Torque/moment of beam in equilibrium

In summary, the conversation discusses an equilibrium problem involving the sum of forces and moments. The individual provides a summary of their attempt at solving the problem, including drawing free body diagrams and formulating equations. They also receive guidance and clarification from another individual. Eventually, they are able to solve the problem and express satisfaction with their understanding and progress.
  • #1
Hyperfluxe
35
0

Homework Statement


http://i.imgur.com/5plXa.png

Homework Equations


ƩFx = 0, ƩFy = 0, ƩM = 0

The Attempt at a Solution


I realize that this is an equilibrium problem and that the sum of the forces in the x and y directions is 0, and the sum of the moments is 0. I draw two free body diagrams, one at point A (the contact point between the left wall and beam), and one at point B (the edge). Each FBD has the forces exerted by the bar on the wall - I call it Fa and Fb respectively. Also, both points have normal forces which are perpendicular to the surface. FNa is perpendicular to the wall while FNb is perpendicular to the rod/edge. Also incorporating the weights (Wa and Wb), I get four equations.

What I can't figure out is how to formulate an equation for the sum of the moments, and subsequently solving the problem for theta. Any help would be tremendously appreciated.
 
Physics news on Phys.org
  • #2
Hi Hyperfluxe! :smile:
Hyperfluxe said:
I realize that this is an equilibrium problem and that the sum of the forces in the x and y directions is 0, and the sum of the moments is 0. I draw two free body diagrams, one at point A (the contact point between the left wall and beam), and one at point B (the edge). Each FBD has the forces exerted by the bar on the wall - I call it Fa and Fb respectively. Also, both points have normal forces which are perpendicular to the surface. FNa is perpendicular to the wall while FNb is perpendicular to the rod/edge. Also incorporating the weights (Wa and Wb), I get four equations.

(you don't need to separate the weight, one W will do :wink:)

You should be able to find two equations for FNa and θ.

What is your moments equation (about B)?
 
  • #3
Ok, what I did instead is making a sum of the moments equation about point A. So, from my FBD, I get 3 equations. F1 is the force between the left contact point and the wall, and F2 is the force from the edge contact point. W is the weight (assuming it acts down at the center of the bar), I get:

1 - F1 = F2sin(theta) ----> I didn't need this equation

2 - F2cos(theta) = W

3 - (F2cos(theta))(0.325)-(W)(x) + (F2sin(theta))(y)
where x=0.5cos(theta) and y=0.325tan(theta) from geometry.

I equate W, cancel out F2, and solve for theta using a some algebra and a trig identity to get theta = 29.98degrees.

Is my method and answer correct? Thanks!
 
  • #4
Hi Hyperfluxe! :smile:
Hyperfluxe said:
2 - F2cos(theta) = W

3 - (F2cos(theta))(0.325)-(W)(x) + (F2sin(theta))(y)
where x=0.5cos(theta) and y=0.325tan(theta) from geometry.

I equate W, cancel out F2, and solve for theta using a some algebra and a trig identity to get theta = 29.98degrees.

Is my method and answer correct? Thanks!

Yes, your method is fine. :smile:

(though you could just have said F20.5secθ, instead of (cosθ + sinθtanθ) :wink:)

(i haven't chekced your numerical result)
 
  • #5
Thank you very much! Turns out I didn't have enough knowledge on moments on Friday, so I wasted 2 hours trying to solve that question. It feels really satisfying right now though!
 
  • #6
(You obviously needed practice on moments, so I didn't suggest this before …)

There are only three external forces, and you know the directions of all of them

so can you see a way to find θ just by drawing some lines on the diagram? :wink:
 

What is meant by torque/moment of beam in equilibrium?

The torque or moment of a beam in equilibrium refers to the force that causes rotation around an axis, often referred to as the pivot point. In other words, it is the measure of the tendency of a force to cause an object to rotate.

How is torque/moment of beam in equilibrium calculated?

The torque/moment of a beam in equilibrium is calculated by multiplying the force applied to the beam by the distance from the pivot point to the point where the force is applied. The formula is T = F x d, where T is torque, F is force, and d is distance.

What are the units for torque/moment of beam in equilibrium?

The units for torque/moment of beam in equilibrium are Newton-meters (N·m) or foot-pounds (ft·lb) in the SI and Imperial systems, respectively. These units represent the product of force and distance, which are Newtons (N) and meters (m) in the SI system, and pounds (lb) and feet (ft) in the Imperial system.

How is the direction of torque/moment of beam in equilibrium determined?

The direction of torque/moment of beam in equilibrium is determined by the right-hand rule. Point your right-hand thumb in the direction of the force and curl your fingers towards the pivot point. The direction your fingers are pointing in is the direction of the torque/moment.

What is the significance of torque/moment of beam in equilibrium in engineering and physics?

The torque/moment of beam in equilibrium is a crucial concept in both engineering and physics. It is used to analyze the stability and strength of structures, as well as to design and optimize machines and mechanical systems. In physics, it is a fundamental principle in understanding rotational motion and the conservation of angular momentum.

Similar threads

Replies
6
Views
684
  • Introductory Physics Homework Help
Replies
13
Views
782
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
988
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
8K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
3K
  • Introductory Physics Homework Help
Replies
6
Views
1K
Back
Top