91Ó°ÊÓ

(a) The tension in the guy wire is 3412 N.

(b) The horizontal and vertical components of the force exerted on the boom at its lower end are 3412.14 N and 7600 N respectively.

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,

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

Substituting the values in the above expression and we get:

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

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.