problem 2.8 - the 400 - kg precast concrete section is hoisted by a vertical force t from a crane. find the…

problem 2.8 - the 400 - kg precast concrete section is hoisted by a vertical force t from a crane. find the tension in cables ab, ac, ad.

problem 2.8 - the 400 - kg precast concrete section is hoisted by a vertical force t from a crane. find the tension in cables ab, ac, ad.

Answer

Explanation:

Step1: Calculate gravitational force

The mass of the pre - cast concrete section is $m = 400$ kg. The gravitational force $F=mg$, where $g = 9.81$ m/s². So $F=400\times9.81=3924$ N. And $T = F=3924$ N.

Step2: Determine position vectors

Let $A=(0,5,0)$, $B=(1,0,1)$, $C=(1,0, - 1)$, $D=( - 3,0,1)$. The position vector $\vec{r}{AB}=(1 - 0)\vec{i}+(0 - 5)\vec{j}+(1 - 0)\vec{k}=\vec{i}-5\vec{j}+\vec{k}$, $|\vec{r}{AB}|=\sqrt{1^{2}+(-5)^{2}+1^{2}}=\sqrt{27}=3\sqrt{3}$ m. The position vector $\vec{r}{AC}=(1 - 0)\vec{i}+(0 - 5)\vec{j}+(-1 - 0)\vec{k}=\vec{i}-5\vec{j}-\vec{k}$, $|\vec{r}{AC}|=\sqrt{1^{2}+(-5)^{2}+(-1)^{2}}=\sqrt{27}=3\sqrt{3}$ m. The position vector $\vec{r}{AD}=(-3 - 0)\vec{i}+(0 - 5)\vec{j}+(1 - 0)\vec{k}=-3\vec{i}-5\vec{j}+\vec{k}$, $|\vec{r}{AD}|=\sqrt{(-3)^{2}+(-5)^{2}+1^{2}}=\sqrt{35}$ m.

Step3: Set up equilibrium equations

At point $A$, $\sum\vec{F}=\vec{0}$. Let the tension in $AB$ be $T_{AB}$, in $AC$ be $T_{AC}$, in $AD$ be $T_{AD}$ and the upward force be $T$. In the $x$ - direction: $T_{AB}\frac{1}{3\sqrt{3}}+T_{AC}\frac{1}{3\sqrt{3}}+T_{AD}\frac{-3}{\sqrt{35}} = 0$. In the $y$ - direction: $T_{AB}\frac{-5}{3\sqrt{3}}+T_{AC}\frac{-5}{3\sqrt{3}}+T_{AD}\frac{-5}{\sqrt{35}}=-T$. In the $z$ - direction: $T_{AB}\frac{1}{3\sqrt{3}}+T_{AC}\frac{-1}{3\sqrt{3}}+T_{AD}\frac{1}{\sqrt{35}} = 0$. Since the structure is symmetric about the $y$ - axis, $T_{AB}=T_{AC}$. Substitute $T_{AB}=T_{AC}$ into the $x$ - direction equation: $2T_{AB}\frac{1}{3\sqrt{3}}+T_{AD}\frac{-3}{\sqrt{35}} = 0$, so $T_{AD}=\frac{2\sqrt{35}}{9\sqrt{3}}T_{AB}$. Substitute $T_{AB}=T_{AC}$ and $T_{AD}=\frac{2\sqrt{35}}{9\sqrt{3}}T_{AB}$ into the $y$ - direction equation: [ \begin{align*} 2T_{AB}\frac{-5}{3\sqrt{3}}+\frac{2\sqrt{35}}{9\sqrt{3}}T_{AB}\frac{-5}{\sqrt{35}}&=-T\ -\frac{10T_{AB}}{3\sqrt{3}}-\frac{10T_{AB}}{27\sqrt{3}}&=-T\ -\frac{90T_{AB}+10T_{AB}}{27\sqrt{3}}&=-T\ -\frac{100T_{AB}}{27\sqrt{3}}&=-T \end{align*} ] Since $T = 3924$ N, $T_{AB}=T_{AC}=\frac{27\sqrt{3}\times3924}{100}\approx 1847$ N. $T_{AD}=\frac{2\sqrt{35}}{9\sqrt{3}}\times\frac{27\sqrt{3}\times3924}{100}=\frac{2\sqrt{35}\times3924}{100}\approx 461$ N.

Answer:

$T_{AB}\approx1847$ N, $T_{AC}\approx1847$ N, $T_{AD}\approx461$ N