Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 8P from Chapter 2 from Hibbeler's Engineering Mechanics.
Problem 8P
Chapter:
Problem:
Resolve the force F2 into components acting along the u and v axes and determine...
Step-by-Step Solution
Step 1
Step 2
Step 3
Step 4
Step 5
Step 6
Step 7
Step 8
Step 9
We are given forces ${F_1} = 4\;{\rm{kN}}$, and ${F_2} = 6\;{\rm{kN}}$.
We are required to determine magnitude of forces in $u$ and $v$ direction.
Step 2
The forces along with all the angles are shown below.
Step 3
By the geometry, the angle $\alpha $ can be calculated as,
\[\begin{array}{c} \alpha = 180^\circ - 75^\circ - 30^\circ \\ = 75^\circ \end{array}\]Step 4
The angle $\beta $ can be calculated as,
\[\begin{array}{c} \beta = 180^\circ - 75^\circ - 30^\circ - 30^\circ \\ = 45^\circ \end{array}\]Step 5
Now, we shall make the triangle of forces ${F_2}$, ${F_v}$ and ${F_u}$.
Step 6
Apply the sine rule for the forces ${F_2}$ and ${F_u}$.
\[\begin{array}{c} \frac{{{F_u}}}{{\sin \alpha }} = \frac{{{F_2}}}{{\sin \left( {180^\circ - \alpha - 30^\circ } \right)}}\\ {F_u} = \frac{{{F_2} \times \sin \alpha }}{{\sin \left( {180^\circ - \alpha - 30^\circ } \right)}} \end{array}\]Step 7
Substitute all the values in the above equation.
\[\begin{array}{c} {F_u} = \frac{{\left( {6\;{\rm{kN}}} \right) \times \sin 75^\circ }}{{\sin \left( {180^\circ - 75^\circ - 30^\circ } \right)}}\\ = 6\;{\rm{kN}} \end{array}\]Step 8
Apply the sine rule for the forces ${F_2}$ and ${F_v}$ .
\[\begin{array}{c} \frac{{{F_v}}}{{\sin 30^\circ }} = \frac{{{F_2}}}{{\sin \left( {180^\circ - \alpha - 30^\circ } \right)}}\\ {F_v} = \frac{{{F_2} \times \sin 30^\circ }}{{\sin \left( {180^\circ - \alpha } \right)}} \end{array}\]Step 9
Substitute all the values in the above equation.
\[\begin{array}{c} {F_v} = \frac{{\left( {6\;{\rm{kN}}} \right) \times \sin 30^\circ }}{{\sin \left( {180^\circ - 75^\circ - 30^\circ } \right)}}\\ = 3.11\;{\rm{kN}} \end{array}\]