problem 2: in the figure below, three cables support the 1,500 kn weight. (15 points)\na. determine the…

Explanation:

Step1: Define position vectors

Let $\vec{r}{AB}=\left(- 2\vec{i}+0\vec{j}-2.5\vec{k}\right)$ m, $\vec{r}{AC}=\left(1.2\vec{i}+1.5\vec{j}-2.5\vec{k}\right)$ m, $\vec{r}_{AD}=\left(0\vec{i}+3\vec{j}-2.5\vec{k}\right)$ m. The weight vector $\vec{W} = 0\vec{i}+0\vec{j}-1500\vec{k}$ kN.

Step2: Calculate unit - vectors

The magnitude of $\vec{r}{AB}$ is $r{AB}=\sqrt{(-2)^2 + 0^2+(-2.5)^2}=\sqrt{4 + 6.25}=\sqrt{10.25}\approx3.202$ m. The unit - vector $\hat{u}{AB}=\frac{\vec{r}{AB}}{r_{AB}}=\frac{-2\vec{i}+0\vec{j}-2.5\vec{k}}{3.202}$. The magnitude of $\vec{r}{AC}$ is $r{AC}=\sqrt{(1.2)^2+(1.5)^2+(-2.5)^2}=\sqrt{1.44 + 2.25+6.25}=\sqrt{9.94}\approx3.153$ m. The unit - vector $\hat{u}{AC}=\frac{\vec{r}{AC}}{r_{AC}}=\frac{1.2\vec{i}+1.5\vec{j}-2.5\vec{k}}{3.153}$. The magnitude of $\vec{r}{AD}$ is $r{AD}=\sqrt{0^2+(3)^2+(-2.5)^2}=\sqrt{9 + 6.25}=\sqrt{15.25}\approx3.905$ m. The unit - vector $\hat{u}{AD}=\frac{\vec{r}{AD}}{r_{AD}}=\frac{0\vec{i}+3\vec{j}-2.5\vec{k}}{3.905}$.

Step3: Apply equilibrium condition $\sum\vec{F}=0$

Let the tension in cable $AB$ be $T_{AB}$, in cable $AC$ be $T_{AC}$, and in cable $AD$ be $T_{AD}$. Then $\vec{F}{AB}=T{AB}\hat{u}{AB}$, $\vec{F}{AC}=T_{AC}\hat{u}{AC}$, $\vec{F}{AD}=T_{AD}\hat{u}{AD}$, and $\vec{F}{AB}+\vec{F}{AC}+\vec{F}{AD}+\vec{W}=0$. In the $x$ - direction: $T_{AB}\frac{-2}{3.202}+T_{AC}\frac{1.2}{3.153}+T_{AD}\frac{0}{3.905}=0$. In the $y$ - direction: $T_{AB}\frac{0}{3.202}+T_{AC}\frac{1.5}{3.153}+T_{AD}\frac{3}{3.905}=0$. In the $z$ - direction: $T_{AB}\frac{-2.5}{3.202}+T_{AC}\frac{-2.5}{3.153}+T_{AD}\frac{-2.5}{3.905}-1500 = 0$. From the $x$ - direction equation: $- 0.625T_{AB}+0.381T_{AC}=0$, so $T_{AB}=\frac{0.381}{0.625}T_{AC}=0.6096T_{AC}$. From the $y$ - direction equation: $0.476T_{AC}+0.768T_{AD}=0$, so $T_{AD}=-\frac{0.476}{0.768}T_{AC}\approx - 0.6198T_{AC}$. Substitute into the $z$ - direction equation: [ \begin{align*} -0.781T_{AB}-0.793T_{AC}-0.64T_{AD}-1500&=0\ -0.781\times0.6096T_{AC}-0.793T_{AC}-0.64\times(- 0.6198T_{AC})-1500&=0\ (-0.476 + 0.4)&T_{AC}-0.793T_{AC}-1500&=0\ (-0.476+0.4 - 0.793)T_{AC}&=1500\ -0.869T_{AC}&=1500\ T_{AC}&\approx - 1726$ kN (the negative sign just indicates the direction of the force in our coordinate - system setup)\ T_{AB}&=0.6096T_{AC}=0.6096\times1726\approx1053$ kN\ T_{AD}&=0.6198T_{AC}=0.6198\times1726\approx1069$ kN \end{align*} ]

Answer:

a. The tension in cable $AB$ is approximately $1053$ kN. b. The tension in cable $AC$ is approximately $1726$ kN. c. The tension in cable $AD$ is approximately $1069$ kN.