Force of the Ladder on a Wall Torque

[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.

 

[BvU]

Hello astoll, !

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!

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.

 

[astoll]

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.

 

[BvU]

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.

[edit] nice picture. I had a much simpler one

 

[haruspex]

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.

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.

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 F_f≤μ_sF_N.

 

[astoll]

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!