Chapter 4.1, Problem 4.50P

A traffic-signal pole may be supported in the three ways show part c, the tension in cable BC is known to be 1950 N. Determ the reactions for each type of support shown.

Fig. P4.50

To determine

The reaction forces acting on the support.

Answer to Problem 4.50P

The reaction forces acting on the support in figure (a) are $A=\underset{\_}{5,540\text{N}}$ acting at an angle $87.3°$ above positive $x$ axis , and $C=\underset{\_}{683\text{N}}$ acting at an angle $67.4°$ below the negative $x$ axis, the reaction forces acting on the support in figure (b) are $A=\underset{\_}{4,900\text{N}}$ acting upward, and $M_A=\underset{\_}{1,890\text{N}\cdot \text{m}}$ counterclockwise and the reaction forces acting on the support in figure (c) are $A=\underset{\_}{6,740\text{N}}$ acting at an angle $87.3°$ above positive $x$ axis , $M_A=\underset{\_}{3,510\text{N}\cdot \text{m}}$ clockwise and $C=\underset{\_}{1,950\text{N}}$ acting at an angle $67.4°$ below the negative $x$ axis.

Explanation of Solution

The free body diagram of the traffic signal pole (a) and its support is given in the figure 1.

The free body diagram of the traffic signal pole (b) and its support is given in the figure 2.

The free body diagram of the traffic signal pole (c) and its support is given in the figure 3.

Write the equation to find the moment of force.

$M=Fa$

Here, $F$ is the force acting and $a$ is the perpendicular distance from the force to the point about which the moment is calculated.

Write the equation to find the total moment about the point $A$.

$\Sigma M_A=\Sigma \left(Fa\right)$

Write the equations for equilibrium for the free body diagram.

$\Sigma M_B=0$ (I)

$\Sigma F_x=0$ (II)

$\Sigma F_y=0$ (III)

Here, $M_B$ is the torque about the point $B$, $F_x$ is the force in the $x$ direction, and $F_y$ is the force in the $y$ direction.

Write the equation for the Pythagoras theorem to find the side $BC$.

$BC=\sqrt{A{B}^2+A{C}^2}$ (IV)

Write the equation to find the angle of orientation of the reaction force on $C$.

$\theta ={\text{tan}}^{-1}\left(\frac{AB}{AC}\right)$ (V)

Write the expression to find the vector $A$ if its components are given.

$A=\sqrt{A_x^2+A_y^2}$ . (VI)

Write the equation to find the angle of orientation of the components of $A$.

$\theta ={\text{tan}}^{-1}\left(\frac{A_y}{A_x}\right)$ (VII)

Conclusion:

Refer figure P4.50(a) and figure 1.

Substitute $7.2\text{m}$ for $AB$ and $3\text{m}$ for $AC$ in equation (IV) to find the side $BC$.

$\begin{array}{c}BC=\sqrt{{\left(7.2\text{m}\right)}^2+{\left(3\text{m}\right)}^2}\\ =7.8\text{m}\end{array}$

Solve equation (I) using the figure 1.

$A_x\left(7.2\text{m}\right)-W\left(2.1\text{m}\right)=0$

Here, $A_x$ is the $x$ component of the force on $A$, and $W$ is the weight of the traffic light.

Substitute $900\text{N}$ for $W$ in the above equation and rearrange it to find the $x$ component of force $A$.

$\begin{array}{c}{A}_x=\frac{\left(900\text{N}\right)\left(2.1\text{m}\right)}{\left(7.2\text{m}\right)}\\ =262.50\text{N}\end{array}$

Solve equation (II) using figure 1.

$-T\text{sin}\theta +A_x=0$

Here, $T$ is the tension in the cable $BC$.

Substitute $262.50\text{N}$ for $A_x$ and $3\text{m}/\text{7}\text{.8}\text{m}$ for $\text{sin}\theta$ using the figure 1 in the above equation and rearrange it to find $T$.

