Question 15.5: Solve Sample Prob. 15.3, using the method of the instantaneo......

Vector Mechanics for Engineers Statics and Dynamics [1686025]

Solve Sample Prob. 15.3, using the method of the instantaneous center of rotation.

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Motion of Crank AB. Referring to Sample Prob. 15.3, we obtain the velocity of point B; $\text v _B=628.3 \text { in. } / s \text { c } 50^{\circ} \text {. }$

Motion of the Connecting Rod BD. We first locate the instantaneous center C by drawing lines perpendicular to the absolute velocities $\text v _B \text { and }\text v _D$ . Recalling from Sample Prob. 15.3 that b = 13.95° and that BD = 8 in., we solve the triangle BCD.

g _B=40^{\circ}+ b =53.95^{\circ} \quad\quad\quad g _D=90^{\circ}- b =76.05^{\circ}

\frac{B C}{\sin 76.05^{\circ}}=\frac{C D}{\sin 53.95^{\circ}}=\frac{8 \text { in. }}{\sin 50^{\circ}}

BC = 10.14 in. CD = 8.44 in.

Since the connecting rod BD seems to rotate about point C, we write

v_B=(B C)\text v _{B D} \\ \\ 628.3\text { in }. / s =(10.14 \text { in } .) \text v _{B D} \\ \\ \text V _{B D}=62.0 ~rad / s~\text l

v_D=(C D)\text v _{B D}=(8.44 \text { in. })(62.0~ rad / s )

= 523 in./s = 43.6 ft/s

\text v _P=\text v _D=43.6~ ft / s~ \text y

15.5.1

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