# Force of the Ladder on a Wall Torque

• astoll
In summary, you first solved the triangle. Hypotenuse is 5, height is 4, the last side is 3 and the angle is 53 degrees. You then solved for force of friction given Ff = usFN or (0.7)(1176) = 823.2 N. But because the ladder is about to slip, you find that the force of friction is not equal to the normal force.
astoll
Homework Statement
A 40-kg uniform ladder that is 5.0 m long is placed against a smooth wall at a height of h = 4.0 m,
as shown in the figure. The base of the ladder rests on a rough horizontal surface whose coefficient
of static friction with the ladder is 0.70. An 80-kg bucket is suspended from the top rung of the
ladder, just at the wall. What is the magnitude of the force that the ladder exerts on the wall?
Relevant Equations
sum of forces in the y direction = Fn - Fbucket - Fearth = 0
Sum of the forces in the X direction = Ff +Fladder -Fn = 0
Ff = usFn
Torque= Flsin(theta)
I've been working on this problem for a couple days now and I'm clearly missing something.
I first went ahead and solved the triangle. Hypotenuse is 5, height is 4, the last side is 3 and the angle is 53 degrees.
I went ahead and did the sum of forces in the y direction = Fn - Fbucket - Fearth = 0 to solve for Fn in the y direction and got (80)(9.8) + (40)(9.8) = Fn = 1176 N.
I then solved for force of friction given Ff = usFN or (0.7)(1176) = 823.2 N

I set the axis of rotation to where the bottom of the ladder meets the ground and set up torque equations.

The bottom of the ladder at the axis of rotation has no torque (tfriction, tFny =0).
Torque of the ladder is given at the center of mass of the ladder (392)(2.5)(sin37) = -589.78 N*m (clockwise)
Torque of the bucket is given by (784)(5)(sin37) = -2359.11 N*m (clockwise)
Torque of the normal Force in the x direction pushing off the wall is given by (FNx)(5)(sin53) = 3.99FN (counterclockwise)
Torque of the force of the ladder on the wall is (Fladder)(5)(sin127) = -3.99Fl (clockwise)

This gives me -2948.99 +3.99Fnx = 3.99Fl
-739 + FN = Fl
FN = - Fl
This will give me half of the answer I am looking for. Is my issue just that while there are equal and opposite forces pushing from the wall on the ladder and the ladder on the wall, they are not equal in torque? As in, the wall on the ladder will have a torque, but the ladder on the wall will not?

The correct answer is supposed to be 740 N. Any help on where I am going wrong would be appreciated.

Last edited by a moderator:
Hello astoll, !
astoll said:
I'm clearly missing something.
Yes, me too: A clear sketch of the situation !
Apart from the skidding at the foot of the ladder (to be checked separately), the torque balance around that foot will give you the desired answer. No need for sines either!

astoll said:
This gives me -2948.99 +3.99Fnx = 3.99Fl
No. righthand should be a zero: you set up a balance equation for the torque on the ladder.

Tip: use symbols and only substitute numbers as the very last step.

Apologizes, I couldn't figure out how to attach the image the first time. If the righthand side was zero, that would mean that there is a torque from the normal force on the ladder, but not a torque from the ladder on the wall? If this is true, I know how to complete the problem.

There is a force from the ladder on the wall, but it is not part of your torque balance. Only the reaction force -- the normal force from the wall on the ladder contributes a torque on the ladder.

The force from the ladder obviously is balanced somewhere (the wall does not move, we may assume), but that's not part of the system we consider.

 nice picture. I had a much simpler one

Last edited:
astoll said:
and the angle is 53 degrees.
It is rarely of any benefit to find the angle. You are only interested in trig functions of the angle, and those can be had by the ratios of the lengths. Going via the angle reduces accuracy.
astoll said:
the sum of forces in the y direction
To what end? Taking moments about the base of the ladder, as you did, gives all you need.
astoll said:
solved for force of friction given Ff = usFN
Wrong. You are not told the ladder is about to slip, so all you know about the relationship between the normal force and the frictional force is Ff≤μsFN.

I figured it out! Thank you everyone! For whatever reason, when I gave this problem to my AP class yesterday we had some discrepancies on the torque and I thought I would get a second opinion. Many thanks!

BvU

## What is the "Force of the Ladder on a Wall Torque"?

The "Force of the Ladder on a Wall Torque" is a physical concept that describes the rotational force or torque applied by a ladder leaning against a wall. It is an important factor to consider in ladder safety and stability.

## How is the "Force of the Ladder on a Wall Torque" calculated?

The torque is calculated by multiplying the force applied by the ladder (usually the weight of the person and any additional weight on the ladder) by the distance between the ladder's base and the point of contact with the wall. This distance is known as the moment arm.

## What factors affect the "Force of the Ladder on a Wall Torque"?

The force of the ladder on a wall torque is affected by several factors, including the weight of the person and any additional weight on the ladder, the angle of the ladder against the wall, and the friction between the ladder and the ground or wall.

## Why is it important to consider the "Force of the Ladder on a Wall Torque"?

It is important to consider the force of the ladder on a wall torque because it can significantly affect the stability of the ladder. If the torque is too high, it can cause the ladder to tip over, potentially resulting in injury. Understanding and properly managing this force is crucial for ladder safety.

## How can the "Force of the Ladder on a Wall Torque" be reduced?

The force of the ladder on a wall torque can be reduced by minimizing the weight on the ladder, using a ladder with a wider base, and increasing the angle of the ladder against the wall. Additionally, using non-slip materials on the ladder and the ground can help to increase friction and reduce the torque.

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