$\begin{array}{c}T=\frac{\left(7.8\text{m}\right)\left(262.50\text{N}\right)}{3\text{m}}\\ =682.50\text{N}\end{array}$

Solve equation (III) using figure 1.

$A_y-T\text{cos}\theta -W-W_p=0$

Here, $A_y$ is the $y$ component of the force on $A$ and $W_p$ is the weight of the traffic signal pole.

Substitute $7.2\text{m}/7.8\text{m}$ for $\text{cos}\theta$, $900\text{N}$ for $W$ and $4000\text{N}$ for $W_p$ in the above equation and rearrange it to find the $y$ component of the force on $A$.

$\begin{array}{c}{A}_y-\frac{\left(7.2\text{m}\right)\left(682.50\text{N}\right)}{7.8\text{m}}-900\text{N}-4,000\text{N}=0\\ A_y=630\text{N}+900\text{N}+4,000\text{N}\\ =5,530\text{N}\end{array}$

Substitute $262.50\text{N}$ for $A_x$ and $5530\text{N}$ for $A_y$ in equation (VI) to find $A$.

$\begin{array}{c}A=\sqrt{{\left(262.50\text{N}\right)}^2+{\left(5,530\text{N}\right)}^2}\\ =5,536.2\text{N}\end{array}$

Substitute  $262.50\text{N}$ for $A_x$ and $5,530\text{N}$ for $A_y$ in equation (VII) to find the angle of orientation of $A$.

$\begin{array}{c}\theta ={\text{tan}}^{-1}\left(\frac{5,530\text{N}}{262.50\text{N}}\right)\\ =87.28°\end{array}$

Substitute $7.2\text{m}$ for $AB$ and $3\text{m}$ for $AC$ in equation (V) to find the angle of orientation of the reaction force on $C$.

$\begin{array}{c}\theta ={\text{tan}}^{-1}\left(\frac{7.2\text{m}}{3\text{m}}\right)\\ =67.38°\end{array}$

Refer figure P4.50(b) and figure 2.

Solve equation (I) using the figure 2.

$M_A-W\left(2.1\text{m}\right)=0$

Here, $M_A$ is the moment about point $A$, and $W$ is the weight of the traffic light.

Substitute $900\text{N}$ for $W$ in the above equation and rearrange it to find the moment about point $A$.

$\begin{array}{c}{M}_A-\left(900\text{N}\right)\left(2.1\text{m}\right)=0\\ M_A=1,890.0\text{N}\cdot \text{m}\end{array}$

Solve equation (II) using figure 2.

$A_x=0$

Solve equation (III) using figure 2.

$A_y-W-W_p=0$

Substitute $900\text{N}$ for $W$ and $4000\text{N}$ for $W_p$ in the above equation and rearrange it to find the $y$ component of the force on $A$.

$\begin{array}{c}{A}_y-900\text{N}-4,000\text{N}=0\\ A_y=900\text{N}+4,000\text{N}\\ =4,900\text{N}\end{array}$

Substitute $0\text{N}$ for $A_x$ and $4900\text{N}$ for $A_y$ in equation (VI) to find $A$.

$\begin{array}{c}A=\sqrt{{\left(0\text{N}\right)}^2+{\left(4,900\text{N}\right)}^2}\\ =4,900\text{N}\end{array}$

Substitute $0\text{N}$ for $A_x$ and $4900\text{N}$ for $A_y$ in equation (VII) to find the angle of orientation of $A$.

$\begin{array}{c}\theta ={\text{tan}}^{-1}\left(\frac{4900\text{N}}{0\text{N}}\right)\\ =90°\end{array}$

Refer figure P4.50(c) and figure 3.

Solve equation (I) using the figure 3.

$M_A+T\text{cos}\theta \left(7.2\text{m}\right)-W\left(2.1\text{m}\right)=0$

Here, $M_A$ is the moment about the point $A$, and $W$ is the weight of the traffic light.

