Static equilibrium diagram help

Click For Summary

Homework Help Overview

The discussion revolves around a static equilibrium problem involving a uniform boom with a specified weight. Participants are analyzing the forces and moments acting on the boom, particularly focusing on the tension in a wire and the reaction forces at the hinge.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to establish the equilibrium conditions by applying the sum of moments and forces. Some participants clarify the application of forces at specific points along the boom and question the setup of the free body diagram.

Discussion Status

Participants are actively discussing the conditions for static equilibrium and have provided insights into resolving forces and moments. There is an ongoing exploration of the relationships between tension, weight, and the forces acting on the boom, but no consensus has been reached regarding the correctness of the calculations presented.

Contextual Notes

There is mention of a diagram that is not visible in the thread, which may limit the clarity of the discussion. The original poster expresses uncertainty about their free body diagram and the implications of the wire's attachment point.

naeblis
Messages
26
Reaction score
0
hi I've been looking at this problem for hours and I am not sure if I am approaching it right.

Q: The boom in the figure [check attachment] is uniform and weighs 2.20kN. The magnitude of the force exerted on the boom by the hinge is...

i knowthe boom has a mass of 224.261 kg, and summation tau = TLsin theta - Mg 2/3 L - mg 1/2 L if my free body diagram is right, not sure though the wire attaching at 2/3 L is really throwing me off hopefully some one can shed some insight on this.

i apologize ahead for my crapy diagrams but its the best i can do on ms paint.
 

Attachments

Physics news on Phys.org
For static equilibrium, the sum of the moment forces = 0.

Apply the weight (mg) of the boom at its center of mass, L/2.

The tension T in the wire is applied at 2/3 L.

The forces, boom weight (mg) and tension (T), must be resolved into components normal to the boom for the sum of the moments.
 
Astronuc said:
For static equilibrium, the sum of the moment forces = 0.
Apply the weight (mg) of the boom at its center of mass, L/2.
The tension T in the wire is applied at 2/3 L.
The forces, boom weight (mg) and tension (T), must be resolved into components normal to the boom for the sum of the moments.


ok so i have:

summation tau = 0 = T 2/3 L sin 60 - mg 1/2 L
T = 1.91 kN

and 0 = Tension + Weight + Force

so F = -4.11 kN

is that right?
 
0 = T 2/3 L sin 60 - mg 1/2 L
the weight of the boom acts downward (vertical), so one must use the Tension component normal to the boom.

In the second case, the reaction at the base of the boom must push (balance) the Tension and Weight components which are parallel to the boom.

-- Clarification

Basically, in this type of statics problem, the sum of the forces in horizontal (x), \Sigma{F_x}=0, sum of forces in vertical (y), \Sigma{F_y}=0, and sum of moments \Sigma{M}=0. A positive moment is counterclockwise, negative moment is clockwise. The moment is the product of the normal force (perpendicular) at point of application and the moment arm (distance between point of application and pivot point).
 
Last edited:

Similar threads

Replies
4
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
14
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
5
Views
5K
  • · Replies 7 ·
Replies
7
Views
4K
Replies
8
Views
2K
  • · Replies 32 ·
2
Replies
32
Views
4K
Replies
9
Views
4K