problem 3: as shown in the figure, a load p is supported by the tripod. if the force in any leg is limited…

Explanation:

Step1: Establish coordinate system

Let the origin be at $O$. The position vectors of points $A(0,0,8)$, $B(6,6,0)$, $C( - 3,6,0)$, $D(-6,0,0)$.

Step2: Find unit - vectors of the members

The vector $\overrightarrow{AD}=\langle-6 - 0,0 - 0,0 - 8\rangle=\langle-6,0,-8\rangle$, and the unit - vector $\hat{u}{AD}=\frac{\overrightarrow{AD}}{\vert\overrightarrow{AD}\vert}=\frac{\langle-6,0,-8\rangle}{\sqrt{(-6)^2+0^2+(-8)^2}}=\langle-\frac{3}{5},0,-\frac{4}{5}\rangle$. The vector $\overrightarrow{AC}=\langle-3 - 0,6 - 0,0 - 8\rangle=\langle-3,6,-8\rangle$, and the unit - vector $\hat{u}{AC}=\frac{\overrightarrow{AC}}{\vert\overrightarrow{AC}\vert}=\frac{\langle-3,6,-8\rangle}{\sqrt{(-3)^2 + 6^2+(-8)^2}}=\langle-\frac{3}{\sqrt{73}},\frac{6}{\sqrt{73}},-\frac{8}{\sqrt{73}}\rangle$. The vector $\overrightarrow{AB}=\langle6 - 0,6 - 0,0 - 8\rangle=\langle6,6,-8\rangle$, and the unit - vector $\hat{u}_{AB}=\frac{\overrightarrow{AB}}{\vert\overrightarrow{AB}\vert}=\frac{\langle6,6,-8\rangle}{\sqrt{6^2+6^2+(-8)^2}}=\langle\frac{3}{5},\frac{3}{5},-\frac{4}{5}\rangle$.

Step3: Set up equilibrium equations

At point $A$, $\sum\vec{F}=\vec{0}$, so $\vec{F}{AD}\hat{u}{AD}+\vec{F}{AC}\hat{u}{AC}+\vec{F}{AB}\hat{u}{AB}-P\hat{k}=\vec{0}$. In the $x$ - direction: $F_{AD}(-\frac{3}{5})+F_{AC}(-\frac{3}{\sqrt{73}})+F_{AB}(\frac{3}{5}) = 0$. In the $y$ - direction: $F_{AC}(\frac{6}{\sqrt{73}})+F_{AB}(\frac{3}{5}) = 0$. In the $z$ - direction: $F_{AD}(-\frac{4}{5})+F_{AC}(-\frac{8}{\sqrt{73}})+F_{AB}(-\frac{4}{5})-P = 0$. Assume the forces in the members are in tension. From the $y$ - direction equation $F_{AC}=-\frac{\sqrt{73}}{10}F_{AB}$. Substitute $F_{AC}$ into the $x$ - direction equation: [ \begin{align*} F_{AD}(-\frac{3}{5})-\frac{\sqrt{73}}{10}F_{AB}(-\frac{3}{\sqrt{73}})+F_{AB}(\frac{3}{5})&=0\ -\frac{3}{5}F_{AD}+\frac{3}{10}F_{AB}+\frac{3}{5}F_{AB}&=0\ -\frac{3}{5}F_{AD}+\frac{9}{10}F_{AB}&=0\ F_{AD}=\frac{3}{2}F_{AB} \end{align*} ] Substitute $F_{AD}$ and $F_{AC}$ into the $z$ - direction equation: [ \begin{align*} \frac{3}{2}F_{AB}(-\frac{4}{5})-\frac{\sqrt{73}}{10}F_{AB}(-\frac{8}{\sqrt{73}})+F_{AB}(-\frac{4}{5})-P&=0\ -\frac{6}{5}F_{AB}+\frac{4}{5}F_{AB}-\frac{4}{5}F_{AB}-P&=0\ -\frac{6}{5}F_{AB}-P&=0\ F_{AB}=-\frac{5}{6}P \end{align*} ] Then $F_{AD}=-\frac{5}{4}P$ and $F_{AC}=\frac{\sqrt{73}}{12}P$. a. Since the force in any leg is limited to $2000$ N. Let's first find $F_{AD}$. We know $F_{AD}=-\frac{5}{4}P$. b. $F_{AC}=\frac{\sqrt{73}}{12}P$. c. $F_{AB}=-\frac{5}{6}P$. d. Since $\vert F_{AD}\vert\leq2000$, $\vert F_{AC}\vert\leq2000$, $\vert F_{AB}\vert\leq2000$. For $F_{AD}$, $\vert-\frac{5}{4}P\vert = 2000$, then $P = 1600$ N. For $F_{AC}$, $\vert\frac{\sqrt{73}}{12}P\vert=2000$, $P=\frac{24000}{\sqrt{73}}\approx2790$ N. For $F_{AB}$, $\vert-\frac{5}{6}P\vert = 2000$, $P = 2400$ N. The maximum load $P$ is determined by the member with the smallest allowable - load capacity. So the maximum load $P = 1600$ N.

a. $F_{AD}=-\frac{5}{4}P$ (in terms of $P$), when considering the limit, if we assume the maximum - force condition for $AD$ and $P = 1600$ N, then $F_{AD}=- 2000$ N. b. $F_{AC}=\frac{\sqrt{73}}{12}P$, when $P = 1600$ N, $F_{AC}=\frac{\sqrt{73}}{12}\times1600\approx1493$ N. c. $F_{AB}=-\frac{5}{6}P$, when $P = 1600$ N, $F_{AB}=-\frac{5}{6}\times1600\approx - 1333$ N. d. # Answer: $1600$ N