Static equilibrium diagram help

In summary, the conversation is discussing a static equilibrium problem involving a boom with a uniform weight. The tension in the wire and the forces acting on the boom must be taken into consideration in order to solve the problem. The individual provides equations and calculations to determine the tension and force exerted on the boom.
  • #1
naeblis
26
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.
 

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  • #2
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.
 
  • #3
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?
 
  • #4
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), [itex]\Sigma{F_x}[/itex]=0, sum of forces in vertical (y), [itex]\Sigma{F_y}[/itex]=0, and sum of moments [itex]\Sigma{M}[/itex]=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).
 
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FAQ: Static equilibrium diagram help

What is a static equilibrium diagram?

A static equilibrium diagram is a visual representation of the forces acting on an object at rest. It shows the magnitude and direction of each force, as well as the object's center of mass.

Why is a static equilibrium diagram important?

A static equilibrium diagram is important because it helps us analyze and understand the forces that keep an object in a state of rest. It allows us to determine if the object is in a stable, unstable, or neutral equilibrium.

How do you draw a static equilibrium diagram?

To draw a static equilibrium diagram, you first need to identify all the forces acting on the object and their respective magnitudes and directions. Then, you can draw a free body diagram, which is a simplified version of the object with all the forces acting on it. Finally, you can use vector addition to draw the equilibrium diagram, with the forces represented by arrows pointing in the direction of the force and their lengths proportional to their magnitudes.

What is the difference between static and dynamic equilibrium diagrams?

The main difference between static and dynamic equilibrium diagrams is that static equilibrium diagrams show forces on an object at rest, while dynamic equilibrium diagrams show forces on an object in motion. In dynamic equilibrium, the forces are balanced, but the object is still moving at a constant velocity.

How can a static equilibrium diagram be used in real-world applications?

A static equilibrium diagram can be used in various real-world applications, such as engineering, architecture, and physics. It can help engineers and architects design stable structures and determine the maximum load that an object can withstand. In physics, it is used to analyze and understand the forces acting on objects in a state of rest.

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