Chapter 2.5, Problem 2.107P

Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that Q = 0, find the value of P for which the tension in cable AD is 305 N.

Fig.P2.107and P2.108

To determine

The value of $P$ shown in figure $\text{P2}\text{.107}$ in the textbook for which tension in the cable shown $AD$ is $305\text{N}$.

Answer to Problem 2.107P

The force $P$ at the point $A$ shown in figure $\text{P2}\text{.107}$ is $\underset{\_}{960\text{N}}$.

Explanation of Solution

The sketch of the cables connected at $A$, where the forces $P$ and $Q$ applied is shown in below figure1.

Free body diagram at $\text{A}$ is shown in figure 2.

Here, $T_{AB}$ is the magnitude of tension in cable $AB$, $T_{AC}$ is the magnitude of tension in cable $AC$, $T_{AD}$ is the magnitude of tension in the cable $AD$ , $P$ is the magnitude of the force exerted in the $x$ direction and $Q$ is the magnitude of force exerted in the $y$ direction.

The force $Q$ exerted at point $A$ is zero and the tension in cable $AD$ is $305\text{N}$.

Let $T_{AB}$, $T_{AC}$, and $P$ are the tension vector in cable $AB$, $AC$, $AD$ and force exerted in the $x$ direction respectively and $Q$ is the force at $A$ along the $y$ direction.

Let $i$ , $j$ and $k$ are the unit vectors along the of $x,y\text{and}z$ direction.

Write the equation of vector distance $AB$.

$\overrightarrow{AB}=\left(x_2-x_1\right)i+(y_2-y_1)j+(z_2-z_1)k$ (I)

Here, $\overrightarrow{AB}$ is the vector distance of the cable $AB$, the variables $x_{1,}{y}_1{\text{and z}}_1$ are the coordinates of point $A$ and $x_2,y_2{\text{and z}}_2$ are the coordinates of point $B$.

Write the vector distance of the cable $AC$.

$\overrightarrow{AC}=\left(x_3-x_1\right)i+(y_3-y_1)j+(z_3-z_1)k$ (II)

Here, $\overrightarrow{AC}$ is the vector distance of the cable $AC$, the variables $x_{1,}{y}_1{\text{and z}}_1$ are the coordinates of point $A$ and $x_3,y_3{\text{and z}}_3$ are the coordinates of point $C$.

Write the vector distance of the cable $AD$.

$\overrightarrow{AD}=\left(x_4-x_1\right)i+(y_4-y_1)j+(z_4-z_1)k$ (III)

Here, $\overrightarrow{AD}$ is the vector distance of the cable $AD$, the variables $x_{1,}{y}_1{\text{and z}}_1$ are the coordinates of point $A$ and $x_4,y_4{\text{and z}}_4$ are the coordinates of point $D$.

Write the equation of tension in the cable $AB$.

$T_{AB}={\lambda }_{AB}{T}_{AB}$ (IV)

Here, $T_{AB}$ is the tension in the cable $AB$ , $T_{AB}$ is the magnitude of the tension in the cable $AB$ and ${\lambda }_{AB}$ is the unit vector in the direction of $AB$.

Write the equation of ${\lambda }_{AB}$.

${\lambda }_{AB}=\frac{\overrightarrow{AB}}{AB}$ (V)

Write the equation of tension in the cable $AC$.

$T_{AC}={\lambda }_{AC}{T}_{AC}$ (VI)

Here, $T_{AC}$ is the tension in the cable $AC$ , $T_{AC}$ is the magnitude of the tension in the cable $AC$ and ${\lambda }_{AC}$ is the unit vector in the direction of $AC$.

Write the equation of ${\lambda }_{AC}$.

${\lambda }_{AC}=\frac{\overrightarrow{AC}}{AC}$ (VII)

Write the equation of tension in the cable $AD$.

$T_{AD}={\lambda }_{AD}{T}_{AD}$ (VIII)

Here, $T_{AD}$ is the tension in the cable $AD$ , $T_{AD}$ is the magnitude of the tension in the cable $AD$ and ${\lambda }_{AD}$ is the unit vector in the direction of $AD$.

Write the equation of ${\lambda }_{AD}$.

${\lambda }_{AD}=\frac{\overrightarrow{AD}}{AD}$ (IX)

Write the equation of force exerting at point $A$ along $x$ direction.

$P=Pj$ (X)

Here, $P$ is the vector notation of weight of the body and $P$ is the magnitude of weight.

Write the equation of force exerting at point $A$ along $y$ direction.

