Why Does This Statics Solution Use These Specific Equations?

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The discussion centers on understanding the equations used in a statics problem involving tension in cords and angles. Participants analyze the calculations for angles and the forces acting on a pulley system, emphasizing the importance of free body diagrams (FBDs) for clarity. The tension in the cords is clarified, noting that tensions are equal on both sides of an ideal pulley, which is crucial for solving the problem. There is confusion about the formation of the equations, with a focus on the equilibrium of the system and the relationship between the tensions and weights. The conversation highlights the necessity of correctly identifying forces and their components to arrive at the correct solution.
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I did arctan(12/5) to get the degrees + 180 to get the angle from the X axis to B, then I just split the angles into components.

I did 100*cos(246) + 100*cos(theta) = 0, and I got 67 degrees.
Then I was doing the Y components, and realized they were equal and opposite, so it would equal 0, which the weight can't be equal to.

So I looked at the solution, and here how the answer is found.

Sum of X: 100cos(theta) = Wa(5/13)
Sum of Y: 100sin(theta) = Wa(12/13)+Wa

Why is the answer figured like this, and when do I do it differently from sum of the components? I'm totally lost, but I have a slight idea.

Thanks.
 
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Why, did you add the 180 degrees? Assuming the x-axis is at the center of the pully the angle is simply tan^-1(12/5)?

Also, draw a simple free body diagram to find the force vector components. You will find the actual numerical value of the angle is not needed.

Thanks
Matt
 
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I meant the positive X axis.

Also, I did draw a free body diagram, but I am not sure why I couldn't add the sum of the components.
 
You seem to be assuming that the tension in cord CD and BC are both 100 lbs. That is not correct. Also, you are not noting that the tension in cords draped around frictionless ideal pulleys are the same on both sides of the wrap (that is, Tension in AC = Tension in BC). Draw an FBD of the hanging weight, and redo your FBD of the pulley at joint C.
 
The tension in AC = BC? Do you mean the tension in BC = DC?

I still don't see how the equation was formed, could someone explain how the equation was determined?
 
akhmed966 said:
The tension in AC = BC?
Yes. The cord ACB is one continuous chord wrapped around an ideal pulley, in which case AC = BC.
Do you mean the tension in BC = DC?
No.
I still don't see how the equation was formed, could someone explain how the equation was determined?
The system is in equilibrium. Draw a FBD of the weight to determine that the tension in AC must be W_a. Then draw an FBD of the pulley at C, and you should be able to get the same equations as the book's to solve for theta and W_a. Which cord do you think has the 100 pound tension load?.
 

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