Chapter 4.2, Problem 4.69P

A 50-kg crate is attached to the trolley-beam system shown. Knowing that a = 1.5 m, determine (a) the tension in cable CD, (b) the reaction at B

Fig.P4.69

To determine

The tension in the cable $CD$.

Answer to Problem 4.69P

The tension in the cable $CD$ is $499\text{N}$

Explanation of Solution

Calculation:

If the rigid body subjected to three forces, such a body is commonly called a three-force body. If a two force body is in equilibrium the lines of action of the three forces must be either concurrent or parallel.

The figure 1 represents the free body diagram of the system. The weight is acting downwards. The $W$ and $T_{CD}$ intersect at $E$. The three forces $W$, $T_{CD}$ and $B$ are intersect at $E$.

Write the expression for the angle $\beta$ from the given figure.

$\tan \beta =\frac{0.749\text{m}}{WB}$ (I)

Here, $\beta$ is the angle.

Write the expression for the weight acting on the system.

$W=mg$

Here, $m$ is the mass, $g$ is the acceleration due to gravity

Figure 2 represents a force triangle as shown in the above figure with its interior angles computed from the known directions of the forces. Then use the law of sines to find the unknown forces.

Write the expression for the law of sines from the force triangle.

$\frac{W}{\sin \left(61.55°\right)}=\frac{T_{CD}}{\sin \left(90-\beta \right)°}=\frac{A}{\sin \left(55°\right)}$ (II)

Conclusion:

Substitute $1.5\text{m}$ for $WB$ in equation I to find $\beta$.

$\begin{array}{c}\tan \beta =\frac{0.7497\text{m}}{1.5\text{m}}\\ =0.499\\ \beta ={\tan }^{-1}\left(0.499\right)\\ =26.55°\end{array}$

Substitute $50\text{kg}$ for $m$ and $9.81\text{m}/{\text{s}}^2$ in equation II to find $g$.

$\begin{array}{c}W=\left(50\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\\ =490.5\text{N}\end{array}$

Substitute $490.5\text{N}$ for $W$, and $26.55°$ for $\beta$ in equation III to find $B$ and $T_{CD}$.

$\begin{array}{l}\frac{490.5\text{N}}{\sin \left(61.55°\right)}=\frac{T_{CD}}{\sin \left(90-26.55\right)°}=\frac{A}{\sin \left(55°\right)}\\ \frac{W}{\sin \left(61.55°\right)}=\frac{T_{CD}}{\sin 63.44°}=\frac{A}{\sin \left(55°\right)}\\ \frac{490.5}{0.879}=\frac{T_{CD}}{0.894}=\frac{A}{0.819}\end{array}$

On rearranging the above equation to find $T$.

$\begin{array}{c}{T}_{CD}\times 0.879=438.50\\ T_{CD}=\frac{438.50}{0.879}\\ =498.87\text{N}\\ =\text{499}\text{N}\end{array}$

Therefore, the tension in the cable $CD$ is $499\text{N}$

To determine

The reaction at $B$.

Answer to Problem 4.69P

The reaction at $B$ is $457\text{N}$.

Explanation of Solution

Calculation:

If the rigid body subjected to three forces, such a body is commonly called a three-force body. If a two force body is in equilibrium the lines of action of the three forces must be either concurrent or parallel.

The figure 1 represents the free body diagram of the system. The weight is acting downwards. The $W$ and $T_{CD}$ intersect at $E$. The three forces $W$, $T_{CD}$ and $B$ are intersect at $E$.

Write the expression for the angle $\beta$ from the given figure.

$\tan \beta =\frac{0.749\text{m}}{WB}$ (I)

Here, $\beta$ is the angle.

Write the expression for the weight acting on the system.

$W=mg$

Here, $m$ is the mass, $g$ is the acceleration due to gravity

Figure 2 represents a force triangle as shown in the above figure with its interior angles computed from the known directions of the forces. Then use the law of sines to find the unknown forces.

Write the expression for the law of sines from the force triangle.

$\frac{W}{\sin \left(61.55°\right)}=\frac{T_{CD}}{\sin \left(90-\beta \right)°}=\frac{B}{\sin \left(55°\right)}$ (II)

Conclusion:

Substitute $1.5\text{m}$ for $WB$ in equation I to find $\beta$.

$\begin{array}{c}\tan \beta =\frac{0.7497\text{m}}{1.5\text{m}}\\ =0.499\\ \beta ={\tan }^{-1}\left(0.499\right)\\ =26.55°\end{array}$

Substitute $50\text{kg}$ for $m$ and $9.81\text{m}/{\text{s}}^2$ in equation II to find $g$.

$\begin{array}{c}W=\left(50\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\\ =490.5\text{N}\end{array}$

Substitute $490.5\text{N}$ for $W$, and $26.55°$ for $\beta$ in equation III to find $B$ and $T_{CD}$.

$\begin{array}{l}\frac{490.5\text{N}}{\sin \left(61.55°\right)}=\frac{T_{CD}}{\sin \left(90-26.55\right)°}=\frac{B}{\sin \left(55°\right)}\\ \frac{W}{\sin \left(61.55°\right)}=\frac{T_{CD}}{\sin 63.44°}=\frac{B}{\sin \left(55°\right)}\\ \frac{490.5}{0.879}=\frac{T_{CD}}{0.894}=\frac{B}{0.819}\end{array}$

Rearrange the above equation to find $B$.

$\begin{array}{c}490.50\text{N}\times \text{0}\text{.819=}B\times 0.879\\ B=457.01\text{N}\end{array}$

Therefore, reaction at $B$ is $457\text{N}$.