Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 212P from Chapter 12 from Hibbeler's Engineering Mechanics.
Problem 212P
Step-by-Step Solution
We have given the following values:
The speed of rope at C is ${v_C} = 6\;{\rm{ft}}/{\rm{s}}$.
The length of AB segment of rope is ${l_{AB}} = 50\;{\rm{ft}}$.
We are asked to determinethat how fast the boat approaches the pier at the instant.
Step 2
Consider the given diagram; the total length l of the rope is given by:
\[\sqrt {{{\left( {8\;{\rm{ft}}} \right)}^2} + x_B^2} + {x_C} = l\]Differentiate the above equation with time t:
\[\begin{array}{c} \frac{d}{{dt}}\left( {\sqrt {{{\left( {8\;{\rm{ft}}} \right)}^2} + x_B^2} + {x_C}} \right) = \frac{{dl}}{{dt}}\\ \frac{1}{2}{\left( {{{\left( {8\;{\rm{ft}}} \right)}^2} + x_B^2} \right)^{ - \frac{1}{2}}}2{x_B}\frac{{d{x_B}}}{{dt}} + \frac{{d{x_C}}}{{dt}} = 0 \end{array}\]Replace $\frac{{d{x_B}}}{{dt}}$ by ${v_B}$ and $\frac{{d{x_C}}}{{dt}}$ by ${v_C}$ in the above equation:
\[\frac{1}{2}{\left( {{{\left( {8\;{\rm{ft}}} \right)}^2} + x_B^2} \right)^{ - \frac{1}{2}}}2{x_B}{v_B} + {v_C} = 0\] … (1)Step 3
Consider the given diagram, the length ${x_B}$ can be calculated as:
\[{x_B} = \sqrt {l_{AB}^2 - {{\left( {8\;{\rm{ft}}} \right)}^2}} \]Substitute the value of ${l_{AB}}$ in the above equation:
\[\begin{array}{c} {x_B} = \sqrt {{{\left( {50\;{\rm{ft}}} \right)}^2} - {{\left( {8\;{\rm{ft}}} \right)}^2}} \\ {x_B} = 49.356\;{\rm{ft}} \end{array}\]Step 4
Substitute the value of ${x_B}$ and ${v_C}$ in equation (1):
\[\begin{array}{c} \frac{1}{2}{\left( {{{\left( {8\;{\rm{ft}}} \right)}^2} + {{\left( {49.356\;{\rm{ft}}} \right)}^2}} \right)^{ - \frac{1}{2}}}2\left( {49.356\;{\rm{ft}}} \right){v_B} + 6\;{\rm{ft}}/{\rm{s}} = 0\\ {v_B} = - 6.08\;{\rm{ft}}/{\rm{s}} \end{array}\]Here, a negative direction shows that the boat is moving to the left.