Question

The square steel plate has a mass of 1800 kg with mass center at its center G. Calculate the tension in each of the three cables with which the plate is lifted while remaining horizontal.

Answer 1

$$ \begin{aligned} &\sum F_{x}=0 \Rightarrow T_{B D-x}+T_{A D-x}-T_{C D-x}=0\\ &\sum F_{y}=0 \Rightarrow T_{B D-y}-T_{A D-y}=0\\ &\sum F_{z}=W \Rightarrow T_{B D-z}+T_{A D-z}+T_{C D-z}=m g=17640\\ &T_{B D-x}=\frac{B D_{x}}{|B D|} T_{B D}=\frac{1200}{\sqrt{(1200)^{2}+(1200)^{2}+(2400)^{2}}} T_{B D}=.41 T_{B D}\\ &T_{A D-x}=\frac{A D_{x}}{|A D|} T_{A D}=\frac{1200}{\sqrt{(1200)^{2}+(1200)^{2}+(2400)^{2}}} T_{B D}=.41 T_{A D}\\ &T_{C D-x}=\frac{C D_{x}}{|C D|} T_{C D}=\frac{1200}{\sqrt{(1200)^{2}+(0)^{2}+(2400)^{2}}} T_{B D}=.45 T_{C D}\\ &T_{B D-y}=\frac{B D_{y}}{|B D|} T_{B D}=\frac{1200}{\sqrt{(1200)^{2}+(1200)^{2}+(2400)^{2}}} T_{B D}=.41 T_{B D}\\ &T_{A D-y}=\frac{A D_{y}}{|A D|} T_{A D}=\frac{1200}{\sqrt{(1200)^{2}+(1200)^{2}+(2400)^{2}}} T_{B D}=.41 T_{A D}\\ &T_{B D-z}=\frac{B D_{z}}{|B D|} T_{B D}=\frac{2400}{\sqrt{(1200)^{2}+(1200)^{2}+(2400)^{2}}} T_{B D}=.82 T_{B D}\\ &T_{A D-z}=\frac{A D_{z}}{|A D|} T_{A D}=\frac{2400}{\sqrt{(1200)^{2}+(1200)^{2}+(2400)^{2}}} T_{B D}=.82 T_{A D}\\ &T_{C D-x}=\frac{C D_{z}}{|C D|} T_{C D}=\frac{2400}{\sqrt{(1200)^{2}+(0)^{2}+(2400)^{2}}} T_{B D}=.89 T_{C D}\\ &\text { Substitue }\\ &\left[\begin{array}{l} \sum F_{x}=0 \Rightarrow 0.41 T_{B D}+0.41 T_{A D}-0.45 T_{C D}=0 \\ \sum F_{y}=0 \Rightarrow 0.41 T_{B D}-0.41 T_{A D}=0 \\ \sum F_{y}=0 \Rightarrow 0.82 T_{B D}+0.82 T_{A D}+0.89 T_{C D}=17640 \end{array}\right] \Rightarrow\left[\begin{array}{l} T_{B D}=5408.1 \\ T_{A D}=5408.1 \\ T_{C D}=9854.7 \end{array}\right] \end{aligned} $$