Chapter 3, Problem 3.147RP

A 300-N force P is applied at point A of the bell crank shown. (a) Compute the moment of the force P about O by resolving it into horizontal and vertical components. (b) Using the result of part a, determine the perpendicular distance from O to the line of action of P.

To determine

The moment of the force P about O

Answer to Problem 3.147RP

The magnitude of the moment is about O is $\underset{\_}{\text{20}\text{.5}\text{N}\cdot \text{m}}$.

Explanation of Solution

The magnitude of P is $300\text{N}$.

Write the expression for the vector from O to A.

$\begin{array}{c}{r}_{OA}=200\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\cos 40°\widehat{\text{i}}+200\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\sin 40°\widehat{\text{j}}\\ =\left(0.153209\widehat{\text{i}}+0.128558\widehat{\text{j}}\right)\text{m}\end{array}$ (I)

Here $r_{OA}$ is the vector from O to A.

Write the expression for the force vector $F$.

$\begin{array}{c}F=300\text{N}\left(\sin 30°\right)\widehat{\text{i}}+300\text{N}\left(\cos 30°\right)\widehat{\text{j}}\\ =\left(\text{15}0\text{N}\right)\widehat{\text{i}}+\left(259.81\text{N}\right)\widehat{\text{j}}\end{array}$ (II)

Write the expression for the moment at O.

$M_O=\left(r_{OA}\times F\right)$

Here $M_O$ is the moment at O.

Conclusion:

Substitute (I) and (II) in the above equation to calculate $M_O$.

$\begin{array}{c}{M}_O=\left(0.153209\widehat{\text{i}}+0.128558\widehat{\text{j}}\right)\text{m}\times \left(\text{15}0\text{N}\widehat{\text{i}}+259.81\text{N}\widehat{\text{j}}\right)\\ =\left(39.805\widehat{\text{k}}-19.2837\widehat{\text{k}}\right)\text{N}\cdot \text{m}\\ =\left(\text{20}\text{.521}\text{N}\cdot \text{m}\right)\widehat{\text{k}}\\ =\text{20}\text{.5}\text{N}\cdot \text{m}\end{array}$

Thus, the magnitude of moment at O is $\underset{\_}{\text{20}\text{.5}\text{N}\cdot \text{m}}$.

To determine

The perpendicular distance from the point O to $P$.

Answer to Problem 3.147RP

The perpendicular distance from the point O to $P$ is $\underset{\_}{68.3\text{mm}}$.

Explanation of Solution

The magnitude of moment at O is $\text{20}\text{.5}\text{N}\cdot \text{m}$.

Write the expression to calculate the perpendicular distance from the point O to force $P$.

$Fd=M_O$

Here $d$ is the perpendicular distance, $F$ is the magnitude of force $P$ and $M_O$ is the magnitude of the moment.

Conclusion:

Substitute $300\text{N}$ for P and $\text{20}\text{.5}\text{N}\cdot \text{m}$ for $M_O$ in the above equation to calculate d.

$\begin{array}{c}\left(300\text{N}\right)d=\text{20}\text{.5}\text{N}\cdot \text{m}\\ d=\frac{\text{20}\text{.5}\text{N}\cdot \text{m}}{300\text{N}}\\ =0.0683\text{m}\left(\frac{1000\text{mm}}{1\text{m}}\right)\\ =68.3\text{mm}\end{array}$

Thus, the perpendicular distance from the point O to $P$ is $\underset{\_}{68.3\text{mm}}$.