Question

The rigid square frame in Fig. 12-79 consists of the four side bars $\mathbf{\text{AB}}$ ,$\mathbf{\text{BC}}$ , $\mathbf{\text{CD}}$ , and $\mathbf{\text{DA}}$ plus two diagonal bars $\mathbf{\text{AC}}$and $\mathbf{\text{BD}}$ , which pass each other freely at E. By means of the turnbuckle G, bar $\mathbf{\text{AB}}$is put under tension, as if its ends were subject to horizontal, outward forces$\overrightarrow{\mathbf{T}}$ of magnitude$\mathbf{535}\mathbf{\text{\hspace{0.17em}N}}$ .

(a) Which of the other bars are in tension? What are the magnitudes of (b) the forces causing the tension in those bars and (c) the forces causing compression in the other bars?

Short answer

1. The other bars which are in tension are$\text{BC}$,$\text{CD}$and $\text{DA}$.
2. The magnitude of the force causing tension,$\text{T}=535\text{\hspace{0.17em}N}$ .
3. The magnitude of the force causing compression on $\text{CA}$ and $\text{DB}$, ${\text{F}}_{\text{d}}=757\text{\hspace{0.17em}N}$.

Step-by-step solution

Understanding the given information

The magnitude of the horizontal outward force,$\text{T}=535\text{\hspace{0.17em}N}$ .

Concept and formula used in the given question

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.

$\begin{array}{c}\sum {\mathrm{F}}_{\mathrm{x}}=0\\ \sum {\mathrm{F}}_{\mathrm{y}}=0\\ \sum \mathrm{\tau }=0\end{array}$

(a) Calculation for the other bars which are in tension

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 $\mathrm{\theta }$with the side of the cube as shown in the figure, then:

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

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 $\mathrm{DA}$.

(b) Calculation for the forces causing the tension in those bars

The magnitude of the horizontal outward force is $535\text{\hspace{0.17em}N}$. This force causes tension. Thus, the magnitude of the force causing tension is $535\text{\hspace{0.17em}N}$.

(c) Calculation for the forces causing compression in the other bars

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

$\begin{array}{c}\mathrm{\Sigma }\overrightarrow{{\mathrm{F}}_{\mathrm{x}\mathrm{net}}}=0\\ {\mathrm{F}}_{\mathrm{dx}}-\mathrm{T}=0\\ {\mathrm{F}}_{\mathrm{dx}}=\mathrm{T}\\ {\mathrm{F}}_{\mathrm{d}}\mathrm{sinsin}\mathrm{\theta }=\mathrm{T}\\ {\mathrm{F}}_{\mathrm{d}}=\frac{\mathrm{T}}{\mathrm{sinsin}\mathrm{\theta }}\\ {\mathrm{F}}_{\mathrm{d}}=\frac{535\mathrm{N}}{\mathrm{sinsin}45°}\\ {\mathrm{F}}_{\mathrm{d}}=757\text{\hspace{0.17em}N}\end{array}$

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