Chapter 3, Problem 97RP

For the shift lever shown, determine the magnitude and the direction of the smallest force P that has a 210-lb-in. clockwise moment about B.

To determine

The direction and magnitude of the smallest force P that has $210\text{ lb}\cdot \text{in}$ close wise moment about B.

Answer to Problem 97RP

The direction and magnitude of the smallest force P that has $210\text{ lb}\cdot \text{in}$ close wise moment about B is $19.98°\text{ and }8.97\text{ lb}$.

Explanation of Solution

Concept used:

Write the expression for angle $\beta$.

$\tan \beta =\frac{AC}{BC}$ …… (1)

Write the expression for $\alpha$.

$\alpha =90-\beta$ …… (2)

Write the expression for the position vector of point A with respect to point B.

${\overrightarrow{d}}_{AB}=\left(8\widehat{i}+22\widehat{j}+0\widehat{k}\right)\text{ in}$

Write the expression for $P$ in vector form.

$P=\left(P\cos \alpha \widehat{i}-P\sin \alpha \widehat{j}+0\widehat{k}\right)\text{ lb}\cdot \text{in}$

Write the expression for the moment of P about B.

$M_B={\overrightarrow{d}}_{AB}\times P$ …… (3)

Calculation:

Substitute $22\text{ in}$ for AC and $8\text{ in}$ for AB in equation (1)

$\begin{array}{c}\tan \beta =\frac{22}{8}\\ \beta ={\tan }^{-1}\left(2.75\right)\\ =70.02°\end{array}$

Substitute $70.02°$ for $\beta$ in equation (2)

$\begin{array}{c}\alpha =90-70.02\\ =19.98°\end{array}$

Substitute $\left(P\cos \alpha \widehat{i}-P\sin \alpha \widehat{j}+0\widehat{k}\right)\text{ lb}\cdot \text{in}$ for $P$

$210\text{ lb}\cdot \text{in}$ for $M_B$ and $\left(8\widehat{i}+22\widehat{j}+0\widehat{k}\right)\text{ in}$ for ${\overrightarrow{d}}_{AB}$ in equation (3)

$\begin{array}{c}210\widehat{k}=\left(8\widehat{i}+22\widehat{j}+0\widehat{k}\right)\times \left(P\cos \alpha \widehat{i}-P\sin \alpha \widehat{j}+0\widehat{k}\right)\\ 210\widehat{k}=\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ 8& 22& 0\\ P\cos \alpha & -P\sin \alpha & 0\end{array}\right|\\ 210\widehat{k}=\left(8P\sin \alpha -22P\cos \alpha \right)\widehat{k}\end{array}$

Substitute $19.98°$ for $\alpha$ in above equation.

$\begin{array}{c}210=\left(-8P\sin \left(19.98°\right)-22P\cos \left(19.98°\right)\right)\\ 210=23.40P\\ P=\frac{210}{23.40}\\ =8.97\text{ lb}\end{array}$

Conclusion:

Thus, the direction and magnitude of the smallest force P that has $210\text{ lb}\cdot \text{in}$ close wise moment about B is $19.98°\text{ and }8.97\text{ lb}$.