problem 3.4 - given the cable tension t = 80 lb, determine a) the moment about point a due to t, and b) the…

problem 3.4 - given the cable tension t = 80 lb, determine a) the moment about point a due to t, and b) the shortest distance from point a to line bc.

problem 3.4 - given the cable tension t = 80 lb, determine a) the moment about point a due to t, and b) the shortest distance from point a to line bc.

Answer

Explanation:

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

The coordinates of point $A=(0,0,0)$ and point $B=(32,10,0)$. So, $\vec{r}_{AB}=32\vec{i}+ 10\vec{j}+0\vec{k}$ (in inches).

Step2: Define unit - vector of the cable $\hat{u}_{BC}$

The coordinates of point $B=(32,10,0)$ and point $C=(0,12 + 10,0)=(0,22,0)$. $\vec{r}{BC}=(0 - 32)\vec{i}+(22 - 10)\vec{j}+(0 - 0)\vec{k}=-32\vec{i}+12\vec{j}+0\vec{k}$. The magnitude of $\vec{r}{BC}$ is $|\vec{r}{BC}|=\sqrt{(-32)^2+12^2+0^2}=\sqrt{1024 + 144}=\sqrt{1168}=4\sqrt{73}$ inches. Then $\hat{u}{BC}=\frac{\vec{r}{BC}}{|\vec{r}{BC}|}=-\frac{32}{4\sqrt{73}}\vec{i}+\frac{12}{4\sqrt{73}}\vec{j}+0\vec{k}=-\frac{8}{\sqrt{73}}\vec{i}+\frac{3}{\sqrt{73}}\vec{j}+0\vec{k}$.

Step3: Find the force vector $\vec{T}$

$\vec{T}=T\hat{u}_{BC}=80\left(-\frac{8}{\sqrt{73}}\vec{i}+\frac{3}{\sqrt{73}}\vec{j}+0\vec{k}\right)=-\frac{640}{\sqrt{73}}\vec{i}+\frac{240}{\sqrt{73}}\vec{j}+0\vec{k}$ (in - lb).

Step4: Calculate the moment $\vec{M}_A$ using $\vec{M}A=\vec{r}{AB}\times\vec{T}$

[ \begin{align*} \vec{M}_A&=\begin{vmatrix} \vec{i}&\vec{j}&\vec{k}\ 32&10&0\ -\frac{640}{\sqrt{73}}&\frac{240}{\sqrt{73}}&0 \end{vmatrix}\ &=\vec{k}\left(32\times\frac{240}{\sqrt{73}}-10\times\left(-\frac{640}{\sqrt{73}}\right)\right)\ &=\vec{k}\left(\frac{7680 + 6400}{\sqrt{73}}\right)\ &=\frac{14080}{\sqrt{73}}\vec{k}\text{ in - lb}\approx1634\vec{k}\text{ in - lb} \end{align*} ]

Step5: Calculate the shortest distance $d$ from point $A$ to line $BC$

The magnitude of the moment $\vec{M}_A$ is related to the force $T$ and the shortest distance $d$ by $|\vec{M}_A| = Td$. We know $|\vec{M}_A|=\frac{14080}{\sqrt{73}}$ in - lb and $T = 80$ lb. So, $d=\frac{|\vec{M}_A|}{T}=\frac{\frac{14080}{\sqrt{73}}}{80}=\frac{176}{\sqrt{73}}\approx20.5$ inches.

Answer:

a) $\vec{M}_A=\frac{14080}{\sqrt{73}}\vec{k}\text{ in - lb}\approx1634\vec{k}\text{ in - lb}$ b) $d=\frac{176}{\sqrt{73}}\text{ inches}\approx20.5\text{ inches}$