Question

2.5 points the 12 - ft boom ab has a fixed end a. a steel cable is stretched from the free end b of the boom to a point c located on the vertical wall. if the tension in the cable is 412 lb, determine the moment about a of the force exerted by the cable at b. the moment about a of the force exerted by the cable at b is ( ) lb - ft)i + ( ) lb - ft)j + ( ) lb - ft)k.

Explanation:

Step1: Define position vector $\vec{r}_{AB}$

The position vector from $A$ to $B$ is $\vec{r}_{AB}=12\vec{i}$.

Step2: Define force vector $\vec{F}_{BC}$

First, find the length of the cable $BC$. Using the Pythagorean - theorem, if the vertical distance from $B$ to $C$ is $8$ ft and the horizontal distance from $B$ to $C$ is $4.8$ ft, the length of $BC$ is $l_{BC}=\sqrt{8^{2}+4.8^{2}}=\sqrt{64 + 23.04}=\sqrt{87.04}=9.33$ ft.
The force vector $\vec{F}_{BC}=F_{BC}\frac{\vec{r}_{BC}}{|\vec{r}_{BC}|}$, where $\vec{r}_{BC}=-4.8\vec{i}-8\vec{j}$ and $F_{BC} = 412$ lb. So $\vec{F}_{BC}=412\times\frac{-4.8\vec{i}-8\vec{j}}{9.33}=-210.6\vec{i}-351\vec{j}$ lb.

Step3: Calculate the moment $\vec{M}_{A}=\vec{r}_{AB}\times\vec{F}_{BC}$

Using the cross - product formula $\vec{M}_{A}=

$$\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\12&0&0\\-210.6&-351&0\end{vmatrix}$$

$.
The $k$ - component of the cross - product is non - zero. $\vec{M}_{A}=(12\times(- 351)-0\times(-210.6))\vec{k}=-4212\vec{k}$ lb - ft.

Answer:

$0$ lb - ft$\vec{i}+0$ lb - ft$\vec{j}-4212$ lb - ft$\vec{k}$