Chapter 5.1, Problem 5.29P

The frame for a sign is fabricated from thin, flat steel bar stock of mass per unit length 4.73 kg/m. The frame is supported by a pin at C and by a cable AB. Determine (a) the tension in the cable, (b) the reaction at C.

To determine

The tension in the cable.

Answer to Problem 5.29P

The tension in the cable is $\underset{\_}{T_{BA}=125.3\text{N}}$.

Explanation of Solution

The diagram for the centre of gravity is depicted below:

Write the equation of x-coordinate of the centre of gravity,

$\overline{X}\Sigma L=\Sigma \overline{x}L$ (I)

Here, the area is $\Sigma \overline{x}L$, the x-coordinate is $\overline{X}$, the total length $\Sigma L$.

Write the equation of weight,

$W=\left(m\text{'}\Sigma L\right)g$ (II)

Here, the weight is $W$, the mass per unit length is $\left(m\text{'}\Sigma L\right)$, the gravitational acceleration is $g$.

Write the equation of momentum at equilibrium,

$r_1\left[\left(3/5\right)T_{BA}\right]-\left[r_{2}W\right]=0$ (III)

Conclusion:

	$L\left(\text{m}\right)$	$\overline{x}\left(\text{m}\right)$	$\overline{x}L\left({\text{m}}^{\text{2}}\right)$
1	1.35	0.675	0.91125
2	0.6	0.3	0.18
3	0.75	0	0
4	0.75	0.2	0.15
5	$\left(\pi /2\right)\left(0.75\right)=1.1781$	1.07746	1.26936
$\Sigma$	4.62810		2.5106

Substitute $4.62810\text{m}$ for $\Sigma L$, $2.5106{\text{m}}^{\text{2}}$ for $\Sigma \overline{x}L$ in Equation (I).

$\begin{array}{c}\overline{X}\left(4.62810\text{m}\right)=2.5106{\text{m}}^2\\ \overline{X}=0.54247\text{m}\end{array}$

Substitute $4.73\text{kg/m}$ for $m\text{'}$, $4.62810\text{m}$ for $\Sigma L$, $9.8{\text{m/s}}^{\text{2}}$ for $g$ in Equation (II).

$\begin{array}{c}W=\left[\left(4.73\text{kg/m}\right)\left(4.62810\text{m}\right)\right]\left(9.8{\text{m/s}}^{\text{2}}\right)\\ =214.75\text{N}\end{array}$

Substitute, $1.55\text{m}$ for $r_1$, $0.54247\text{m}$ for $r_2$, $214.75\text{N}$ for $W$.

$\begin{array}{c}\left(1.55\text{m}\right)\left[\left(3/5\right)T_{BA}\right]-\left[\left(0.54247\text{m}\right)\left(214.75\text{N}\right)\right]=0\\ T_{BA}=125.3\text{N}\end{array}$

Thus, The tension in the cable is $\underset{\_}{T_{BA}=125.3\text{N}}$.

To determine

The reaction at C.

Answer to Problem 5.29P

The reaction at C is $\underset{\_}{137.0\text{N at an angle 56}\text{.7}°}$.

Explanation of Solution

The diagram for the reaction is depicted below:

Refer fig 2.

Write the equation of reaction along x-axis.

$C_x-\frac{3}{5}\left(T_{BA}\right)=0$ (IV)

Here, the reaction along x-axis is $C_x$.

Write the equation of reaction along y-axis.

$C_y+\frac{4}{5}\left(T_{BA}\right)-W=0$ (V)

Here, the reaction along y-axis is $C_y$.

Write the expression for the magnitude of the reaction at C,

$C=\sqrt{C_x^2+C_y^2}$ (VI)

Here, $C$ is the magnitude of reaction.

Write the equation of the direction of the reaction,

$\theta ={\tan }^{-1}\left(\frac{C_y}{C_x}\right)$ (VII)

Conclusion:

Substitute, $125.264\text{N}$ for $T_{BA}$ in equation (IV),

$\begin{array}{c}{C}_x-\frac{3}{5}\left(125.64\text{N}\right)=0\\ C_x=75.158\text{N}\end{array}$

Substitute, $125.264\text{N}$ for $T_{BA}$, $\left(27\sqrt{5}/25\right)P$ for $R$ in equation (II)

$\begin{array}{c}{C}_y+\frac{4}{5}\left(125.264\text{N}\right)-\left(214.75\text{N}\right)=0\\ C_y=114.539\text{N}\end{array}$

Substitute, $75.158\text{N}$ for $C_x$, $114.539\text{N}$ for $C_y$ in equation (VI),

$\begin{array}{c}C=\sqrt{{\left(75.158\text{N}\right)}^2+{\left(114.539\text{N}\right)}^2}\\ =137.0\text{N}\end{array}$

Substitute, $75.158\text{N}$ for $C_x$, $114.539\text{N}$ for $C_y$ in equation (VII)

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{114.539\text{N}}{75.158\text{N}}\right)\\ =56.7°\end{array}$

Thus, the reaction at C is $\underset{\_}{137.0\text{N at an angle 56}\text{.7}°}$.