91Ó°ÊÓ

The weight of the crane is: W = 15,000 N .

The angle made by cable with the crane is:${25}^{∘}$.

The length of the crane is: 16 m .

The distance of the center of gravity from the axle is: 7 m .

The distance of cable attached from the upper end of the crane is: 3 m .

The angle of the crane is raised above the horizontal is: ${55}^{∘}$.

The weight of the pallet of bricks held by the crane is:$W_1=11,000N$.

The length of the very light cord holding the pallet of bricks is: 2.2 m .

When a force acts on an inclined component at an angle then it can be resolved into two components, horizontal and vertical components.

The value of the horizontal and vertical force components relies on the magnitude of the inclined force and its angle of inclination.

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

Here, T is the tension in the cable, $F_x$is the horizontal and $F_y$is the vertical components of the force that the axle exerts on the crane.

Taking moments about the axle of the crane,

$W×7\text{m}×\text{sin35}{}^{∘}+{\text{W}}_1×16\text{m}×\text{sin35}{}^{∘}-T×13\text{m}×\text{sin25}{}^{∘}=0$

Putting the values:

$$\begin{gathered} 15000\text{N}×7\text{m}×\text{0.574}+11000\text{N}×16\text{m}×\text{0.574}-T×13\text{m}×\text{0.423}=0 \\ 60225.53+100949.45-5.5T=0 \\ T=\frac{161174.98}{5.5} \\ T=29304.54\text{N} \end{gathered}$$

Hence, the tension in the cable is 29304.54 N .

Balancing all the forces in the vertical direction,

$$\begin{gathered} F_y-W-W_1-T\sin \text{30}{}^{∘}=0 \\ {F}_y-15000\text{N}-11000\text{N}-29304.54\text{N}×0.5=0 \\ {F}_y=26000\text{N}+\text{14652.27}\text{N} \\ {F}_y=40652.27\text{N} \end{gathered}$$

And, balancing all the forces in the horizontal direction,

$$\begin{gathered} F_x-T\cos 30{}^{∘}=0 \\ {F}_x-29304.54\text{N}×\text{0.866}=0 \\ {F}_x=25378.48\text{N} \end{gathered}$$

Hence, the horizontal and vertical components of the force are 25378.48 N and 40652.27 N respectively.