Chapter 3.2, Problem 3.64P

In Prob. 3.55, determine the perpendicular distance between a line joining points O and D and the line of action of P.

To determine

The perpendicular distance between a line joining points $O$ and $D$.

Answer to Problem 3.64P

The perpendicular distance between a line joining points $O$ and $D$ is $d=9.50\text{in}$.

Explanation of Solution

Refer the Problem 3.55

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 $520\text{lb}$ for $P$ and $0.923\widehat{i}-0.231\widehat{j}+0.308\widehat{k}$ for ${\lambda }_{EH}$ in equation (I).

$\begin{array}{c}\overrightarrow{P}=\left(520\text{lb}\right)\left(0.923\widehat{i}-0.231\widehat{j}+0.308\widehat{k}\right)\\ =\left(480.0\text{lb}\right)\widehat{i}-\left(120.0\text{lb}\right)\widehat{j}+\left(160.0\text{lb}\right)\widehat{k}\end{array}$

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

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

The magnitude of the line joining points $O$ and $D$ is,

$\begin{array}{c}OD=\sqrt{{\left(30\text{in}\right)}^2+{\left(15\text{in}\right)}^2+{\left(10\text{in}\right)}^2}\\ =35\text{in}\end{array}$

The unit vector at line joining points $O$ and $D$ is,

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

Here, the unit vector at line joining points $O$ and $D$ is ${\lambda }_{OD}$.

Substitute $\left(30\text{in}\right)\widehat{i}+\left(15\text{in}\right)\widehat{j}+\left(10\text{in}\right)\widehat{k}$ for $\overrightarrow{OD}$ and $35\text{in}$ for $OD$.

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

Write the equation of the moment of $P$ about a line joining points $O$ and $D$ is

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

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

Substitute $\left(480.0\text{lb}\right)\widehat{i}-\left(120.0\text{lb}\right)\widehat{j}+\left(160.0\text{lb}\right)\widehat{k}$ for $\overrightarrow{P}$ and $0.857\widehat{i}+0.429\widehat{j}+0.286\widehat{k}$ for ${\lambda }_{OD}$ in equation (II).

$\begin{array}{c}{P}_{\text{parallel}}=\left(\left(480.0\text{lb}\right)\widehat{i}-\left(120.0\text{lb}\right)\widehat{j}+\left(160.0\text{lb}\right)\widehat{k}\right)\left(0.857\widehat{i}+0.429\widehat{j}+0.286\widehat{k}\right)\\ =411.4\text{lb}-51.5\text{lb}+45.8\text{lb}\\ =405.7\text{lb}\end{array}$

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 $P_{\text{perpendicular}}$.

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

Substitute $405.7\text{lb}$ for $P_{\text{parallel}}$ and $520\text{lb}$ for $P$.

$\begin{array}{c}{P}_{\text{perpendicular}}=\sqrt{{\left(520\text{lb}\right)}^2-{\left(405.7\text{lb}\right)}^2}\\ =325.28\text{lb}\end{array}$

Write the equation for the moment about a line joining points $O$ and $D$.

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

Here, the moment about a line joining points $O$ and $D$ is $M_{OD}$ and the perpendicular distance is $d$.

Rewrite the relation in terms of $d$.

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

Refer the problem 77268-3.2-3.55P

The value of the moment line joining points $F$ and $B$ is $M_{FB}$ is $3090\text{lb}\cdot \text{in}$

Substitute $3090\text{lb}\cdot \text{in}$ for $M_{OD}$ and $325.28\text{lb}$ for ${\left(P\right)}_{\text{perpendicular}}$.

$\begin{array}{c}d=\frac{3090\text{lb}\cdot \text{in}}{325.28\text{lb}}\\ =9.50\text{in}\end{array}$

Therefore, the perpendicular distance between a line joining points $O$ and $D$ is $d=9.50\text{in}$.