# Forces exerted on the ladder by the wall and ground

• moenste
In summary: N?2. Then we multiply the F force (506 N) by the perpendicular line (8 m). This gives us 4050 N.In summary, The wall exerts a force of 506 N at 90 degrees to the ladder, and a force of 1121 N at 63 degrees to the ground.
moenste

## Homework Statement

A uniform ladder which is 10 m long and weights 300 N leans with its upper end against a smooth vertical wall and its lower end on rough horizontal ground. The bottom of the ladder is 6 m from the base of the wall. A man weighting 700 N stands on the ladder at a point 6 m above the ground. Calculate the magnitudes and directions of the forces exerted on the ladder by (a) the wall, (b) the ground.

Answers: (a) 506 N at 90 degrees to the wall, (b) 1121 N at 63 degrees to the ground.

2. The attempt at a solution
At first I did a graph of the situation:

R = normal reaction to the wall, S = to the ground, F = friction (rough ground).

Find 8 by Pythagoras. S is = 700 N + 300 N = 1000 N.

And after this point I am a bit stuck. From a different book example I get:
(F * 8) + (300 * 3) + (700 * 1.5) = (S * 6)
8 F = 4050
F = 506.25 N

As I understand, the body is in equilibrium so in that case: normal reaction S (anti-clockwise) should be multipled by the perpendicular line which is 6 m -> (S * 6), then all of other forces which are facing clockwise direction should be used: we have F and we sort of move it to the left so the perpendicular line is 8 m (F * 8), two other clockwise forces are 700 and 300. 300 * 3 and 700 * 1.5 (the vertical side of the triangle is 2 m, the hypotenuse is 2.5 m and by Pythagoras the horizontal line is 1.5). So we get F = 506 N. But how to find the angle? Because F is a force directed towards the wall that is the reason why it is equal to 90? And why don't we take into account the R force?

As I understand, the body is in equilibrium so in that case: normal reaction S (anti-clockwise) should be multipled by the perpendicular line which is 6 m -> (S * 6), (1) then all of other forces which are facing clockwise direction should be used: we have F and we sort of move it to the left so the perpendicular line is 8 m (F * 8), two other clockwise forces are 700 and 300. 300 * 3 and 700 * 1.5 (the vertical side of the triangle is 2 m, the hypotenuse is 2.5 m and by Pythagoras the horizontal line is 1.5). So we get F = 506 N. But how to find the angle? (2) Because F is a force directed towards the wall that is the reason why it is equal to 90? And why don't we take into account the R force?

(1) Here you calculate a contribution to the total torque ##\Sigma \vec \tau## which must be 0 for equilibrium. Apparently with the top of the ladder as axis of rotation, considering you move F to the left over 6 m. And that's why R doesn't contribute. That gives you F. And sum of forces horizontal = 0 gives you R

(2) The wall is smooth, so it can only exercise a force perpendicular to itself.

CWatters and moenste
BvU said:
(1) Here you calculate a contribution to the total torque ##\Sigma \vec \tau## which must be 0 for equilibrium. Apparently with the top of the ladder as axis of rotation, considering you move F to the left over 6 m. And that's why R doesn't contribute. That gives you F. And sum of forces horizontal = 0 gives you R
(2) The wall is smooth, so it can only exercise a force perpendicular to itself.
Vertically: we find S = 700 + 300 = 1000 N

Then we take moments at the point where the ladder touches the wall: (700 * 1.5) + (300 * 3) + (F * 8) = (1000 * 6) -> F = 506 N. In the answer 506 N at 90 degrees to the wall but isn't 506 the F force? I mean the friction with the ground? Why is it 90 degrees then?

If we take the horizontal forces R = F (similarly as the vertical forces) then R = 506 N and then ok the force is at 90 degrees to the wall. But in that case what is the 1121 N at 63 degrees to the ground answer? Shouldn't it be F? But in that case we have already found F and it is equal to 506 N.

Really lost on this one.

