Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 12P from Chapter 5 from Hibbeler's Engineering Mechanics.
Problem 12P
Step-by-Step Solution
We are given the triangular force per unit length $P = 4\;{\rm{kN}}$ at $a = 6\;{\rm{m}}$ from point $A$ and $b = 2\;{\rm{m}}$ from $B$, and the slope of plane for point $B$ is $\theta = 60^\circ $.
We are asked to determine the reactions at pin $A$ and reaction at rocker $B$ on the beam.
Step 2
First draw the free body diagram for the given beam.
Figure-(1)
Step 3
Now find the reaction at $B$ taking moment at $A$ equal to zero using the following relation.
\[{R_B}\sin \theta \times \left( {a + b} \right) - P \times a = 0\]Substitute the known values in the above expression.
\[\begin{array}{c} {R_B}\sin 60^\circ \times \left( {6\;{\rm{m}} + 2\;{\rm{m}}} \right) - \left( {4\;{\rm{kN}}} \right) \times \left( {6\;{\rm{m}}} \right) = 0\\ {R_B} = 3.46\;{\rm{kN}} \end{array}\]Step 4
Find the sum of all the forces acting in $x$ direction using the following relation.
\[{A_x} - {R_B}\cos \theta = 0\]Substitute the known values in the above expression.
\[\begin{array}{c} {A_x} - {R_B}\cos \theta = 0\\ {A_x} - \left( {3.46\;{\rm{kN}}} \right)\cos 60^\circ = 0\\ {A_x} = 1.73\;{\rm{kN}} \end{array}\]Step 5
Find the sum of all the forces acting in $y$ direction using the following relation.
\[{A_y} - P + {R_B}\sin \theta = 0\]Substitute the known values in the above expression.
\[\begin{array}{c} {A_y} - 4\;{\rm{kN}} + \left( {3.46\;{\rm{kN}}} \right)\sin 60^\circ = 0\\ {A_y} = 1.0\;{\rm{kN}} \end{array}\]