91Ó°ÊÓ

The magnitude of the horizontal outward force,$\text{T}=535\text{ N}$ .

You draw the free body diagram. The system is at equilibrium. For such a system, the vector sum of the forces acting on it is zero. You can apply this concept along the x and y axes separately and find out${\mathbf{T}}_{\mathbf{1}}$and ${\mathbf{T}}_{\mathbf{2}}$, The equations are given below.

$\begin{array}{c}∑{\mathrm{F}}_{\mathrm{x}}=0\\ ∑{\mathrm{F}}_{\mathrm{y}}=0\\ ∑\mathrm{τ}=0\end{array}$

From the free body diagram, $\mathrm{GA}$ exerts a force $\mathrm{T}$at the point $\mathrm{A}$ in the left direction. Similarly, $\mathrm{GB}$exerts a force at a point $\mathrm{B}$ in the right direction. The diagonal force ${\mathrm{F}}_{\mathrm{d}}$makes an angle $\mathrm{θ}$with the side of the cube as shown in the figure, then:

$\begin{array}{l}{\mathrm{F}}_{\mathrm{dx}}={\mathrm{F}}_{\mathrm{d}}\sin \mathrm{θ}\\ {\mathrm{F}}_{\mathrm{dy}}={\mathrm{F}}_{\mathrm{d}}\cos \mathrm{θ}\end{array}$

Due to the symmetry of the cube equal force acts on$\mathrm{DB}$ . These two diagonal bars are pulled by the horizontal bottom of the cube similar to the top bar of the cube. Hence$\mathrm{CD}$is also tension. According to the figure, the forces acting at a point$\mathrm{A}$shows that the$\mathrm{AD}$and$\mathrm{AB}$are the vertical and horizontal components of${\mathrm{F}}_{\mathrm{d}}$. Due to the symmetry, they are also in tension.

The other bars which are in tension are $\mathrm{BC}$, $\mathrm{CD}$ and $\mathrm{DA}$.

The magnitude of the horizontal outward force is $535\text{ N}$. This force causes tension. Thus, the magnitude of the force causing tension is $535\text{ N}$.

According to the free body diagram, you can apply static equilibrium conditions along the x-axis as

$\begin{array}{c}\mathrm{Î\pounds }\stackrel{→}{{\mathrm{F}}_{\mathrm{x}\mathrm{net}}}=0\\ {\mathrm{F}}_{\mathrm{dx}}−\mathrm{T}=0\\ {\mathrm{F}}_{\mathrm{dx}}=\mathrm{T}\\ {\mathrm{F}}_{\mathrm{d}}\mathrm{sinsin}\mathrm{θ}=\mathrm{T}\\ {\mathrm{F}}_{\mathrm{d}}=\frac{\mathrm{T}}{\mathrm{sinsin}\mathrm{θ}}\\ {\mathrm{F}}_{\mathrm{d}}=\frac{535\mathrm{N}}{\mathrm{sinsin}45°}\\ {\mathrm{F}}_{\mathrm{d}}=757\text{ N}\end{array}$

The magnitude of the force causing compression on $\mathrm{CA}$ and $\mathrm{DB}$ is $757\mathrm{N}$.