A nonuniform beam 4.50 m long and weighing 1.40 kN makes an angle of $\mathbf{25}\mathbf{.}{\mathbf{0}}^{\mathbf{\circ }}$below the horizontal. It is held in position by a frictionless pivot at its upper right end and by a cable 3.00 m farther down the beam and perpendicular to it (Fig. E 11.20). The center of gravity of the beam is 2.00 m down the beam from the pivot. Lighting equipment exerts a 5.00 kN downward force on the lower left end of the beam. Find the tension T in the cable and the horizontal and vertical components of the force exerted on the beam by the pivot. Start by sketching a free-body diagram of the beam.

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

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

The angle made by the nonuniform beam below the horizontal is:$\theta ={25}^{\circ }$.

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,

$F\times L\cos \theta +W\times b\cos \theta =T\times a$

Putting the values,

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

Hence, the tension in the cable is $7.65\text{kN}$.

Balancing all the forces in the vertical direction,

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

Putting the values,

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

The negative sign indicates that the force is acting downwards.

Balancing all the forces in the horizontal direction,

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

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.