How Do You Calculate Tension in Angled Cables?

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SUMMARY

The discussion focuses on calculating the tension in two angled cables supporting a 21.0 kg loudspeaker suspended 2.40 m below the ceiling using the equation T = mg/2sin(theta). The user initially misapplied the angle in degrees instead of radians, which is critical for accurate trigonometric calculations. The correct approach involves using the sine of the angle derived from a 3-4-5 triangle, yielding a tension value of approximately 171.675 N. The importance of understanding trigonometric functions and their applications in static equilibrium is emphasized.

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  • Ability to construct and interpret free body diagrams
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[SOLVED] Tension in two cables

Homework Statement
21.0 kg loudspeaker is suspended 2.40 m below the ceiling by two 3.90 m long cables that angle outward at equal angles.

The attempt at a solution
So my homework comes from the MasteryingPhysics online homework.

What I first start out with is my free body diagram. I have 3 arrows, one pointing down for weight of the speakers by the Earth (Wse) and two arrows pointing up diagonally (Ts1 and Ts2). Since the speaker is static you have Ts1 + Ts2 + Wse = 0.

That was about as far as I could go. I then looked up information in my book but couldn't find much and so I resorted to the internet. I came across this page: http://science.palomar.edu/physics/Physics230/Tutor/4-31.html
I used the equation they give T = mg/2sin(theta)

I set a x-y axis starting with the speaker as the origin. I find the triangle that one string forms with the y-axis and the ceiling.

I find the angle that corresponds with the picture on the page I found and plug in the information I have into the equation: T = (21*9.81)/(2sin(37.98))

Putting this data into the answer box, I get the feedback: "Remember to use radians, not degrees, in arguments to trigonometric functions."

I don't lose any attempts with the answer so I think I'm on the right track but I don't know what to do at this point.
 
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It sounds to me like the angle is a 3-4-5 triangle. Don't use 2sin(37.98) in the denominator, it is at best an approximation.
Use 2(3/5). When you are given those little reference triangles (dont know actual names), just remember your trig SOH CAH TOA. Since you are looking for the sin, which is S=O/H, that should give you 3/5. Thats all you need. Multiply it bye the magnitude 2.

For your answer I am getting 171.675.

Mastering physics is a good HW site, but just be aware that they are occasional mistakes too.
Good luck and I hope this helped.

Chris
 

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