$Q=Qj$

Write the equilibrium condition for the forces at $A$.

${\displaystyle \sum F}=0$

Here, F is the force

The above equation implies that at equilibrium, total force acting on the cable at $A$ is zero.

Refer figure 2 and write the equation of total force.

$T_{AB}+T_{AC}+T_{AD}+P+Q=0$

Since $Q=0\text{N}$, write the above equation in simplified form

$T_{AB}+T_{AC}+T_{AD}+P=0$ (XI)

Conclusion:

Substitute $960\text{mm}$ for $x_1$ , $240\text{mm}$ for $y_1$ , $0\text{mm}$ for $z_1$ , $0\text{mm}$ for $x_2$ , $0\text{mm}$ for $y_2$ and $380\text{mm}$ for $z_2$ in equation (I) to get $\overrightarrow{AB}$.

$\overrightarrow{AB}=-\left(960\text{mm}\right)i-(240\text{mm})j+(380\text{mm})k$

Calculate the magnitude of $\overrightarrow{AB}$.

$\begin{array}{c}AB=\sqrt{{\left(-960\text{mm}\right)}^2+{(-240\text{mm})}^2+{(-320\text{mm})}^2}\\ =\text{1060}\text{mm}\end{array}$

Substitute $960\text{mm}$ for $x_1$ , $240\text{mm}$ for $y_1$ , $0\text{mm}$ for $z_1$ , $0\text{mm}$ for $x_3$ , $0\text{mm}$ for $y_3$ and $-320\text{mm}$ for $z_3$ in equation (II) to get $\overrightarrow{AC}$.

$\overrightarrow{AC}=-\left(960\text{mm}\right)i-(240\text{mm})j-(320\text{mm})k$

Calculate the magnitude of $\overrightarrow{AC}$.

$\begin{array}{c}AC=\sqrt{{\left(-960\text{mm}\right)}^2+{(-240\text{mm})}^2+{(-320\text{mm})}^2}\\ =1040\text{mm}\end{array}$

Substitute $960\text{mm}$ for $x_1$ , $240\text{mm}$ for $y_1$ , $0\text{mm}$ for $z_1$ ,$0\text{mm}$ for $x_4$ , $960\text{mm}$ for $y_4$ and $0\text{mm}$ for $z_4$ in equation (III) to get $\overrightarrow{AD}$.

$\overrightarrow{AD}=\left(-960\text{mm}\right)i+(720\text{mm})j-(220\text{mm})k$

Calculate the magnitude of $\overrightarrow{AD}$.

$\begin{array}{c}AD=\sqrt{{\left(-960\text{mm}\right)}^2+{(720\text{mm})}^2+{(-220\text{mm})}^2}\\ =1220\text{mm}\end{array}$

Substitute $-\left(960\text{mm}\right)i-(240\text{mm})j+(380\text{mm})k$ for $\overrightarrow{AB}$ and $1060\text{mm}$ for $AB$ in equation (V) to get ${\lambda }_{AB}$.

$\begin{array}{c}{\lambda }_{AB}=\frac{-\left(960\text{mm}\right)i-(240\text{mm})j+(380\text{mm})k}{1060\text{mm}}\\ =-\frac{48}{53}i-\frac{12}{53}j+\frac{19}{53}k\end{array}$

Substitute $-\frac{48}{53}i-\frac{12}{53}j+\frac{19}{53}k$ for ${\lambda }_{AB}$ in equation (IV) to get $T_{AB}$.

$T_{AB}=\left(-\frac{48}{53}i-\frac{12}{53}j+\frac{19}{53}k\right)T_{AB}$

Substitute $-\left(960\text{mm}\right)i-(240\text{mm})j-(320\text{mm})k$ for $\overrightarrow{AC}$ and $7.40\text{mm}$ for $AC$ in equation
(VII) to get ${\lambda }_{AC}$.

$\begin{array}{c}{\lambda }_{AC}=\frac{-\left(960\text{mm}\right)i-(240\text{mm})j-(320\text{mm})k}{1040\text{mm}}\\ =-\frac{12}{13}i-\frac{3}{13}j-\frac{4}{13}k\end{array}$

Substitute $-\frac{12}{13}i-\frac{3}{13}j-\frac{4}{13}k$ for ${\lambda }_{AC}$ in equation (VI) to get $T_{AC}$.

$T_{AC}=\left(-\frac{12}{13}i-\frac{3}{13}j-\frac{4}{13}k\right)T_{AC}$

Substitute $\left(-960\text{mm}\right)i+(720\text{mm})j-(220\text{mm})k$ for $\overrightarrow{AD}$ and $\text{6}\text{.50}\text{mm}$ for $AD$ in equation (IX) to get ${\lambda }_{AD}$.

