Chapter 4.3, Problem 4.129P

Frame ABCD is supported by a ball-and-socket joint at A and by three cables. For a = 150 mm, determine the tension in each cable and the reaction at A.

Fig. P4.129 and P4.130

To determine

The tension in each cable and the reaction at $A$.

Answer to Problem 4.129P

The tension in each cable and the reaction at $A$ are $625\text{N}$ for $T_{DG}$, $600\text{N}$ for $T_{CF}$ and $975\text{N}$ for $T_{BE}$ and $A=\left(2100\text{N}\right)i+\left(175.0\text{N}\right)j-\left(375\text{N}\right)k$ respectively.

Explanation of Solution

The tension in the cable $CF$ can be represented as $T_{CF}$, in the cable $BE$ can be represented as $T_{BE}$ and in the cable $DG$ can be represented as $T_{DG}$. The free body diagram of the given arrangement is given by Figure 1.

Figure 1

The position vector of the point $B$ is given as,

$r_B=\left(0.48\text{m}\right)i$

The position vector of the point $H$ is given as,

$r_H=\left(0.48\text{}\text{ m}\right)i-\left(0.15\text{m}\right)k$

The position vector of the point $C$ is given as,

$r_C=\left(0.48\text{m}\right)i-\left(0.14\text{}\text{m}\right)k$

The position vector of the point $D$ is given as,

$r_D=\left(0.48\text{m}\right)i-\left(0.3\text{}\text{m}\right)k$

The vector $\overrightarrow{BE}$ is given as $\overrightarrow{BE}=\left(-0.48\text{ m}\right)i+\left(0.2\text{m}\right)j$, vector $\overrightarrow{DG}$ is given as $\overrightarrow{DG}=\left(-0.48\text{ m}\right)i+\left(0.14\text{m}\right)k$ and $\overrightarrow{CF}$ is given as $\overrightarrow{CF}=\left(\text{0}\text{.48 m}\right)i$.

The tension across the cable $BE$ as,

$T_{BE}=T_{BE}\frac{\overrightarrow{BE}}{BE}$

Here, $T_{BE}$ is the tension across the cable $BE$, $\overrightarrow{BE}$ is the vector along the path of the points, $BE$ is the magnitude of the vector.

Substitute $\left(-0.48\text{ m}\right)i+\left(0.2\text{m}\right)j$ for $\overrightarrow{BE}$ and $0.50\text{m}$ for $BE$ to get $T_{BE}$.

$\begin{array}{c}{T}_{BE}=\frac{T_{BE}}{0.5\text{m}}\left(\left(-0.48\text{ m}\right)i+\left(0.2\text{m}\right)j\right)\\ =\frac{T_{BE}}{25}\left(-24i+7j\right)\end{array}$

The tension across the cable $DG$ as,

$T_{DG}=T_{DG}\frac{\overrightarrow{DG}}{DG}$

Here, $T_{DG}$ is the tension across the cable $DG$, $\overrightarrow{DG}$ is the vector along the path of the points, $DG$ is the magnitude of the vector.

Substitute $\left(-0.48\text{ m}\right)i+\left(0.14\text{m}\right)k$ for $\overrightarrow{DG}$ and $0.52\text{m}$ for $DG$ to get $T_{DG}$.

$\begin{array}{c}{T}_{DG}=\frac{T_{DG}}{0.52\text{m}}\left(\left(-0.48\text{ m}\right)i+\left(0.14\text{m}\right)k\right)\\ =\frac{T_{DG}}{13}\left(-12j+5k\right)\end{array}$

The tension across the cable $CF$ as,

$T_{CF}=T_{CF}\frac{\overrightarrow{CF}}{CF}$

Here, $T_{CF}$ is the tension across the cable $CF$, $\overrightarrow{CF}$ is the vector along the path of the points, $CF$ is the magnitude of the vector.

Substitute $\left(\text{0}\text{.48 m}\right)i$ for $\overrightarrow{CF}$ and $0.48\text{m}$ for $CF$ to get $T_{CF}$.

$\begin{array}{c}{T}_{CF}=\frac{T_{CF}}{0.48\text{m}}\left(\left(\text{0}\text{.48 m}\right)i\right)\\ =T_{CF}i\end{array}$

The weight is given as,

$W=-mgj$

Here, $W$ is the weight, $m$ is the mass and $g$ is the acceleration due to gravity.

Substitute $\text{350}\text{N}$ for $mg$.

$W=-\left(\text{350}\text{N}\right)j$

The force on the point is zero. This means that the sum of the moments of the force will also be zero.

