Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 50P from Chapter 13 from Hibbeler's Engineering Mechanics.
Problem 50P
Step-by-Step Solution
We have given that the total mass of the elevator is $m = 1\;{\rm{Mg}}$.
We have given that the initial velocity of the elevator is $u = 0\;{\rm{m}}/{\rm{s}}$.
We are given that the tension in each cable is $T = 4\;{\rm{kN}}$.
We are asked to calculate the velocity of the elevator when it has moved a distance of $s = 6\;{\rm{m}}$ upward.
Step 2
Draw a free-body diagram of the elevator.
Here, ${N_1}$ is the normal reaction force acting on the upper right corner, ${N_2}$ is the normal reaction force acting on the lower-left corner, $a$ is the acceleration of the elevator, and $W$ is the weight of the elevator.
Step 3
Calculate the weight of the elevator:
\[W = mg\]Substitute the value of $m$, and $9.81\;{\rm{m}}/{{\rm{s}}^2}$ for g (acceleration due to gravity) in the above equation:
\[\begin{array}{c} W = \left( {\left( {1\;{\rm{Mg}}} \right) \times \left( {9.81\;{\rm{m}}/{{\rm{s}}^2}} \right)} \right) \times \frac{{1000\;{\rm{kg}}}}{{1\;{\rm{Mg}}}} \times \frac{{1\;{\rm{N}}}}{{1\;{\rm{kg}} \cdot {\rm{m}}/{{\rm{s}}^2}}}\\ = 9810\;{\rm{N}} \end{array}\]Step 4
Consider the free-body diagram and apply the equilibrium equation of motion in the vertical direction:
\[\begin{array}{c} \sum {{F_y}} = 0\\ T + T + T - W = ma\\ 3T - W = ma \end{array}\]Substitute the value of $T$, $W$, and $m$ in the above equation:
\[\begin{array}{c} 3\left( {4\;{\rm{kN}} \times \frac{{1000\;{\rm{N}}}}{{1\;{\rm{kN}}}}} \right) - \left( {9810\;{\rm{N}}} \right) = \left( {1\;{\rm{Mg}} \times \frac{{1000\;{\rm{kg}}}}{{1\;{\rm{Mg}}}}} \right)a\\ 2190\;{\rm{N}} \times \frac{{1\;{\rm{kg}} \cdot {\rm{m}}/{{\rm{s}}^2}}}{{1\,{\rm{N}}}} = 1000\;{\rm{kg}} \times a\\ a = 2.19\;{\rm{m}}/{{\rm{s}}^2} \end{array}\]Step 5
Apply the third equation of motion to calculate the velocity of the elevator when it has moved a distance of $s = 6\;{\rm{m}}$ upward:
\[\begin{array}{c} {v^2} = {u^2} + 2as\\ v = \sqrt {{u^2} + 2as} \end{array}\]Substitute the value of $u$, $a$, and $s$ in the above equation:
\[\begin{array}{c} v = \sqrt {{{\left( {0\;{\rm{m}}/{\rm{s}}} \right)}^2} + 2\left( {2.19\;{\rm{m}}/{{\rm{s}}^2}} \right)\left( {6\;{\rm{m}}} \right)} \\ = \sqrt {\left( {26.28\;{{\rm{m}}^2}/{{\rm{s}}^2}} \right)} \\ = 5.13\;{\rm{m}}/{\rm{s}} \end{array}\]