kansas state university ce 333 kyle larson problem #3 a thin uniform square plate, 3 m on each side, has a…

Explanation:

Step1: Define position vectors

Let the origin be (O). Point (A=(0,0,4)), (B=( - 1.5,-1.5,0)), (C=(1.5,-1.5,0)), (D=(0,1.5,0)). The position vectors (\vec{r}{AB}=\left\langle-1.5,-1.5,-4\right\rangle), (\vec{r}{AC}=\left\langle1.5,-1.5,-4\right\rangle), (\vec{r}{AD}=\left\langle0,1.5,-4\right\rangle). The weight vector (\vec{W}=\left\langle0,0, - 10\right\rangle). Let the forces in the cables (AB), (AC), (AD) be (\vec{F}{AB}=F_{AB}\frac{\vec{r}{AB}}{\vert\vec{r}{AB}\vert}), (\vec{F}{AC}=F{AC}\frac{\vec{r}{AC}}{\vert\vec{r}{AC}\vert}), (\vec{F}{AD}=F{AD}\frac{\vec{r}{AD}}{\vert\vec{r}{AD}\vert}), where (\vert\vec{r}{AB}\vert=\sqrt{(-1.5)^{2}+(-1.5)^{2}+(-4)^{2}}=\sqrt{2.25 + 2.25+16}=\sqrt{20.5}), (\vert\vec{r}{AC}\vert=\sqrt{(1.5)^{2}+(-1.5)^{2}+(-4)^{2}}=\sqrt{2.25 + 2.25+16}=\sqrt{20.5}), (\vert\vec{r}_{AD}\vert=\sqrt{0^{2}+(1.5)^{2}+(-4)^{2}}=\sqrt{2.25 + 16}=\sqrt{18.25}).

Step2: Apply equilibrium equations

(\sum\vec{F}=\vec{F}{AB}+\vec{F}{AC}+\vec{F}{AD}+\vec{W}=\vec{0}). In the (x -)direction: (F{AB}\frac{-1.5}{\sqrt{20.5}}+F_{AC}\frac{1.5}{\sqrt{20.5}}+0 = 0), which implies (F_{AB}=F_{AC}). In the (y -)direction: (F_{AB}\frac{-1.5}{\sqrt{20.5}}+F_{AC}\frac{-1.5}{\sqrt{20.5}}+F_{AD}\frac{1.5}{\sqrt{18.25}}=0). In the (z -)direction: (F_{AB}\frac{-4}{\sqrt{20.5}}+F_{AC}\frac{-4}{\sqrt{20.5}}+F_{AD}\frac{-4}{\sqrt{18.25}}-10 = 0). Substitute (F_{AB}=F_{AC}) into the (y -)direction and (z -)direction equations. From the (y -)direction equation: (-\frac{3F_{AB}}{\sqrt{20.5}}+\frac{1.5F_{AD}}{\sqrt{18.25}} = 0), so (F_{AD}=2F_{AB}\frac{\sqrt{18.25}}{\sqrt{20.5}}). Substitute into the (z -)direction equation: (-\frac{8F_{AB}}{\sqrt{20.5}}-\frac{4F_{AD}}{\sqrt{18.25}}=10). Replace (F_{AD}) in the (z -)direction equation: (-\frac{8F_{AB}}{\sqrt{20.5}}-\frac{4\times2F_{AB}\frac{\sqrt{18.25}}{\sqrt{20.5}}}{\sqrt{18.25}}=10), (-\frac{8F_{AB}}{\sqrt{20.5}}-\frac{8F_{AB}}{\sqrt{20.5}}=10), (-\frac{16F_{AB}}{\sqrt{20.5}}=10), (F_{AB}=F_{AC}=-\frac{10\sqrt{20.5}}{16}\approx - 2.85\mathrm{kN}), (F_{AD}=2F_{AB}\frac{\sqrt{18.25}}{\sqrt{20.5}}\approx - 5.7\mathrm{kN}). The negative sign just indicates the direction (tension, pulling upwards). So (F_{AB}\approx2.85\mathrm{kN}), (F_{AC}\approx2.85\mathrm{kN}), (F_{AD}\approx5.7\mathrm{kN}).

Answer:

The force in cable (AB) is approximately (2.85\mathrm{kN}), the force in cable (AC) is approximately (2.85\mathrm{kN}), and the force in cable (AD) is approximately (5.7\mathrm{kN}).