What is the tension in each cable holding up a 100kg Halloween decoration?

AI Thread Summary
The problem involves calculating the tension in three cables supporting a 100kg Halloween decoration, each making a 30-degree angle with the building. The vertical component of the tension is expressed as Tcos30, leading to the equation 3Tcos30 = 100(9.81) for equilibrium. After some calculation errors were acknowledged, the correct tension was determined to be approximately 283.2N. The length of the cables is likely included to emphasize the symmetry of the setup. The discussion concludes with a focus on ensuring accurate calculations and understanding the problem's symmetry.
hamzaarfeen
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Homework Statement


A huge Halloween decoration of mass 100kg is suspended by 3 cables. Each cable is exactly 25m long and is attached to each of the three building equidistant from each other. Each cable makes an angle of 30 degrees with the side of the building to which it is attached. What is the magnitude of the tension in each cable?

Homework Equations


fnet=ma

The Attempt at a Solution


vertical component for all cables= Tcos30
fnet in y direction = 0
fnet=0
3(Tcos30)=Fg
3Tcos30=100(9.81)
T= 283.2N
 
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Looks reasonable apart from the numerical value you got in the end, what is your question? Did you multiply the RHS with cos30 instead of dividing when solving for T?
 
Orodruin said:
Looks reasonable apart from the numerical value you got in the end, what is your question? Did you multiply the RHS with cos30 instead of dividing when solving for T?

oh you sorry i made a mistake in calculations...but i still don't know if my answer is right. and what is the purpose for giving the length of the cable?
 
They probably just wanted to make really really sure that you treated the problem as symmetric under 120 degree rotations.

So what do you get now?
 
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