Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 205P from Chapter 12 from Hibbeler's Engineering Mechanics.
Problem 205P
Chapter:
Problem:
If block A of the pulley system is moving...
Step-by-Step Solution
Step 1
Step 2
Step 3
Step 4
Step 5
Step 6
Step 7
We are given that the speed of block A is ${v_A} = 6\;{\rm{ft/s}}$ and the speed of block C is ${v_C} = 18\;{\rm{ft/s}}$.
We are asked to calculate the relative velocity of block B with respect to block C.
Step 2
The diagram of the pulley system is shown as:
Step 3
The total length of the cable in the diagram is given as:
\[l = {s_A} + 2{s_B} + 2{s_C}\]Step 4
Differentiate the above equation with respect to time:
\[\begin{array}{c} \frac{{dl}}{{dt}} = \frac{{d\left( {{s_A}} \right)}}{{dt}} + \frac{d}{{dt}}\left( {2{s_B}} \right) + \frac{d}{{dt}}\left( {2{s_C}} \right)\\ 0 = {v_A} + 2{v_B} + 2{v_C}\\ 2{v_B} = - \left( {{v_A} + 2{v_C}} \right)\\ {v_B} = - \frac{{\left( {{v_A} + 2{v_C}} \right)}}{2} \end{array}\]Step 5
Substitute the known value in equation:
\[\begin{array}{c} {v_B} = - \frac{{\left( {{\rm{6}}\;{\rm{ft/s}} + 2\left( {18\;{\rm{ft/s}}} \right)} \right)}}{2}\\ = - \frac{{\left( {6\;{\rm{ft/s}} + 36\;{\rm{ft/s}}} \right)}}{2}\\ = - \frac{{\left( {42\;{\rm{ft/s}}} \right)}}{2}\\ = - 21\;{\rm{ft/s}} \end{array}\]Step 6
To calculate the relative velocity of block B we use the formula:
\[{v_{B/C}} = {v_B} - {v_C}\]Step 7
Substitute the known values in the formula:
\[\begin{array}{c} {v_{B/C}} = - 21\;{\rm{ft/s}} - 18\;{\rm{ft/s}}\\ = - 39\;{\rm{ft/s}} \end{array}\]Here, the negative sign is due to opposite motion of the body.