Struggling Engineering Student Needs Help with Tension in Cable Homework

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Discussion Overview

The discussion revolves around a homework problem related to tension in a cable within the context of statics, specifically focusing on the equilibrium of forces and torques acting on a beam. Participants explore the application of torque equations and the placement of force vectors in a diagram to solve the problem.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant states the equations for tension components in the cable, Tx = Tcos150 and Ty = Tsin150, and expresses uncertainty about finding the moment arm.
  • Another participant suggests summing the torques about point A, noting that the beam's weight creates a clockwise torque while the wire must create an anticlockwise torque, leading to the conclusion that these torques must sum to zero.
  • A participant attempts to simplify the problem by replacing the diagonal wire with a vertical wire and calculates T = 200kN based on this assumption, but later questions the discrepancy with the expected answer of T = 100kN.
  • One participant asks where the weight of the beam acts and encourages the drawing of its force vector on the diagram.
  • A later reply indicates that the participant realized the weight's force vector should be drawn from the middle of the beam, leading to a better understanding of the moment arm relationship.
  • Another participant points out that the equations previously written were for forces, not torques, prompting a reevaluation of the moment arms involved.

Areas of Agreement / Disagreement

The discussion reflects a progression of understanding among participants, with some clarifications made regarding the application of torques and forces. However, there is no explicit consensus on the correct approach to the problem, as participants continue to explore different aspects of the solution.

Contextual Notes

Participants express uncertainty regarding the moment arms and the placement of force vectors, indicating that the solution may depend on these factors. The discussion also highlights the distinction between forces and torques, which remains a point of confusion for some participants.

CheesyPeeps
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Homework Statement


2013-Q4.jpg


Homework Equations


∑MA = 0, where A is the pin.
Tx = Tcos150
Ty = Tsin150

The Attempt at a Solution


I've drawn in the 100kN force of the beam's weight, and identified that ∑MA = 0. I attempted to find the moment arm, but really wasn't sure how do to it, or if it would even help.

I'm a first year engineering student who's struggling with statics. Any help is appreciated, but please do your best to keep it as easy to understand as possible.
 

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CheesyPeeps said:

Homework Statement


View attachment 232980

Homework Equations


∑MA = 0, where A is the pin.

Tx = Tcos150
Ty = Tsin150

150?

The Attempt at a Solution


I've drawn in the 100kN force of the beam's weight, and identified that ∑MA = 0. I attempted to find the moment arm, but really wasn't sure how do to it, or if it would even help.

I'm a first year engineering student who's struggling with statics. Any help is appreciated, but please do your best to keep it as easy to understand as possible.

Try summing the torques about A.

The mass of the beam creates a clockwise torque. The wire must create an anticlockwise torque. It's not accelerating so these torques must sum to zero.

Perhaps temporarily replace the diagonal wire with a vertical wire?
 
CWatters said:
150?
Try summing the torques about A.

The mass of the beam creates a clockwise torque. The wire must create an anticlockwise torque. It's not accelerating so these torques must sum to zero.

Perhaps temporarily replace the diagonal wire with a vertical wire?

Thank you very much for your advice.
I replaced the wire with a vertical wire equal to Tsin30, and then wrote that Tsin30-100kN = 0. That means T = 200kN, but I checked the answer to the question, and it's T = 100kN. Have I missed something? I have the moments/torque chapter of my textbook open in front of me, but it's not helping much :confused:
 
Where does the weight of the beam act? If you were to draw its force vector, where would you place it on the diagram?
 
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gneill said:
Where does the weight of the beam act? If you were to draw its force vector, where would you place it on the diagram?

Thank you so much! I got it!
I drew the weight's force vector down from the middle of the beam, and when I did that, it finally clicked that the moment arm of the weight would be half the moment arm of Tsin30. I had been stressing over this question since yesterday - I'm so glad I understand it now. Thank you again!
 
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CheesyPeeps said:
Thank you very much for your advice.
I replaced the wire with a vertical wire equal to Tsin30, and then wrote that Tsin30-100kN = 0.

Those aren't torques they are forces.
 
CWatters said:
Those aren't torques they are forces.

Yeah, I see that now. I wasn't thinking at all about the moment arm, but once I spotted the fact that Tsin30 and the weight have different moment arms, I got it. Thank you again for your help! Statics has been driving me mad, so I'm really glad I understand this now :oldbiggrin:
 

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