${\lambda }_{AD}=\frac{\left(-960\text{mm}\right)i+(720\text{mm})j-(220\text{mm})k}{\text{1220}\text{mm}}$

Substitute $\frac{\left(-960\text{mm}\right)i+(720\text{mm})j-(220\text{mm})k}{\text{1220}\text{mm}}$ for ${\lambda }_{AD}$ and $305\text{N}$ for $T_{AD}$ in equation (VIII) to get $T_{AD}$.

$\begin{array}{c}{T}_{AD}=\left(\frac{\left(-960\text{mm}\right)i+(720\text{mm})j-(220\text{mm})k}{\text{1220}\text{mm}}\right)\left(305\text{N}\right)\\ =-\left(240\text{N}\right)i+\left(180\text{N}\right)j-\left(55\text{N}\right)k\end{array}$

Substitute $\left(-\frac{48}{53}i-\frac{12}{53}j+\frac{19}{53}k\right)T_{AB}$ for $T_{AB}$ , $\left(-\frac{12}{13}i-\frac{3}{13}j-\frac{4}{13}k\right)T_{AC}$ for $T_{AC}$ , $-\left(240\text{N}\right)i+\left(180\text{N}\right)j-\left(55\text{N}\right)k$ and $Pi$ for $P$ in the in equation (XI) to get force $P$ exerted at point $A$.

$\begin{array}{l}\left(-\frac{48}{53}i-\frac{12}{53}j+\frac{19}{53}k\right)T_{AB}+\left(-\frac{12}{13}i-\frac{3}{13}j-\frac{4}{13}k\right)T_{AC}+-\left(240\text{N}\right)i+\left(180\text{N}\right)j-\left(55\text{N}\right)k+Pj=0\\ \left(-\frac{48}{53}{T}_{AB}-\frac{12}{13}{T}_{AC}-240\text{N+}P\right)i+\left(-\frac{12}{53}{T}_{AB}-\frac{3}{13}{T}_{AC}+180\text{N}\right)j+\left(+\frac{19}{53}{T}_{AB}-\frac{4}{13}{T}_{AC}-55\text{N}\right)k=0\end{array}$

Since total force is zero. Equate force along each direction as zero.

$-\frac{48}{53}{T}_{AB}-\frac{12}{13}{T}_{AC}-240\text{N+}P=0$ (XII)

$-\frac{12}{53}{T}_{AB}-\frac{3}{13}{T}_{AC}+180\text{N}\text{=}\text{0}$ (XIII)

$\frac{19}{53}{T}_{AB}-\frac{4}{13}{T}_{AC}-55\text{N}\text{=}\text{0}$ (XIV)

Multiply equation (XIII) by $-\frac{4}{3}$ to modify the equation.

$\frac{48}{159}{T}_{AB}-\frac{4}{13}{T}_{AC}+180\text{N}\text{=}\text{0}$ (XV)

Add above equation with equation (XIV) to get $T_{AB}$.

$\begin{array}{c}\frac{48}{159}{T}_{AB}+\frac{4}{13}{T}_{AC}-240\text{N}+\left(\frac{19}{53}{T}_{AB}-\frac{4}{13}{T}_{AC}-55\text{N}\right)\text{=}\text{0}\\ \frac{48}{159}{T}_{AB}-240\text{N}+\left(\frac{19}{53}{T}_{AB}-55\text{N}\right)=0\\ T_{AB}=446.71\text{N}\end{array}$

Substitute $446.71\text{N}$ for $T_{AB}$ in equation (XIV) to get $T_{AC}$.

$\begin{array}{c}\frac{19}{53}\left(446.71\text{N}\right)-\frac{4}{13}{T}_{AC}-55\text{N}\text{=}\text{0}\\ {\text{T}}_{AC}=\text{341}\text{.71}\text{N}\end{array}$

Substitute $\text{341}\text{.71}\text{N}$ for $T_{AC}$, $446.71\text{N}$ for $T_{AB}$ in equation (XII) to get $P$.

$\begin{array}{c}-\frac{48}{53}\left(446.71\text{N}\right)-\frac{12}{13}\left(\text{341}\text{.71}\text{N}\right)-240\text{N+}P=0\\ P=960\text{N}\end{array}$

Therefore, the force $P$ exerted at $A$ is $\underset{\_}{960\text{N}i}$.