Chapter 5.7, Problem 38E

To determine

How far is the position from the rim of the crankshaft after 0.01 second, if the crankshaft rotates 20 revolutions per second and P is at the horizontal position when it begins to rotate?

Answer to Problem 38E

10.8 cm

Explanation of Solution

Given information: Mechanics: The blades of a power lawn mower are rotated by a two-stroke engine with a piston sliding back and forth in the engine cylinder. As the piston moves back and forth, the connecting rod rotates the circular crankshaft. Suppose the crankshaft is 5 centimeters long and the connecting rod is 15 centimeters.

Calculation:

P turns $20\cdot \left({360}^{\circ }\right){\text{ or 7200}}^{\circ }$ every second.

The angle (at O) after 0.01 second equals ${72}^{\circ }$ .

Using the triangle (and the Law of Sines), we find the angles at Q and P

  $\begin{array}{l}\frac{PQ}{\sin O}=\frac{OP}{\sin Q}\\ \frac{15}{\sin {72}^{\circ }}=\frac{5}{\sin Q}\\ \sin Q=\frac{5\sin {72}^{\circ }}{15}\\ Q={\sin }^{-1}\left(\frac{5\sin {72}^{\circ }}{15}\right)\\ Q\approx 18.48273235\\ P\approx {180}^{\circ }-\left({72}^{\circ }+{18.5}^{\circ }\right)\approx {89.5}^{\circ }\end{array}$

We are looking to find QO, then subtract the radius,

  $\begin{array}{l}\frac{QO}{\sin P}=\frac{PQ}{\sin O}\\ \frac{QO}{\sin {89.5}^{\circ }}\approx \frac{15}{\sin {72}^{\circ }}\\ QO\approx \frac{15\sin {89.5}^{\circ }}{\sin {72}^{\circ }}\\ QO\approx 15.77133282\\ QO-5\approx 15.8-5\\ \approx 10.8\text{ cm}\end{array}$