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 $V_{B}$.
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