Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 3RP from Chapter 14 from Hibbeler's Engineering Mechanics.
Problem 3RP
Step-by-Step Solution
We are given the weight of the block is $W = 1.5\;{\rm{lb}}$.
We are asked to determine the speed of block at which it slides off at B.
Step 2
The diagram of the system is shown as:
We have the coordinates of point A is $A = \left( {5\;{\rm{ft, 0, 10 ft}}} \right)$.
We have the coordinates of point B is $B = \left( {{\rm{0, 8 ft, 0}}} \right)$.
We have the velocity at point A is ${v_A} = 0\;{\rm{ft/s}}$.
Consider that the datum is located at point B.
We have the height of point A from datum is ${h_A} = 10\;{\rm{ft}}$.
We have the height of point B from datum is ${h_B} = 0\;{\rm{ft}}$.
Step 3
The formula to calculate the speed of block at which it slides off at point B by using energy conservation is,
\[\begin{array}{c} {T_A} + {V_A} = {T_B} + {V_B}\\ \frac{1}{2}m{\left( {{v_A}} \right)^2} + W{h_A} = \frac{1}{2}m{\left( {{v_B}} \right)^2} + W{h_B}\\ \frac{1}{2}\frac{W}{g}{\left( {{v_A}} \right)^2} + W{h_A} = \frac{1}{2}\frac{W}{g}{\left( {{v_B}} \right)^2} + W{h_B}\\ \frac{{{{\left( {{v_A}} \right)}^2}}}{{2g}} + {h_A} = \frac{{{{\left( {{v_B}} \right)}^2}}}{{2g}} + {h_B} \end{array}\]Here, g is the acceleration due to gravity, and its standard value is $32.2\;{\rm{ft/}}{{\rm{s}}^2}$.
Step 4
Substitute the values in the above expression.
\[\begin{array}{c} \frac{{{{\left( {0\;{\rm{ft/s}}} \right)}^2}}}{{2\left( {32.2\;{\rm{ft/}}{{\rm{s}}^2}} \right)}} + \left( {10\;{\rm{ft}}} \right) = \frac{{{{\left( {{v_B}} \right)}^2}}}{{2\left( {32.2\;{\rm{ft/}}{{\rm{s}}^2}} \right)}} + \left( {0\;{\rm{ft}}} \right)\\ {\left( {{v_B}} \right)^2} = \left( {20\;{\rm{ft}}} \right)\left( {32.2\;{\rm{ft/}}{{\rm{s}}^2}} \right)\\ {v_B} = 25.37\;{\rm{ft/s}} \end{array}\]