# Tension / Equilibrium Question

## Homework Statement

A load on which the force of gravity is 46 N hands from the centre of a frictionless clothesline, pulling its centre down so that each half of the line makes an angle of 25° with the horizontal. What is the force of tension in the line?

## Homework Equations

$$\Sigma$$Y = FT - FG
= 2FTsin25° - FG

## The Attempt at a Solution

I found that FT is equal to 46 N / 2sin25° $$\approx$$ 54 N. However, I don't understand if this is the force of tension in each half of the line, or through the whole line. Can someone help me understand this?

Thanks.

Michael

Last edited:

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I think you've calculated the tension in one of the lines.
You can calculate both of them by removing the 2 from the denominator.

I approached it like this: If the weight is 46N, then each side of the line is contributing 23N to supporting it. That means that the vertical component of the tension in one side of the line is 23N, so 23/sin(25) = F_T in one side. This is clearly the same as your expression.

PhanthomJay