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

[hamzaarfeen]

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

 

[Orodruin]

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?

 

[hamzaarfeen]

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?

 

[Orodruin]

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?

 

[HalfLostAndSearching]

I would first clarify that the given information is assuming ideal conditions, where the cables are perfectly horizontal and the building walls are perfectly vertical. In reality, there may be some slight variations in the angles and lengths of the cables, which could affect the tension values.

Using the given information, I would approach this problem by first drawing a free body diagram for the Halloween decoration, with the three cables as the only forces acting on it. I would then use the equation Fnet=ma to set up an equation for the forces in the vertical direction, where the net force is equal to zero since the decoration is not moving up or down. This would lead to the equation 3Tcos30=Fg, where T is the tension in each cable and Fg is the force of gravity acting on the decoration (mass 100kg multiplied by acceleration due to gravity, 9.81 m/s^2).

Solving for T, I would get T=Fg/(3cos30). Plugging in the values, I get T= 283.2N. This means that the tension in each cable is 283.2N. However, as mentioned earlier, this value may vary slightly depending on the actual angles and lengths of the cables. Additionally, it is important to note that this is the magnitude of the tension, and the direction of the tension would be along the cables, pulling upwards to counteract the force of gravity on the decoration.