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

• Elias Waranoi
In summary, the upward force on the 3 meter ladder is 360 N and the upward force on the 4 meter ladder is 448 N.
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

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 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?

#### Attachments

• Namnlös.png
21.2 KB · Views: 747
Elias Waranoi said:
Fx*sin(53.13)*3
Where's Fy ?

BvU said:
Where's Fy ?
But Fy is pararell to the length of the torque so it shouldn't contribute to the torque right?

#### Attachments

• Namnlös.png
20.1 KB · Views: 805
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.

BvU
TomHart said:
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 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.

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.

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.

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.

TomHart

## 1. What is the definition of "sum of torques in equilibrium"?

The sum of torques in equilibrium refers to the balanced forces acting on an object that is not rotating. In other words, the net torque on the object is equal to zero, meaning there is no rotational acceleration.

## 2. How is the sum of torques in equilibrium calculated?

The sum of torques in equilibrium is calculated by multiplying the force applied to an object by the distance from the pivot point (known as the moment arm). This calculation takes into account both the magnitude and direction of the force.

## 3. What is the significance of the sum of torques in equilibrium?

The sum of torques in equilibrium is significant because it allows us to predict the behavior of an object in a state of rotational equilibrium. It helps us understand how forces acting on an object can affect its rotation or lack thereof.

## 4. How does the sum of torques in equilibrium relate to Newton's First Law of Motion?

The sum of torques in equilibrium is related to Newton's First Law of Motion, also known as the Law of Inertia. In this law, an object at rest or in motion will remain in its state unless acted upon by an external force. In the case of the sum of torques in equilibrium, the object will remain stationary if the net torque is equal to zero.

## 5. What are some real-life applications of the sum of torques in equilibrium?

The sum of torques in equilibrium is applicable in various real-life situations, such as balancing a see-saw or a ladder against a wall, opening a door, and even maintaining balance while riding a bike. It is also crucial in engineering and construction, where structures and machines need to be designed to maintain equilibrium and prevent unwanted rotations.

• Introductory Physics Homework Help
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
3K
• Introductory Physics Homework Help
Replies
13
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
952
• Introductory Physics Homework Help
Replies
11
Views
3K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
4K
• Introductory Physics Homework Help
Replies
1
Views
6K