does not support this.

... and so on.
Context is everything.

6. Feb 3, 2015

### haruspex

I think you are missing the point of the imaginary painter argument. It demonstrates that there is insufficient information to determine which way the frictional force from the wall points. There is a continuum of possibilities consistent with the statics equations.

7. Feb 3, 2015

### Nathanael

If you treat the ladder as rigid, then would it be impossible for the frictional forces to point the other way?

8. Feb 3, 2015

### haruspex

It doesn't solve the problem that there is a range of solutions to the equations. If there is sufficient coefficient of static friction from the ground, that range will include a downwards frictional force from the wall. In the real world (where nothing is completely rigid), this is resolved by the history (Simon's 'context').