Why Isn't the Sum of Torques Zero in This Ladder Equilibrium Problem?

[Elias Waranoi]

Homework Statement

Two ladders, 4.00 m and 3.00 m long, are hinged at point A and tied together by a horizontal rope 0.90 m above the floor (Fig. P11.89). The ladders weigh 480 N and 360 N, respectively, and the center of gravity of each is at its center. Assume that the floor is freshly waxed and frictionless. (a) Find the upward force at the bottom of each ladder. (b) Find the tension in the rope. (c) Find the magnitude of the force one ladder exerts on the other at point A

Homework Equations

τ = F×L

The Attempt at a Solution

So with the help of my textbook I was able to solve (c) by taking the sum of vertical forces of the right ladder F_normal - F_gravity = F_y and sum of horizontal forces F_tension = F_x and correctly got the magnitude of the force one ladder exerts on the other at point A to be 335 Newton.

This got me thinking though, since the ladder is in static equilibrium and we have all the forces acting on the right ladder then the sum of torque τ on that ladder should be zero right? I did the maths and got F_gravity*cos(53.13)*1.5 + F_tension*0.9 - F_x*sin(53.13)*3 = -160 where 53.13 is the angle in degrees between the floor and right ladder, 1.5 is half the length of the ladder and 3 is the length of the ladder. This is not zero, why not? Is my math wrong? Are there other forces affecting the torque about the point between the right ladder and the floor that I did not account for?

 

[BvU]

F_x*sin(53.13)*3

Where's F_y ?

 

[Elias Waranoi]

Where's F_y ?

But F_y is pararell to the length of the torque so it shouldn't contribute to the torque right?

 

[TomHart]

The perpendicular distance of Fy to the moment point is not zero. If you draw a vertical line downward from Fy, the perpendicular distance from the moment will be where that line intersects the ground.

 

[Elias Waranoi]

The perpendicular distance of Fy to the moment point is not zero. If you draw a vertical line downward from Fy, the perpendicular distance from the moment will be where that line intersects the ground.

oops, thank you. But it's still weird, with counter-clockwise rotation being the positive torque I get F_gravity*cos(53.13)*1.5 + F_tension*0.9 - F_x*sin(53.13)*3 - F_y*cos(53.13)*3 = -321. If I add F_y as a positive torque I get zero for some reason. F_gravity is 360N, F_tension is 322N, F_x is F_normal - F_gravity = 89N and F_y is F_tension = 322N.

 

[TomHart]

Did you find Fy to be an upward force acting on the 3 meter ladder? I thought it was a downward force as it acted on the 3 meter ladder, and an upward force on the 4 meter ladder.

 

[Elias Waranoi]

Well the answer for (a) quoted from the book "449 N (3.00-m ladder)" and from how I interpret the question, the 3.00m ladder is the one weighing 360 N. To get vertical force equilibrium I can't make this into anything but an upward force on the 3 meter ladder.

 

[TomHart]

The normal force (an upward force) on the 3 meter ladder is 449 N. Is that what you are saying? If so, I agree. The weight of the 3 meter ladder is 360 N. At this point your upward force is greater than your downward force by 89 N. Don't you need an addition 89 N downward to achieve equilibrium?

 

[Elias Waranoi]

...!
You're right! Everything adds up now :'D Thank you very much for helping me out.