Question

A 15,000 N crane pivots around a friction-free axle at its base and is supported by a cable making a 25° angle with the crane (Fig. E 11.18). The crane is 16 m long and is not uniform, its center of gravity being 7.0 m from the axle as measured along the crane. The cable is attached 3.0 m from the upper end of the crane. When the crane is raised to 55° above the horizontal holding an 11,000 N pallet of bricks by a 2.2 m, very light cord, find (a) the tension in the cable and (b) the horizontal and vertical components of the force that the axle exerts on the crane. Start with a free-body diagram of the crane.

Short answer

(a) The tension in the cable is 29304.54 N .

(b) The horizontal and vertical components of the force that the axle exerts on the crane are 25378.48 N and 40652.27 N respectively.

Step-by-step solution

Given information:

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 .

Force components:

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.

(a) The tension in the cable:

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,

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

Putting the values:

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

Hence, the tension in the cable is 29304.54 N .

(b) The horizontal and vertical components of the force:

Balancing all the forces in the vertical direction,

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

And, balancing all the forces in the horizontal direction,

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

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