How Does Ladder Stability Change with Different Forces and Distances?

In summary, the problem involves finding the force at the top and bottom of a ladder leaning against a frictionless rain gutter. The normal force and friction are assumed to be zero due to the choice of pivot point at the bottom of the ladder. The force at the top of the ladder is found to be 133N and the force at the bottom is found to be 795N. It is determined that the normal force acts perpendicular to the surface of contact, which in this case is the edge of the gutter. The direction of the force is dependent on the differentiable surface, with the gutter exerting a force perpendicular to the ladder and the wall exerting a force parallel to it. If there were friction, it would act parallel to
  • #1
Jazz
103
5

Homework Statement


image.jpg


Where it says ''from the bottom'' I assumed it's referring to a distance along the ladder. So:
image.jpg


Data:

##w_{ladder} = 98.0\ N##

##w_{person} = 686\ N##

##d_1 = 4\sqrt(2)\ m##

##d_2 = 1\ m##

##d_3 = 2/3\ m##

Homework Equations



##\sum \tau = 0##

##\sum F = 0##

The Attempt at a Solution


[/B]
Diagram:

image.jpg
image.jpg


I chose my pivot point to be the bottom of the ladder. Hence, torques due to normal force ##N## and friction ##f## are zero. ##ccw## is positive and ##cw## is negative.

##0 = w_pd_2 + w_ld_3 - F_wd_1##

##F_wd_1 = w_pd_2 + w_ld_3##

##F_w = \frac{w_pd_2 + w_ld_3}{d_1} = \frac{686\ N \cdot 1.00\ m + 98.0\ N \cdot 2/3\ m}{4\sqrt{2}\ m} = 133\ N##

##F_w = f = 133\ N##

Then:

##N = w_p + w_l = 686\ N + 98.0\ N = 784\ N##

The force at the top of the ladder is just ##133\ N##.
The force at the bottom is ##\small{\sqrt{N^2+ f^2} = \sqrt{ (784\ N)^2+ (133\ N)^2} = 795\ N}##. Doubt: this magnitude, whether it's correct or not, acts along the ladder, right? (as a tension would?).

The textbook's solutions are ##126\ N## and ##751\ N##, respectively. It seems I haven't missed anything, so I don't know where the mistake is.
 
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  • #2
Which direction does Fw point ?
 
  • #3
BvU said:
Which direction does Fw point ?

To the right.

And in a cw direction about the pivot point.
 
  • #4
No vertical component ?
 
  • #5
BvU said:
No vertical component ?

Components of ##F_w##? Mmm, It didn't occur to me that there could be vertical components there. Has that of a ''frictionless rain gutter'' something to do with it?
 
  • #6
It seems that I also forgot to take ##F_w## into account when finding the force at the bottom.
 
  • #7
Jazz said:
Components of ##F_w##? Mmm, It didn't occur to me that there could be vertical components there. Has that of a ''frictionless rain gutter'' something to do with it?
Frictionless says there will only be a normal force. Normal means perpendicular to the surface of contact. What direction will that be? You should assume the ladder projects slightly beyond the point of contact with the gutter.
 
  • #8
Jazz said:
Components of ##F_w##? Mmm, It didn't occur to me that there could be vertical components there. Has that of a ''frictionless rain gutter'' something to do with it?
Consider a more extreme case where the ladder is at 45 degrees (or even less) wrt the ground. Would it lean ON the gutter, as well as against the gutter ?
 
  • #9
I think I'm getting the idea. It would be something like this:

gutter.jpg
However I'm not sure what is the correct direction that ##F_w## takes.

I think the sharp corner of the gutter can be viewed as a fulcrum. Should ##F_w## act perpendicular to the edge of the gutter, as bisecting it? If that were the case, my drawing wouldn't be totally accurate. Would it be in that direction, independent of the angle that the ladder makes with the ground?.
 
  • #10
Jazz said:
Should ##F_w## act perpendicular to the edge of the gutter, as bisecting it?
No. It might help to rotate the problem around to a more familiar set-up. Imagine a table on flat level ground, but the legs of the table are sharp wedges, meeting the ground at some odd angle. Which way would the normal force act?
More generally, as I wrote, normal means perpendicular to the contact plane. I'm not sure how contact planes are defined generally, but if either surface is differentiable then it will be the tangent plane which passes through the point of contact. Taking the ladder uprights to be rectangular in section, they have a differentiable surface. Or if you take them to be arbitrarily thin rods then it must be a plane containing that rod and containing the edge line of the gutter - there is only one such. Sounds complicated, but the underlying idea is fairly intuitive.
 
  • #11
Another approach: ladder pushes against gutter. In which direction ? Gutter pushes back, opposite direction (no friction).

If there were a rubber hose between the two, how would it compress ? To a vertical ellipse ? Or to an ellipse with the long axis parallel to the ladder ?
 
  • #12
I would say to an ellipse with the long axis parallel to the ladder, which would mean that the force exerted by the gutter on the ladder is perpendicular to it.

In the other case (leaning against a wall), the wall becomes the differentiable surface and the rubber hose would compress parallel to it. It makes sense.

If the gutter weren't frictionless, friction would act parallel to the ladder's structure, right?
 
  • #13
Jazz said:
I would say to an ellipse with the long axis parallel to the ladder, which would mean that the force exerted by the gutter on the ladder is perpendicular to it.

In the other case (leaning against a wall), the wall becomes the differentiable surface and the rubber hose would compress parallel to it. It makes sense.

If the gutter weren't frictionless, friction would act parallel to the ladder's structure, right?
Yes, I agree with all of that.
 

Related to How Does Ladder Stability Change with Different Forces and Distances?

1. What is the "Ladder Problem" in terms of stability?

The "Ladder Problem" is a basic physics problem that involves determining the stability of a ladder when it is leaning against a wall. The ladder is considered stable if it does not tip over when a person is climbing up or down it.

2. What factors affect the stability of a ladder?

The stability of a ladder is affected by several factors, including the weight of the ladder itself, the weight of the person climbing on it, the angle at which the ladder is leaning against the wall, and the friction between the ladder and the ground or wall.

3. How do you calculate the stability of a ladder?

The stability of a ladder can be calculated using the "ladder rule," which states that the base of the ladder should be one-quarter of the working length of the ladder away from the wall. This ensures that the ladder is at a safe and stable angle.

4. What happens if the ladder is not stable?

If the ladder is not stable, it can tip over and cause injury to the person climbing on it. It can also damage surrounding objects or structures and potentially lead to further accidents.

5. Can stability be improved for a ladder?

Yes, there are several ways to improve the stability of a ladder. These include using a ladder with non-slip feet, securing the top of the ladder to the wall or a sturdy object, and making sure the ground is level and free of debris. It is also important to follow proper ladder safety guidelines and not exceed the weight limit for the ladder.

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