Chapter 3, Problem 3.156RP

(a)

To determine

The magnitude of d.

Answer to Problem 3.156RP

The value of d is $\underset{\_}{135\text{mm}}$.

Explanation of Solution

The z-component of the replaced equivalent couple $M_{2z}$ is zero, magnitude of $F_1$ is $77\text{N}$ and moment $M_1$ is $31\text{N}\cdot \text{m}$.

Write the expression for the vector from B to H.

$\begin{array}{c}{r}_{BH}=310\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\widehat{\text{i}}-23.3\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\widehat{\text{j}}\\ =0.31\widehat{\text{i}}-0.0233\widehat{\text{j}}\end{array}$ (I)

Here $r_{BH}$ is the vector from B to H.

Write the expression for the vector from E to H.

$r_{EH}=60\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\widehat{\text{i}}+60\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\widehat{\text{j}}-70\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\widehat{\text{k}}$

Here $r_{EH}$ is the vector from E to H.

Write the expression for the force $F_1$ in the direction of E to H.

$\begin{array}{c}{F}_1=\left(77\text{N}\right)\frac{60\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\widehat{\text{i}}+60\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\widehat{\text{j}}-70\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\widehat{\text{k}}}{\sqrt{{\left(60\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2+{\left(60\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2+{\left(-70\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2}}\\ =\left(42\widehat{\text{i}}+42\widehat{\text{j}}-49\widehat{\text{k}}\right)\text{N}\end{array}$ (III)

Write the expression for the vector from E to J.

$r_{EJ}=-d\widehat{\text{i}}+30\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\widehat{\text{j}}-70\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\widehat{\text{k}}$

Write the expression for the force $F_1$ in the direction of E to J.

$M_1=\left(31\text{N}\cdot \text{m}\right)\frac{-d\widehat{\text{i}}+30\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\widehat{\text{j}}-70\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\widehat{\text{k}}}{\sqrt{{\left(d\right)}^2+{\left(30\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2+{\left(-70\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2}}$ (II)

Write the expression for the z-component of the moment.

$\widehat{\text{k}}\cdot \left(r_{BH}\times F_1\right)+M_{1z}=0$

Conclusion:

Substitute $0.31\widehat{\text{i}}-0.0233\widehat{\text{j}}$ for $r_{BH}$, $\left(42\widehat{\text{i}}+42\widehat{\text{j}}-49\widehat{\text{k}}\right)\text{N}$ for $F_1$ and the z-component of $M_1$ in the above equation to calculate d.

$\begin{array}{c}\widehat{\text{k}}\cdot \left|\begin{array}{ccc}\widehat{\text{i}}& \widehat{\text{j}}& \widehat{\text{k}}\\ 0.31& -0.0233& 0\\ 42& 42& -49\end{array}\right|+\frac{-70\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\left(31\text{N}\cdot \text{m}\right)}{\sqrt{{\left(d\right)}^2+{\left(30\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2+{\left(-70\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2}}=0\\ \left(13.02+0.9786\right)\text{N}\cdot \text{m}+\frac{\left(-0.07\text{m}\right)\left(31\text{N}\cdot \text{m}\right)}{\sqrt{d^2+0.0058}}=0\\ 13.9986\text{N}\cdot \text{m}=\frac{2.17\text{N}\cdot {\text{m}}^2}{\sqrt{d^2+0.0058}}\\ \left(d^2+0.0058\right){\text{m}}^{\text{2}}={\left(\frac{2.17\text{N}\cdot {\text{m}}^2}{13.9986\text{N}\cdot \text{m}}\right)}^2\\ \left(d^2+0.0058\right){\text{m}}^{\text{2}}=0.024{\text{m}}^{\text{2}}\\ d^2=0.0182{\text{m}}^{\text{2}}\\ d=0.135\text{m}\left(\frac{1000\text{mm}}{1\text{m}}\right)\\ =135\text{mm}\end{array}$

Thus, the value of d is $\underset{\_}{135\text{mm}}$.

(b)

To determine

The vectors $F_2$ and $M_2$.

Answer to Problem 3.156RP

The vectors $F_2$ and $M_2$ are respectively $\underset{\_}{\left(42\widehat{\text{i}}+42\widehat{\text{j}}-49\widehat{\text{k}}\right)\text{N}}$ and $\underset{\_}{-25.858\widehat{\text{i}}+21.19\widehat{\text{j}}}$.

Explanation of Solution

Since the force $F_1$ and $F_2$ is same the vector $F_2$ is $\left(42\widehat{\text{i}}+42\widehat{\text{j}}-49\widehat{\text{k}}\right)\text{N}$.

Write the expression to calculate $M_2$.

$M_2=\left(r_{BH}\times F_1\right)+M_1$

Conclusion:

Substitute (I), (II), (III) and $135\text{mm}$ for d in the above equation to calculate $M_2$.

$\begin{array}{c}{M}_2=\left|\begin{array}{ccc}\widehat{\text{i}}& \widehat{\text{j}}& \widehat{\text{k}}\\ 0.31& -0.0233& 0\\ 42& 42& -49\end{array}\right|+\frac{\left(-135\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\widehat{\text{i}}+30\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\widehat{\text{j}}-70\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\widehat{\text{k}}\right)\left(31\text{N}\cdot \text{m}\right)}{\sqrt{{\left(135\text{mm}\right)}^2+{\left(30\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2+{\left(-70\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2}}\\ =\left(1.1417\widehat{\text{i}}+15.19\widehat{\text{j}}+14\widehat{\text{k}}\right)\text{N}\cdot \text{m}+\frac{\left(-0.135\widehat{\text{i}}+0.03\widehat{\text{j}}-0.07\widehat{\text{k}}\right)\text{m}\left(31\text{N}\cdot \text{m}\right)}{0.155\text{m}}\\ =\left(1.1417\widehat{\text{i}}+15.19\widehat{\text{j}}+14\widehat{\text{k}}\right)\text{N}\cdot \text{m}+\left(-27\widehat{\text{i}}+6\widehat{\text{j}}-14\widehat{\text{k}}\right)\text{N}\cdot \text{m}\\ =-25.858\widehat{\text{i}}+21.19\widehat{\text{j}}-0\widehat{\text{k}}\\ \text{=}-25.858\widehat{\text{i}}+21.19\widehat{\text{j}}\end{array}$

Thus, the vectors $F_2$ and $M_2$ are respectively $\underset{\_}{\left(42\widehat{\text{i}}+42\widehat{\text{j}}-49\widehat{\text{k}}\right)\text{N}}$ and $\underset{\_}{-25.858\widehat{\text{i}}+21.19\widehat{\text{j}}}$.