Chapter 3.2, Problem 3.65P

In Prob. 3.56, determine the perpendicular distance between a line joining points F and B and the line of action of P.

To determine

The perpendicular distance between a line joining points $F$ and $B$.

Answer to Problem 3.65P

The perpendicular distance between a line joining points $F$ and $B$ is $d=4.97\text{in}$.

Explanation of Solution

Refer the Problem 3.56P

Write the expression for force $P$ in terms of unit vector at line joining points $E$ and $H$:

$\overrightarrow{EH}=\left(30\text{in}\right)\widehat{i}-\left(7.5\text{in}\right)\widehat{j}+\left(10\text{in}\right)\widehat{k}$

The magnitude of the line joining points $E$ and $H$ is,

$\begin{array}{c}EH=\sqrt{{\left(30\text{in}\right)}^2+{\left(-7.5\text{in}\right)}^2+{\left(10\text{in}\right)}^2}\\ =32.5\text{in}\end{array}$

The unit vector at line joining points $E$ and $H$ is,

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

Here, the unit vector at line joining points $E$ and $H$ is ${\lambda }_{EH}$.

Substitute $\left(30\text{in}\right)\widehat{i}-\left(7.5\text{in}\right)\widehat{j}+\left(10\text{in}\right)\widehat{k}$ for $\overrightarrow{EH}$ and $32.5\text{in}$ for $EH$.

$\begin{array}{c}{\lambda }_{EH}=\frac{\left(30\text{in}\right)\widehat{i}-\left(7.5\text{in}\right)\widehat{j}+\left(10\text{in}\right)\widehat{k}}{32.5\text{in}}\\ =0.923\widehat{i}-0.231\widehat{j}+0.308\widehat{k}\end{array}$

Write the equation of the moment of $P$ about a line joining points $E$ and $H$ is

$\overrightarrow{P}=P{\lambda }_{EH}$ (I)

Here, the moment of force is $P$ and the force.

Conclusion:

Substitute $910\text{lb}$ for $P$ and $0.923\widehat{i}-0.231\widehat{j}+0.308\widehat{k}$ for ${\lambda }_{EH}$ in equation (I).

$\begin{array}{c}\overrightarrow{P}=\left(910\text{lb}\right)\left(0.923\widehat{i}-0.231\widehat{j}+0.308\widehat{k}\right)\\ =\left(840.0\text{lb}\right)\widehat{i}-\left(210.0\text{lb}\right)\widehat{j}+\left(280\text{lb}\right)\widehat{k}\end{array}$

Write the expression for force $P$ in terms of unit vector at line joining points $F$ and $B$:

$\overrightarrow{FB}=\left(30\text{in}\right)\widehat{i}+\left(15\text{in}\right)\widehat{j}-\left(10\text{in}\right)\widehat{k}$

The magnitude of the line joining points $F$ and $B$ is,

$\begin{array}{c}FB=\sqrt{{\left(30\text{in}\right)}^2+{\left(15\text{in}\right)}^2+{\left(-10\text{in}\right)}^2}\\ =35.0\text{in}\end{array}$

The unit vector at line joining points $F$ and $B$ is,

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

Here, the unit vector at line joining points $F$ and $B$ is ${\lambda }_{FB}$.

Substitute $\left(30\text{in}\right)\widehat{i}+\left(15\text{in}\right)\widehat{j}-\left(10\text{in}\right)\widehat{k}$ for $\overrightarrow{FB}$ and $35.0\text{in}$ for $FB$.

$\begin{array}{c}{\lambda }_{FB}=\frac{\left(30\text{in}\right)\widehat{i}+\left(15\text{in}\right)\widehat{j}-\left(10\text{in}\right)\widehat{k}}{35\text{in}}\\ =0.857\widehat{i}+0.429\widehat{j}-0.286\widehat{k}\end{array}$

Write the equation of the moment of $P$ about a line joining points $F$ and $B$ is

$P_{\text{parallel}}=\overrightarrow{P}{\lambda }_{FB}$ (II)

Here, the perpendicular component contribute to the moment of force about the line $FB$ is $\overrightarrow{P}$

Substitute $\left(840.0\text{lb}\right)\widehat{i}-\left(210.0\text{lb}\right)\widehat{j}+\left(280\text{lb}\right)\widehat{k}$ for $\overrightarrow{P}$ and $0.857\widehat{i}+0.429\widehat{j}-0.286\widehat{k}$ for ${\lambda }_{FB}$ in equation (II).

$\begin{array}{c}{P}_{\text{parallel}}=\left(\left(840.0\text{lb}\right)\widehat{i}-\left(210.0\text{lb}\right)\widehat{j}+\left(280\text{lb}\right)\widehat{k}\right)\left(0.857\widehat{i}+0.429\widehat{j}-0.286\widehat{k}\right)\\ =720\text{lb}-90.1\text{lb}-80.1\text{lb}\\ =550\text{lb}\end{array}$

The relation between perpendicular and parallel component contribute to the moment of force is,

$P^2=P_{{}_{\text{parallel}}}^2+P_{\text{perpendicular}}^2$ (III)

Here, the parallel component of moment is $P_{\text{parallel}}$ and perpendicular component of moment is $P_{\text{perpendicular}}$.

Rewrite the equation (III) to get $P_{\text{perpendicular}}$.

$P_{\text{perpendicular}}=\sqrt{P^2-P_{{}_{\text{parallel}}}^2}$

Refer the Problem 3.56

The moment of $P$ about axis $FB$ is $910\text{lb}$.

Substitute $550\text{lb}$ for $P_{\text{parallel}}$ and $910\text{lb}$ for $P$.

$\begin{array}{c}{P}_{\text{perpendicular}}=\sqrt{{\left(910\text{lb}\right)}^2-{\left(550\text{lb}\right)}^2}828100-302500\\ =724.98\text{lb}\end{array}$

Write the equation for the moment about a line joining points $F$ and $B$.

$M_{FB}=d{\left(P\right)}_{\text{perpendicular}}$

Here, the moment about a line joining points $F$ and $B$ is $M_{FB}$ and the perpendicular distance is $d$.

Rewrite the relation in terms of $d$.

$d=\frac{M_{FB}}{{\left(P\right)}_{\text{perpendicular}}}$

Refer the Problem 3.56

The value of the moment line joining points $F$ and $B$ is $M_{FB}$ is $300\text{lb}\cdot \text{ft}$

Substitute $300\text{lb}\cdot \text{ft}$ for $M_{FB}$ and $724.98\text{lb}$ for ${\left(P\right)}_{\text{perpendicular}}$.

$\begin{array}{c}d=\frac{300\text{lb}\cdot \text{ft}\left(\frac{12\text{lb}\cdot \text{in}}{1\text{lb}\cdot \text{ft}}\right)}{724.98\text{lb}}\\ =\frac{3600\text{lb}\cdot \text{in}}{724.98\text{lb}}\\ =4.97\text{in}\end{array}$

Therefore, the perpendicular distance between a line joining points $F$ and $B$ is $d=4.97\text{in}$.