91Ó°ÊÓ

The length of the nonuniform beam is, L = 4.50m .

The weight of the nonuniform beam is, W = 1.40 kN .

The angle made by the nonuniform beam below the horizontal is, $\mathrm{θ}=25°$.

The distance of the cable from the pivot is, a = 3 m .

The distance of the center of gravity of the beam from the pivot is, b = 2 m .

The downwards force exerted by the lighting equipment on the lower left end of the beam is, F = 5 kN .

The weight of a rigid body can be determined by multiplying body’s mass with the acceleration due to gravity.

For a rigid body located at any plane, the direction of the weight of a body always acts directly downwards.

The free-body diagram of the beam is given below,

Here, T is the tension in the cable, $F_x$is the horizontal force and $F_y$is the vertical force exerted on the beam by the pivot.

Calculating torque about the pivot point of the beam,

$F×L\cos \mathrm{θ}+W×b\cos \mathrm{θ}=T×a$

Putting the values,

$$\begin{gathered} 5\text{kN}×\text{4.50}\text{m}×\text{cos25}{}^{∘}+1.40\text{kN}×2\text{m}×\text{cos25}{}^{∘}=T×3\text{m} \\ 20.4\text{kN}·\text{m}+2.54\text{kN}·\text{m}=T×3\text{m} \\ T=\frac{22.94\text{kN}·\text{m}}{3\text{m}} \\ T=7.65\text{kN} \end{gathered}$$

Hence, the tension in the cable is 7.65 kN.

Balancing all the forces in the vertical direction,

$$\begin{gathered} F+W=T\cos 25{}^{∘}+F_y \\ F+W=T\cos 25{}^{∘}+F_y \end{gathered}$$

Putting the values,

$$\begin{gathered} 5\text{kN}+1.40\text{kN}=7.65\text{kN}×0.906+F_y \\ 6.4\text{kN}=7.65\text{kN}×0.906+F_y \\ {F}_y=6.4\text{kN}-6.93\text{kN} \\ =-0.53\text{kN} \end{gathered}$$

The negative sign indicates that the force is acting downwards.

Balancing all the forces in the horizontal direction,

$$\begin{gathered} F_x=T\sin {25}^{∘} \\ =7.65kN×0.423 \\ =3.23kN \end{gathered}$$

Hence, the horizontal and vertical components of the force exerted on the beam by the pivot are 3.23 kN (towards left) and 0.53 kN (downwards) respectively.