Chapter 8.2, Problem 48AYU

To determine

How far is the piston from the centre of crankshaft?

Answer to Problem 48AYU

  $\boxed{6.80inches}$

Explanation of Solution

Given information:

On a certain automobile, the crankshaft is $3inches$ long and the connecting rod is $9inches$ long. At the time when $\angle OPQ$ is $15°$ , how far is the piston from the centre of crankshaft?

  

Calculation:

Consider the figure,

In the oblique triangle, $\Delta OPQ$ we have,

  $\angle OPQ=15°,PQ=9inch,QO=3inch$

Using law of sine,

  $\begin{array}{l}\frac{\sin O}{PQ}=\frac{\sin P}{QO}=\frac{\sin Q}{OP}\\ now,\\ \frac{\sin O}{9}=\frac{\sin 15°}{3}=\frac{\sin Q}{OP}\end{array}$

Now take,

  $\begin{array}{l}\frac{\sin O}{9}=\frac{\sin 15°}{3}\\ \sin O=3\times \sin 15°\\ \sin O=3\times 0.25882\\ \sin O=0.77646\end{array}$

  $\begin{array}{l}\angle QOP={\sin }^{-1}\left(0.77646\right)\\ or\\ 180°-{\sin }^{-1}\left(0.77646\right)\\ \angle QOP=50.93733°or129.06267°\end{array}$

So $\angle QOP$ is an obtuse angle,

  $\begin{array}{l}\angle QOP+\angle PQO+\angle QOP=180°\\ \angle PQO=180°-\left(\angle QOP+\angle PQO\right)\end{array}$

  $\begin{array}{l}\angle PQO=180°-\left(15°+129.0626°\right)\\ \angle PQO=35.93733\\ \angle PQO\approx 36°\end{array}$

  $\begin{array}{l}\frac{\sin 15°}{3}=\frac{\sin \left(36°\right)}{OP}\\ OP=3\times \frac{\sin \left(36°\right)}{\sin 15°}\\ OP=3\times \frac{0.5868999998}{0.25882}\\ OP\approx 6.80282inches\end{array}$

Hence, the piston from the centre of crankshaft is, $\boxed{6.80inches}$