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## Homework Statement

The boom in the figure below (Figure 1) weighs 2450 N and is attached to a frictionless pivot at its lower end. It is not uniform; the distance of its center of gravity from the pivot is 35.5 % of its length.

Find the tension of the guy wire.

## Homework Equations

Στ=0

∑F=0

τ=F*Dsin(θ)

## The Attempt at a Solution

I made a free body diagram where the boom is the object and i drew three forces acting on it.

The three forces acting on the boom are weight of the boom, weight of the box, and the tension in the rope.

I then tried to use the sum of the torques to find tension:

Στ=0=w

_{boom}sin(60°)(.355L) + w

_{box}sin(60°)L - T

_{1}sin(30°)L

then rearranged

w

_{boom}sin(60°)(.335L) + w

_{box}sin(60°)L = T

_{1}sin(30°)L

then i cancelled out the L variable, and plugged in and simplified and got this:

T

_{1}= 10,167 N => 10200 N.

This is wrong.

I peaked at the yahoo answer and they used cos(θ) instead of sin(θ) which does not make sense to make because then the force calculated would move the boom linearly rather than creating a torque.

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