# Sum of torques in equilibrium

Tags:
1. Mar 23, 2017

### Elias Waranoi

1. The problem statement, all variables and given/known data
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

2. Relevant equations
τ = F×L

3. 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 Fnormal - Fgravity = Fy and sum of horizontal forces Ftension = Fx 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 Fgravity*cos(53.13)*1.5 + Ftension*0.9 - Fx*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?

#### Attached Files:

• ###### Namnlös.png
File size:
30.2 KB
Views:
32
2. Mar 23, 2017

### BvU

Where's Fy ?

3. Mar 23, 2017

### Elias Waranoi

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

#### Attached Files:

• ###### Namnlös.png
File size:
30 KB
Views:
31
4. Mar 23, 2017

### 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.

5. Mar 23, 2017

### Elias Waranoi

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

6. Mar 23, 2017

### 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.

7. Mar 23, 2017

### 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.

8. Mar 23, 2017

### 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?

9. Mar 23, 2017

### Elias Waranoi

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