91Ó°ÊÓ

The weight of the boom is, 2600 N.

The length of the boom is, L.

The distance of the center of gravity of the boom from the pivot is, 0.35 L.

When all the forces acting on a body are represented on a diagram by using arrows showing its magnitude and direction, then the diagram is described as the ‘free-body diagram’.

Using the free-body diagram, the forces acting on a body can be balanced, and also the unknown force values can be determined.

The free-body diagram of the boom is given by:

Here, T is the tension in the guy wire, $F_V$is vertical and $F_H$is the horizontal component of the force exerted on the boom at its lower end.

From the right-angled triangle, we can determine,

$$\begin{gathered} AB=L\cos 60{}^{∘} \\ AB=L×\frac{1}{2} \end{gathered}$$

And,

$$\begin{gathered} OA=L\sin 60{}^{∘} \\ OA=L×\frac{\sqrt{3}}{2} \end{gathered}$$

Taking moments about point O, we get,

$5000\text{N}×AB+2600\text{N}×\left(0.35L\cos 60°\right)-T×OA=0$

Substituting the values in the above expression and we get:

$$\begin{gathered} 5000\text{N}×\frac{L}{2}+2600\text{N}×\left(\frac{0.35L}{2}\right)-T×\frac{\sqrt{3}L}{2}=0 \\ T×\frac{\sqrt{3}L}{2}=5000\text{N}×\frac{L}{2}+2600\text{N}×\left(\frac{0.35L}{2}\right) \end{gathered}$$

Solving further as:

$$\begin{gathered} \sqrt{3}T=5000\text{N}+910\text{N} \\ T=\frac{5910\text{N}}{\sqrt{3}} \\ T=3412.14\text{N} \end{gathered}$$

Hence, the tension in the guy wire is 3412.14 N.

Balancing all the horizontal forces on the boom, we get,

$$\begin{gathered} F_H=T \\ {F}_H=3412.14\text{N} \end{gathered}$$

Balancing all the vertical forces on the boom and we get,

$$\begin{gathered} F_V=2600\text{N}+5000\text{N} \\ {F}_V=7600\text{N} \end{gathered}$$

Hence, the horizontal and vertical components of the force exerted on the boom at its lower end are 3412.14 N and 7600 N respectively.