Update:

1. First we find 8 m the vertical side by Pythagoras. Then 3 m at the 300 N because it's the midpoint of the triangle and 1.5 m at 700 N because it's also a midpoint of the smaller traingle.
2. Vertically: S = 700 + 300 = 1000 N.
3. Moments about ladder touching the wall: (700 * 1.5) + (300 * 3) + (F * 8) = (1000 * 6) -> F = 506 N.
4. Horizontally: F = R -> R = 506 N.
5. R is the normal reaction to the wall and that's why it's at 90 degrees.
6. Now we need to find the force exerted on the ladder by the ground. We take it as RG and it's the force which on noted as "5 m" on the graph. By Pythagoras: 506.252 + 10002 = 11212 N.
7. Now we're looking for the angle which is below the 5 m line (the angle which the line makes with the horizontal / ground). In that case: tan θ (1000 / 506.25) -> θ = 63.2 degrees.

Looks logical to me. Could't understand that the force exerted on the ladder by the wall is not the friction (F) or the normal reaction (S) but the hypotenuse (the ladder) of these forces. And of course F = R and R is 90 degrees and it's logical.

Is everything right now? I am a bit unsure of the first part where I find 1.5 m of 700 N. The smaller traingle has 4, 5 and 3 (ver, hyp, hor) as it's side. If the smallest triangle has a 2 m vertical side, which is the middle point of the vertical side of the small triangle, then, as I understand, I can conclude that the smallest triangle has 1.5 as a horizontal. side.

Last edited:
ladder by the wall is not the friction (F) or the normal reaction (S) but the hypotenuse (the ladder) of these forces.
You mean floor. That force is the vector sum of F and S. Is not in the direction of the ladder.

moenste
BvU said:
You mean floor. That force is the vector sum of F and S. Is not in the direction of the ladder.
Oh yes, I meant floor. The force exerted on the ladder by the floor is not the fliction (F) or the normal reaction (S) but the vertor of these forces.

One quick question:

1. First we find 8 m the vertical side by Pythagoras. Then 3 m at the 300 N because it's the midpoint of the triangle and 1.5 m at 700 N because it's also a midpoint of the smaller traingle.

I am a bit unsure of the first part where I find 1.5 m of 700 N. The smaller traingle has 4, 5 and 3 (ver, hyp, hor) as it's side. If the smallest triangle has a 2 m vertical side, which is the middle point of the vertical side of the small triangle, then, as I understand, I can conclude that the smallest triangle has 1.5 as a horizontal side.

This logic is correct?

You can check for yourself: make your drawing a bit better to scale !

moenste

## 1. What is the difference between the forces exerted on the ladder by the wall and ground?

The force exerted on the ladder by the wall is known as the normal force, which is perpendicular to the surface of the wall. The force exerted by the ground is the weight of the ladder, which is directed downwards towards the ground.

## 2. How do these forces affect the stability of the ladder?

The forces exerted by the wall and ground work together to keep the ladder in a stable position. If the normal force from the wall is greater than the weight of the ladder, the ladder will stay in place. However, if the weight of the ladder is greater than the normal force from the wall, the ladder may slip or topple over.

## 3. Can the forces exerted on the ladder change during use?

Yes, the forces exerted on the ladder can change during use. If the person on the ladder shifts their weight or moves the ladder, the forces exerted by the wall and ground may also change. This is why it is important to always properly secure the ladder and be mindful of your movements while using it.

## 4. How do the materials of the wall and ground impact the forces exerted on the ladder?

The materials of the wall and ground can impact the forces exerted on the ladder. For example, a smooth and flat wall will provide more friction and therefore a greater normal force compared to a rough and uneven wall. Similarly, soft ground may not be able to support the weight of the ladder as well as hard ground.

## 5. Are there any safety precautions to consider when using a ladder?

Yes, there are several safety precautions to consider when using a ladder. Always make sure the ladder is on a stable and level surface, and that it is properly secured to prevent slipping. Do not exceed the weight capacity of the ladder and avoid leaning too far to one side. Lastly, always use caution and common sense while on a ladder to prevent accidents or injuries.

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