Tension on two cables holding a weight

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SUMMARY

The discussion focuses on calculating the tension in two cables supporting an 18.0 kg spotlight suspended 2.40 m below a horizontal pole. The cables, each 3.90 m long, form equal angles with the horizontal. The participant initially referenced an incorrect equation involving static friction, but was advised to utilize a free body diagram to analyze the forces acting on the spotlight. The correct approach involves resolving the tension forces into their horizontal and vertical components and applying the equilibrium condition that the sum of forces equals zero.

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  • Study how to create and analyze free body diagrams
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Homework Statement



A 18.0 kg spotlight is suspended 2.40m below a horizontal pole by two 3.90m long cables that angle upward at equal angles and reach from the top of the light to either end of the horizontal pole.


Homework Equations



The only equation I have for tension is T=[tex]\mu[/tex]*m*g/(cos[tex]\vartheta[/tex]+[tex]\mu[/tex]sin[tex]\vartheta[/tex])

The Attempt at a Solution


My problem is [tex]\mu[/tex] stands for the static friction, which there is none. Do I still need to use this equation or is there a different one I don't know about. Thanks!
 
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Welcome to PF!
Tension is just the force that a string or cable pulls with. You need a "free body diagram" showing the 3 forces acting on the load, with angles. Then you write
"sum of the forces = 0" for the vertical direction (and perhaps for the horizontal direction separately). The tension forces must be separated into their horizontal and vertical components to fit into these equations.
 
Thanks, I was making it way to difficult.
 

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