Chapter 2, Problem 2.136RP

To determine

The magnitude of P and Q.

Answer to Problem 2.136RP

The magnitude of P and Q are $\underset{\_}{131.2\text{N}}$ and $\underset{\_}{29.6\text{N}}$ respectively.

Explanation of Solution

Given information:

The weight (W) is 376 N.

Calculation:

Draw the free body diagram of point A as in Figure (1).

The force applied at the point A is $T_{AB}$,$T_{AC}$,W, $Q$, and P.

Write vector form of P as below:

$P=Pi$

Write vector form of Q as below:

$Q=Qk$

Write vector form of W as below:

$W=-\left(367\text{N}\right)j$

Since the tension in cable AB and AC is equal,

$T_{AB}=T_{AC}=\text{T}$

Express the vector $\overrightarrow{AB}$ as below:

$\overrightarrow{AB}=\left(-130\text{mm}\right)i+\left(400\text{mm}\right)j+\left(160\text{mm}\right)k$

Calculate distance of the member AB as below:

$\begin{array}{c}AB=\sqrt{{\left(-130\right)}^2+{\left(400\right)}^2+{\left(160\right)}^2}\\ =\sqrt{16,900+160,000+25,600}\\ =450\text{mm}\end{array}$

Calculate the unit vector $\left({\lambda }_{AB}\right)$ using the relation:

${\lambda }_{AB}=\frac{\overrightarrow{AB}}{AB}$

Substitute $\left(-130\text{mm}\right)i+\left(400\text{mm}\right)j+\left(160\text{mm}\right)k$ for $\overrightarrow{AB}$ and $450\text{mm}$ for AB.

$\begin{array}{c}{\lambda }_{AB}=\frac{\left(-130\text{mm}\right)i+\left(400\text{mm}\right)j+\left(160\text{mm}\right)k}{450\text{mm}}\\ =-0.2889i+0.889j+0.3556k\end{array}$

Calculate the tension force $\left(T_{AB}\right)$ in cable AB using the relation:

$T_{AB}=T{\lambda }_{AB}$

Substitute $-0.2889i+0.889j+0.355k$ for ${\lambda }_{AB}$.

$T_{AB}=\left(-0.2889i+0.889j+0.3556k\right)T$

Express the vector $\overrightarrow{AC}$ as below:

$\overrightarrow{AC}=\left(-150\text{mm}\right)i+\left(400\text{mm}\right)j+\left(-240\text{mm}\right)k$

Calculate distance of the member AC as below:

$\begin{array}{c}AC=\sqrt{{\left(-150\right)}^2+{\left(400\right)}^2+{\left(-240\right)}^2}\\ =\sqrt{22,500+160,000+57,600}\\ =490\text{mm}\end{array}$

Calculate the unit vector $\left({\lambda }_{AC}\right)$ using the relation:

${\lambda }_{AC}=\frac{\overrightarrow{AC}}{AC}$

Substitute $\left(-150\text{mm}\right)i+\left(400\text{mm}\right)j+\left(-240\text{mm}\right)k$ for $\overrightarrow{AC}$ and 490 mm for AC.

$\begin{array}{c}{\lambda }_{AC}=\frac{\left(-150\text{mm}\right)i+\left(400\text{mm}\right)j+\left(-240\text{mm}\right)k}{490\text{mm}}\\ =-0.3061i+0.8163j-0.4898k\end{array}$

Calculate the tension force $\left(T_{AC}\right)$ in Cable AC using the relation:

$T_{AC}=T{\lambda }_{AC}$

Substitute $-0.3061i+0.8163j-0.4898k$ for ${\lambda }_{AC}$.

$T_{AC}=\left(-0.3061i+0.8163j-0.4898k\right)T$

Calculate the tension in the cable using the relation:

Use the equilibrium condition,

${\displaystyle \sum F}=0$

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

Substitute $\left(-0.2889i+0.889j+0.3556k\right)T$ for $T_{AB}$, $\left(-0.3061i+0.8163j-0.4898k\right)T$ for $T_{AC}$, $Pi$ for P, $Qk$ for Q, and $\left(-376\text{N}\right)j$ for W.

$\begin{array}{l}\left\{\begin{array}{l}\left(-0.2889i+0.889j+0.3556k\right)T\\ +\left(-0.3061i+0.8163j-0.4898k\right)T\\ +Pi+Qk+\left(-376\text{N}\right)j\end{array}\right\}=0\\ \left\{\begin{array}{c}\left(-0.2889T-0.3061T+P\right)i\\ +\left(0.889T+0.8163T-376\right)j\\ +\left(0.3556T-0.4898T+Q\right)k\end{array}\right\}=0\\ \left\{\begin{array}{l}\left(-0.595T+P\right)i+\left(1.7053T-376\right)j\\ +\left(-0.1342T+Q\right)k\end{array}\right\}=0\end{array}$

Equate the component i, j and k to zero,

$-0.595T+P=0$ (1)

$1.7053T-376=0$ (2)

$-0.1342T+Q=0$ (3)

Calculate tension in cable (T) as below:

Solve Equation (2).

$\begin{array}{l}1.7053T-367=0\\ T=\frac{376}{1.7053}\\ T=220.5\text{N}\end{array}$

Calculate magnitude of P as below:

Substitute 220.5 N for T in Equation (1).

$\begin{array}{l}-0.595\left(220.5\right)+P=0\\ P=131.2\text{N}\end{array}$

Calculate magnitude of Q as below:

Substitute 220.5 N for T in Equation (2).

$\begin{array}{l}-0.1342\left(220.5\right)+Q=0\\ Q=29.6\text{N}\end{array}$

Thus, the magnitude P and Q are $\underset{\_}{131.2\text{N}}$ and $\underset{\_}{29.6\text{N}}$ respectively.