Chapter 4.3, Problem 4.117P

(a)

To determine

The tension on the cable.

Answer to Problem 4.117P

The tension on the cable is $\text{345}\text{N}$.

Explanation of Solution

The tension in the cable $DCE$ can be represented as $T_{CD}$ and $T_{CE}$ respectively. The total tension on the cable is given by $T$. The free body diagram of the given arrangement is given by Figure 1.

Figure 1

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

$\begin{array}{c}{r}_{B/A}=\left(960-180\right)i\\ =780i\end{array}$

The position vector of the point $A$ from $G$ is given as,

$\begin{array}{c}{r}_{G/A}=\left(\frac{960}{2}-90\right)i+\frac{450}{2}k\\ =390i+225k\end{array}$

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

$r_{C/A}=600i+450k$

The vector $\overrightarrow{CD}$ is given as $\overrightarrow{CD}=-690i+675j-450k$ and the vector $\overrightarrow{CE}$ is given as $\overrightarrow{CE}=270i+675j-450k$.

The tension across the cable from $D$ to $C$ is given as,

$T_{CD}=T\frac{\overrightarrow{CD}}{CD}$

Here, $T_{CD}$ is the tension across the cable from $D$ to $C$, $\overrightarrow{CD}$ is the vector along the path of the points, $CD$ is the magnitude of the vector.

Substitute $-690i+675j-450k$ for $\overrightarrow{CD}$ and $1065$ for $CD$ to get $T_{CD}$.

$T_{CD}=\frac{T}{1065}\left(-690i+675j-450k\right)$

The tension across the cable from $E$ to $C$ is given as,

$T_{CE}=T\frac{\overrightarrow{CE}}{CE}$

Here, $T_{CE}$ is the tension across the cable from $E$ to $C$, $\overrightarrow{CE}$ is the vector along the path of the points, $CE$ is the magnitude of the vector.

Substitute $270i+675j-450k$ for $\overrightarrow{CE}$ and $855$ for $CE$ to get $T_{CE}$.

$T_{CE}=\frac{T}{855}\left(270i+675j-450k\right)$

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 $100\text{kg}$ for $m$ and $9.81{\text{m/s}}^2$ for $g$.

$\begin{array}{c}W=-\left(100\text{kg}\right)\left(9.81{\text{m/s}}^2\right)j\\ =-\left(981\text{N}\right)j\end{array}$

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

$r_{C/A}\times T_{CD}+r_{C/A}\times T_{CE}+r_{G/A}\times W+r_{B/A}\times B=0$

Here, $B$ is the reaction at $B$.

Substitute the vector values and determine the cross product.

$\begin{array}{l}\left|\begin{array}{ccc}i& j& k\\ 600& 0& 450\\ -690& 675& -450\end{array}\right|\frac{T}{1065}+\left|\begin{array}{ccc}i& j& k\\ 600& 0& 450\\ 270& 675& -450\end{array}\right|\frac{T}{855}+\left|\begin{array}{ccc}i& j& k\\ 390& 0& 225\\ 0& -981& 0\end{array}\right|+\left|\begin{array}{ccc}i& j& k\\ 780& 0& 0\\ 0& B_y& B_z\end{array}\right|=0\\ \left(-\left(675\right)\left(450\right)\frac{T}{1065}-\left(675\right)\left(450\right)\frac{T}{855}+220.73\times {10}^3\right)i+\left(\left(-690\times 450+600\times 450\right)\frac{T}{1065}+\left(270\times 450+600\times 450\right)\frac{T}{855}-780B_z\right)j\\ +\left(\left(600\times 675\right)\frac{T}{1065}+\left(600\times 675\right)\frac{T}{855}-382.59\times {10}^3+780B_y\right)k=0\end{array}$ (I)

Conclusion:

Equate the coefficients of $i$ in (I) and solve for $T$.

$\begin{array}{c}\left(-\left(675\right)\left(450\right)\frac{T}{1065}-\left(675\right)\left(450\right)\frac{T}{855}+220.73\times {10}^3\right)=0\\ T\left(\frac{\left(675\right)\left(450\right)}{1065}+\frac{\left(675\right)\left(450\right)}{855}\right)=220.73\times {10}^3\\ T=344.64\text{N}\\ \approx \text{345}\text{N}\end{array}$

Therefore, the tension on the cable is $\text{345}\text{N}$.

