Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 120P from Chapter 2 from Hibbeler's Engineering Mechanics.
Problem 120P
Step-by-Step Solution
We are given that the forces are ${F_1} = 30\;{\rm{lb}}$, and ${F_2} = 25\;{\rm{lb}}$.
We are required to determine the magnitude of the projected components of ${F_1}$ along the line of action ${F_2}$.
Step 2
The equation for the vector form of force ${F_1}$ can be written as,
\[{F_{1V}} = {F_1} \times \left( {\cos 30^\circ \sin 30^\circ i + \cos 30^\circ \cos 30^\circ j - \sin 30^\circ k} \right)\]Step 3
Substitute all the values in the above equation.
\[\begin{array}{c} {F_{1V}} = \left( {30\;{\rm{lb}}} \right) \times \left( {\cos 30^\circ \sin 30^\circ i + \cos 30^\circ \cos 30^\circ j - \sin 30^\circ k} \right)\\ = \left( {13i + 22.5j - 15k} \right)\;{\rm{lb}} \end{array}\]Step 4
The angle between the force ${F_2}$ and $x$-axis can be calculated as,
\[\begin{array}{c} {\cos ^2}\phi + {\cos ^2}60^\circ + {\cos ^2}60^\circ = 1\\ {\cos ^2}\phi = 1 - {\cos ^2}60^\circ - {\cos ^2}60^\circ \\ \cos \phi = \sqrt {1 - {{\cos }^2}60^\circ - {{\cos }^2}60^\circ } \\ \phi = 135^\circ \end{array}\]Step 5
The unit vector of the force ${F_2}$ can be calculated as,
\[\begin{array}{c} {u_{{F_2}}} = \cos 135^\circ i + \cos 60^\circ j + \cos 60^\circ k\\ = - 0.707i + 0.5j + 0.5k \end{array}\]Step 6
The equation to calculate the magnitude of the projected components of ${F_1}$ along the line of action ${F_2}$ is given by,
\[{F_{12}} = \left| {{F_{1V}} \cdot {u_{{F_2}}}} \right|\]Step 7
Substitute all the values in the above equation.
\[\begin{array}{c} {F_{12}} = \left| {\left( {13i + 22.5j - 15k} \right)\;{\rm{lb}} \cdot \left( { - 0.707i + 0.5j + 0.5k} \right)} \right|\\ = \left| { - 9.194 + 11.25 - 7.5} \right|\;{\rm{lb}}\\ = \left| { - 5.44\;{\rm{lb}}} \right|\\ = 5.44\;{\rm{lb}} \end{array}\]