Engineering Mechanics: Statics & Dynamics (14th Edition)

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

Chapter 12 - Kinematics of a Particle - Section 12.10 - Relative-Motion of Two Particles using Translating Axes - Problems - Page 99: 205


$V_{B/C} = -39\frac{ft}{s}$

Work Step by Step

Only one line is present in this problem. Summate each respective segment to obtain the total length of moving rope: (1) $l = S_{A}+2\times S_{B}+2\times S_{C}$ Take the derivative of equation (1) to obtain the respective velocities of components A, B, and C: (2) $0 = V_{A}+2\times V_{B}+2\times V_{C}$ The problem tells us that $V_{A} = 6\frac{ft}{s}$ downwards and $V_{C} = 18\frac{ft}{s}$ downards. Plug these values into eqation (2) to calculate $V_{C}$: $0 = 6+2\times V_{B}+2\times 18$ $0 = 42 + 2\times V_{B}$ $2\times V_{B} = -42$ $V_{B} = -21\frac{ft}{s}$ Since we treated downward velocity as positive, a negative velocity indicates that rope segment B is moving upwards. Note that $V_{B} = -21\frac{ft}{s}$ is not the final answer. The question wants the relative velocity of $V_{B}$ in terms of $V_{C}$: (3) $V_{B/C} = -21 - 18 = -39\frac{ft}{s}$ $V_{C}$ takes the same sign as $V_{B}$ since rope segment C is moving in the opposite direction of $V_{B}$.
Small 1551733579
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.