Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 89P from Chapter 6 from Hibbeler's Engineering Mechanics.
Problem 89P
Step-by-Step Solution
We are given the load $F = 600\;{\rm{N}}$ and $F' = 300\;{\rm{N}}$.
We are asked to determine the horizontal and vertical component of force which pin $C$ exerts on $ABC$.
Step 2
The free body diagram of the member is shown below:
The vertical length from point $F$ to $E$ is ${d_1} = 4.5\;{\rm{m}}$.
The horizontal length to point $E$ is ${d_2} = 4\;{\rm{m}}$.
The vertical length from point $C$ to $E$ is ${d_3} = 3\;{\rm{m}}$.
To find the reaction force ${A_x}$ we will find the moment about $E$ by using relation,
\[\begin{array}{c} \Sigma {M_E} = 0\\ F' \times {d_1} + F \times {d_2} - {A_x} \times {d_3} = 0 \end{array}\]On plugging the values in the above relation, we get,
\[\begin{array}{c} \left( {300\;{\rm{N}}} \right) \times \left( {4.5\;{\rm{m}}} \right) + \left( {600\;{\rm{N}}} \right) \times \left( {4\;{\rm{m}}} \right) - {A_x} \times \left( {3\;{\rm{m}}} \right) = 0\\ {A_x} = 1250\;{\rm{N}} \end{array}\]Step 3
The free body diagram of the member is shown below:
The distance between $B$ and $D$ is calculated as:
\[\begin{array}{l} {d_{BD}} = \frac{1}{2} \times {d_3}\\ {d_{BD}} = \frac{1}{2} \times \left( {3\;{\rm{m}}} \right)\\ {d_{BD}} = 1.5\;{\rm{m}} \end{array}\]To find the required force ${B_x}$ we will find the moment about D by using relation,
\[\begin{array}{c} \Sigma {M_D} = 0\\ - {B_x} \times {d_{BD}} + F' \times {d_3} = 0 \end{array}\]On plugging the values in the above relation, we get,
\[\begin{array}{c} - {B_x}\left( {1.5\;{\rm{m}}} \right) - \left( {300\;{\rm{N}}} \right)\left( {3\;{\rm{m}}} \right) = 0\\ {B_x} = 600\;{\rm{N}} \end{array}\]Step 4
The free body diagram of the member is shown below:
The length from point $B$ to $C$ is ${d_4} = 2\;{\rm{m}}$.
To find the reactive force at $C$ in $y$ direction we will use the relation.
\[\begin{array}{c} \Sigma {M_B} = 0\\ - {C_y} \times {d_4} = 0\\ - {C_y} \times 2\;{\rm{m}} = 0\\ {C_y} = 0 \end{array}\]To find the reactive force at $C$ in $x$ direction we will use the relation.
\[\begin{array}{c} \Sigma {F_x} = 0\\ {A_x} - F - {C_x} = 0\\ 1250\;{\rm{N}} - 600\;{\rm{N}} - {C_x} = 0\\ {C_x} = 650\;{\rm{N}} \end{array}\]