Question 15.7: Crank AB of the engine system of Sample Prob. 15.3 has a con......
Crank AB of the engine system of Sample Prob. 15.3 has a constant clockwise angular velocity of 2000 rpm. For the crank position shown, determine the angular acceleration of the connecting rod BD and the acceleration of point D.
Motion of Crank AB. Since the crank rotates about A with constant \text v_{AB} = 2000 rpm = 209.4 rad/s, we have a_{AB} = 0. The acceleration of B is therefore directed toward A and has a magnitude
Motion of the Connecting Rod BD. The angular velocity V_{BD} and the value of b were obtained in Sample Prob. 15.3:
\mathrm{V}_{B D}=62.0 ~\mathrm{rad} / \mathrm{s~l} \quad\quad\quad \mathrm{b}=13.95^{\circ}The motion of BD is resolved into a translation with B and a rotation about B. The relative acceleration a_{D/B} is resolved into normal and tangential components:
While \left({a}_{D / B}\right)_{t} must be perpendicular to BD, its sense is not known.
Noting that the acceleration a_D must be horizontal, we write
Equating x and y components, we obtain the following scalar equations: \overset{+}{\text y} x components:
-a_{D}=-10,962 ~\cos 40^{\circ}-2563 ~\cos 13.95^{\circ}+0.6667~\mathrm{a}_{B D}~ \sin 13.95^{\circ}+xy components:
0=-10,962~ \sin 40^{\circ}+2563 ~\sin 13.95^{\circ}+0.6667~ \mathrm{a}_{B D}~ \cos 13.95^{\circ}Solving the equations simultaneously, we obtain a_{BD} = +9940 rad/s² and a_D = +9290 ft/s². The positive signs indicate that the senses shown on the vector polygon are correct; we write
\begin{aligned}{a}_{B D} & =9940 ~\mathrm{rad} / \mathrm{s}^{2}~ \text l\\{a}_{D} & =9290~ \mathrm{ft} / \mathrm{s}^{2}~ \mathrm{z}\end{aligned}