$r_H\times W+r_B\times T_{BE}+r_D\times T_{DG}+r_C\times T_{CF}=0$

Substitute the vector values and determine the cross product.

$\begin{array}{l}\left|\begin{array}{ccc}i& j& k\\ 0.48& 0& -0.15\\ 0& -350& 0\end{array}\right|+\left|\begin{array}{ccc}i& j& k\\ 0.48& 0& 0\\ -24& 7& 0\end{array}\right|\frac{T_{BE}}{25}+\\ \left|\begin{array}{ccc}i& j& k\\ 0.48& 0& -0.3\\ 0& -12& 5\end{array}\right|\frac{T_{DG}}{13}+\left|\begin{array}{ccc}i& j& k\\ 0.48& 0& -0.14\\ 1& 0& 0\end{array}\right|T_{CF}=0\\ \left(0.0875T_{DG}-52.5\text{N}\cdot \text{m}\right)i+\left(-0.1846T_{BE}+\left(\frac{24}{25}\times T_{DG}\times 0.3\text{m}\right)\right)j\\ +\left(-168\text{N}\cdot \text{m}+\left(\frac{7}{25}{T}_{DG}\times 0.48\text{m}\right)+\left(0.14\text{m}\right)T_{CF}\right)k=0\end{array}$ (I)

Since the force at the point is zero, the $x$ component is given as,

$A_x-T_{CF}-\frac{12}{13}{T}_{BE}-\frac{24}{25}{T}_{DG}=0$ (II)

Here, $A_x$ is the $x$ coefficient of the reaction at $A$.

Since the force at the point is zero, the $y$ component is given as,

$A_y+\frac{7}{25}{T}_{DG}-W=0$ (III)

Here, $A_y$ is the $y$ coefficient of the reaction at $A$.

Since the force at the point is zero, the $z$ component is given as,

$A_z+\frac{5}{13}\left(T_{BE}\right)=0$ (IV)

Here, $A_z$ is the $z$ coefficient of the reaction at $A$.

Conclusion:

Equate the coefficients of $i$ in (I) and solve for $T_{DG}$.

$\begin{array}{c}0.0875T_{DG}-52.5\text{N}\cdot \text{m}=0\\ T_{DG}=625\text{N}\end{array}$

Equate the coefficients of $j$ in (I) and solve for $T_{BD}$ by substituting $625\text{N}$ for $T_{DG}$.

$\begin{array}{c}-0.1846T_{BE}+\left(\frac{24}{25}\times \left(625\text{N}\right)\times 0.3\text{m}\right)=0\\ T_{BE}=975\text{N}\end{array}$

Equate the coefficients of $k$ in (I) and solve for $T_{CF}$ by substituting $625\text{N}$ for $T_{DG}$.

$\begin{array}{c}-168\text{N}\cdot \text{m}+\left(\frac{7}{25}\left(625\text{N}\right)\times 0.48\text{m}\right)+\left(0.14\text{m}\right)T_{CF}=0\\ T_{CF}=600\text{N}\end{array}$

Substitute $625\text{N}$ for $T_{DG}$, $600\text{N}$ for $T_{CF}$ and $975\text{N}$ for $T_{BE}$ in (II) to get $A_x$.

$\begin{array}{c}{A}_x-\left(600\text{N}\right)-\frac{12}{13}\left(975\text{N}\right)-\frac{24}{25}\left(625\text{N}\right)=0\\ A_x=2100\text{N}\end{array}$ (V)

Substitute $625\text{N}$ for $T_{DG}$ and $\text{350}\text{N}$ for $W$ in (III) to get $A_y$.

$\begin{array}{c}{A}_y+\frac{7}{25}\left(625\text{N}\right)-\left(\text{350}\text{N}\right)=0\\ A_y=175.0\text{N}\end{array}$ (VI)

Substitute $975\text{N}$ for $T_{BE}$ in (IV) to get $A_z$.

$\begin{array}{c}{A}_z+\frac{5}{13}\left(975\text{N}\right)=0\\ A_z=-375\text{N}\end{array}$ . (VII)

From (V), (VI) and (VII), the reaction on A is given as,

$A=\left(2100\text{N}\right)i+\left(175.0\text{N}\right)j-\left(375\text{N}\right)k$

Therefore, the tension in each cable and the reaction at $A$ are $625\text{N}$ for $T_{DG}$, $600\text{N}$ for $T_{CF}$ and $975\text{N}$ for $T_{BE}$ and $A=\left(2100\text{N}\right)i+\left(175.0\text{N}\right)j-\left(375\text{N}\right)k$ respectively.