Static equilibrium diagram help

AI Thread Summary
The discussion focuses on solving a static equilibrium problem involving a boom weighing 2.20 kN, with tension applied at 2/3 L. The participant calculates the tension (T) as 1.91 kN and the force exerted by the hinge as -4.11 kN, using the equations for summation of moments and forces. Key points include the need to resolve forces into components normal to the boom and the importance of balancing horizontal and vertical forces. The participant seeks confirmation on their calculations and understanding of static equilibrium principles. Overall, the thread emphasizes the application of static equilibrium equations to analyze forces and moments in the given scenario.
naeblis
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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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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).
 
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