(b)

To determine

The reactions at $A$ and $B$.

Answer to Problem 4.117P

The reactions at $A$ and $B$ are $A=\left(114.454\text{N}\right)i+\left(377.30\text{N}\right)j\text{+}\left(141.496\right)k$ and $B=\left(113.178\text{N}\right)j\text{+}\left(\text{185}\text{.516 N}\right)k$ respectively.

Explanation of Solution

The free body diagram of the given arrangement is given in Figure 1.

Equate the coefficients of $j$ in (I) and solve for $B_z$ by substituting $344.64\text{N}$ for $T$.

$\begin{array}{c}\left(\left(-690\times 450+600\times 450\right)\frac{344.64\text{N}}{1065}+\left(270\times 450+600\times 450\right)\frac{344.64\text{N}}{855}-780B_z\right)=0\\ -13106.028+157808.8421-780B_z=0\\ B_z=\frac{144702.8141}{780}\\ =185.516\text{N}\end{array}$ (II)

Equate the coefficients of $k$ in (I) and solve for $B_y$ by substituting $344.64\text{N}$ for $T$.

$\begin{array}{c}\left(\left(600\times 675\right)\frac{344.64\text{N}}{1065}+\left(600\times 675\right)\frac{344.64\text{N}}{855}-382.59\times {10}^3+780B_y\right)=0\\ 780B_y=382.59\times {10}^3-131060.28-163250.53\\ B_y=\frac{88279.19}{780}\\ =113.178\text{N}\end{array}$ (III)

The net force acting on the point is zero.

$A+B+T_{CD}+T_{CE}+W=0$

Here, $A$ is the reaction at the point $A$.

Substitute $113.178\text{N}j\text{+185}\text{.516 N}k$ for $B$, $\frac{T}{1065}\left(-690i+675j-450k\right)$ for $T_{CD}$ and $\frac{T}{855}\left(270i+675j-450k\right)$ for $T_{CE}$ with $344.64\text{N}$ for $T$ and $-\left(981\text{N}\right)j$ for $W$ to get $A$.

$\left(A_{x}i+A_{y}j+A_{z}k\right)+\left(113.178\text{N}j\text{+185}\text{.516 N}k\right)+\frac{344.64\text{N}}{1065}\left(-690i+675j-450k\right)+\left(\frac{344.64\text{N}}{855}\left(270i+675j-450k\right)\right)+-\left(981\text{N}\right)j=0$ (IV)

Equate the coefficients of $i$ in (IV) and solve for $A_x$.

$\begin{array}{c}{A}_x-\frac{690}{1065}\left(344.64\right)+\frac{270}{855}\left(344.64\right)=0\\ A_x=114.454\text{N}\end{array}$ (V)

Equate the coefficients of $j$ in (IV) and solve for $A_y$.

$\begin{array}{c}{A}_y+113.178+\frac{675}{1065}\left(344.64\right)+\frac{675}{855}\left(344.64\right)-981=0\\ A_y=377.30\text{N}\end{array}$ (VI)

Equate the coefficients of $k$ in (IV) and solve for $A_z$.

$\begin{array}{c}{A}_z+185.516-\frac{450}{1065}\left(344.64\right)-\frac{450}{855}\left(344.64\right)=0\\ A_z=141.496\text{N}\end{array}$ (VII)

Conclusion:

From (II) and (III) the vector $B$ is given as,

$B=\left(113.178\text{N}\right)j\text{+}\left(\text{185}\text{.516 N}\right)k$

And, from (V), (VI) and (VII) the vector $A$ is given as,

$A=\left(114.454\text{N}\right)i+\left(377.30\text{N}\right)j\text{+}\left(141.496\right)k$

Therefore, the reactions at $A$ and $B$ are $A=\left(114.454\text{N}\right)i+\left(377.30\text{N}\right)j\text{+}\left(141.496\right)k$ and $B=\left(113.178\text{N}\right)j\text{+}\left(\text{185}\text{.516 N}\right)k$ respectively