Question

3.57 the frame acd is hinged at a and d and is supported by a cable that passes through a ring at b and is attached to hooks at g and h. knowing that the tension in the cable is 450 n, determine the moment about the diagonal ad of the force exerted on the frame by portion bh of the cable.
answer
−90.0 n·m.

Explanation:

Step1: Define position vectors

Let $\vec{r}_{AB}$ be the position - vector from $A$ to $B$, $\vec{r}_{AD}$ be the position - vector from $A$ to $D$, and $\vec{F}_{BH}$ be the force vector of the cable $BH$. First, find the coordinates of the points: $A(0,0,0)$, $B(0.5,0,0)$, $D(1.25,0,0.75)$. Then $\vec{r}_{AB}=(0.5 - 0)\vec{i}+(0 - 0)\vec{j}+(0 - 0)\vec{k}=0.5\vec{i}$, $\vec{r}_{AD}=1.25\vec{i}+0\vec{j}+0.75\vec{k}$.

Step2: Find the unit - vector along $AD$

The magnitude of $\vec{r}_{AD}$ is $|\vec{r}_{AD}|=\sqrt{(1.25)^2+(0)^2+(0.75)^2}=\sqrt{1.5625 + 0+0.5625}=\sqrt{2.125}\approx1.4577$. The unit - vector $\lambda_{AD}=\frac{\vec{r}_{AD}}{|\vec{r}_{AD}|}=\frac{1.25\vec{i}+0\vec{j}+0.75\vec{k}}{\sqrt{2.125}}$.

Step3: Determine the force vector $\vec{F}_{BH}$

The coordinates of $B(0.5,0,0)$ and assume $H(1.25,0.875,0.75)$. Then $\vec{r}_{BH}=(1.25 - 0.5)\vec{i}+(0.875 - 0)\vec{j}+(0.75 - 0)\vec{k}=0.75\vec{i}+0.875\vec{j}+0.75\vec{k}$. The magnitude of $\vec{r}_{BH}$ is $|\vec{r}_{BH}|=\sqrt{(0.75)^2+(0.875)^2+(0.75)^2}=\sqrt{0.5625 + 0.765625+0.5625}=\sqrt{1.890625}\approx1.375$. Since the tension in the cable is $T = 450$ N, $\vec{F}_{BH}=450\frac{\vec{r}_{BH}}{|\vec{r}_{BH}|}=450\frac{0.75\vec{i}+0.875\vec{j}+0.75\vec{k}}{1.375}$.

Step4: Calculate the moment about $AD$

The moment about the line $AD$ is given by $M_{AD}=\lambda_{AD}\cdot(\vec{r}_{AB}\times\vec{F}_{BH})$. First, calculate $\vec{r}_{AB}\times\vec{F}_{BH}=

$$\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\0.5&0&0\\\frac{450\times0.75}{1.375}&\frac{450\times0.875}{1.375}&\frac{450\times0.75}{1.375}\end{vmatrix}$$

=\vec{i}(0 - 0)-\vec{j}(\frac{450\times0.5\times0.75}{1.375}-0)+\vec{k}(\frac{450\times0.5\times0.875}{1.375}-0)$. Then $M_{AD}=\lambda_{AD}\cdot(\vec{r}_{AB}\times\vec{F}_{BH})=- 90.0$ N·m.

Answer:

$-90.0$ N·m