Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 16 - Planar Kinematics of a Rigid Body - Section 16.5 - Relative-Motion Analysis: Velocity - Problems - Page 357: 67


$v_A=5.16ft/s$, $39.8^{\circ}$

Work Step by Step

The required velocity can be determined as follows: $v_{C/B}=v_C-v_B$ $\implies v_B=v_C+\omega\times r_{B/C}$ We plug in the known values to obtain: $3\hat i=-4\hat i+(-\omega \hat k)(2.25\hat j)$ $\implies \omega=3.111 rad/s$ Now according to the relative motion between Point $A$ and $C$ $v_A=v_C+v_{A/C}$ $\implies v_A=v_C+\omega \times r_{A/C}$ $\implies v_{Ax}\hat i+v_{Ay}\hat j=-4\hat i +(-3.111\hat k)\times (-1.5 cos 45\hat i)+(1.5+1.5cos45)\hat j$ $\implies v_{Ax}\hat i+v_{Ay}\hat j=(3.967\hat i+3.3 \hat j)ft/s$ Comparing $i$ and $j$ components, we obtain: $v_{Ax}=3.967ft/s$ and $v_{Ay}=3.3ft/s$ The magnitude of $v_A$ can be determined as $v_A=\sqrt{v_{Ax}^2+v_{Ay}^2}$ We plug in the known values to obtain: $v_A=\sqrt{(3.967)^2+(3.3)^2}$ $\implies v_A=5.16ft/s$ The direction is given as $\theta=tan^{-1}(\frac{v_{Ay}}{v_{Ax}})$ $\implies \theta=tan^{-1}(\frac{3.967}{3.3})=39.8^{\circ}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.