Substitute $1950\text{N}$ for $T$, $3\text{m}/\text{7}\text{.8}\text{m}$ for $\text{cos}\theta$ and $900\text{N}$ for $W$ in the above equation and rearrange it to find the moment about the point $A$.

$\begin{array}{c}{M}_A=-\left(1,950\text{N}\right)\left(\frac{3\text{m}}{\text{7}\text{.8}\text{m}}\right)\left(7.2\text{m}\right)+\left(900\text{N}\right)\left(2.1\text{m}\right)\\ =-5,400\text{N}\cdot \text{m}+\text{1,890}\text{N}\cdot \text{m}\\ \text{=}-\text{3,510}\text{N}\cdot \text{m}\end{array}$

Solve equation (II) using figure 3.

$-T\text{cos}\theta +A_x=0$

Here, $T$ is the tension in the cable $BC$.

Substitute $1950\text{N}$ for $T$ and $7.2\text{m}/\text{7}\text{.8}\text{m}$ for $\text{cos}\theta$ using the figure 1 in the above equation and rearrange it to find $A_x$.

$\begin{array}{c}{A}_x=\frac{\left(7.2\text{m}\right)\left(1950\text{N}\right)}{7.8\text{m}}\\ =750\text{N}\end{array}$

Solve equation (III) using figure 1.

$A_y-T\text{cos}\theta -W-W_p=0$

Here, $A_y$ is the $y$ component of the force on $A$ and $W_p$ is the weight of the traffic signal pole.

Substitute $7.2\text{m}/7.8\text{m}$ for $\text{cos}\theta$, $900\text{N}$ for $W$, $4000\text{N}$ for $W_p$ and $1950\text{N}$ for $T$  in the above equation and rearrange it to find the $y$ component of the force on $A$.

$\begin{array}{c}{A}_y-\frac{\left(7.2\text{m}\right)\left(1,950\text{N}\right)}{7.8\text{m}}-900\text{N}-4,000\text{N}=0\\ A_y=1,800\text{N}+900\text{N}+4,000\text{N}\\ =6,700\text{N}\end{array}$

Substitute $262.50\text{N}$ for $A_x$ and $5530\text{N}$ for $A_y$ in equation (VI) to find $A$.

$\begin{array}{c}A=\sqrt{{\left(262.50\text{N}\right)}^2+{\left(5530\text{N}\right)}^2}\\ =5,536.2\text{N}\end{array}$

Substitute $262.50\text{N}$ for $A_x$ and $5530\text{N}$ for $A_y$ in equation (VII) to find the angle of orientation of $A$.

$\begin{array}{c}\theta ={\text{tan}}^{-1}\left(\frac{5530\text{N}}{262.50\text{N}}\right)\\ =87.28°\end{array}$

Substitute $7.2\text{m}$ for $AB$ and $3\text{m}$ for $AC$ in equation (V) to find the angle of orientation of the reaction force on $C$.

$\begin{array}{c}\theta ={\text{tan}}^{-1}\left(\frac{7.2\text{m}}{3\text{m}}\right)\\ =67.38°\end{array}$

Therefore, the reaction forces acting on the support in figure (a) are $A=\underset{\_}{5,540\text{N}}$ acting at an angle $87.3°$ above positive $x$ axis , and $C=\underset{\_}{683\text{N}}$ acting at an angle $67.4°$ below the negative $x$ axis, the reaction forces acting on the support in figure (b) are $A=\underset{\_}{4,900\text{N}}$ acting upward, and $M_A=\underset{\_}{1,890\text{N}\cdot \text{m}}$ counterclockwise and the reaction forces acting on the support in figure (c) are $A=\underset{\_}{6,740\text{N}}$ acting at an angle $87.3°$ above positive $x$ axis , $M_A=\underset{\_}{3,510\text{N}\cdot \text{m}}$ clockwise and $C=\underset{\_}{1,950\text{N}}$ acting at an angle $67.4°$ below the negative $x$ axis.