Question 6.5: Determine the components of the forces acting on each member...
Determine the components of the forces acting on each member of the frame shown.
STRATEGY: The approach to this analysis is to consider the entire frame as a free body to determine the reactions, and then consider separate members. However, in this case, you will not be able to determine forces on one member without analyzing a second member at the same time.
MODELING and ANALYSIS: The external reactions involve only three unknowns, so compute the reactions by considering the free-body diagram of the entire frame (Fig. 1).
Now dismember the frame. Since only two members are connected at each joint, force components are equal and opposite on each member at each joint (Fig. 2).
Free Body: Member BCD.
+\circlearrowleft \sum{M_B}=0: \quad \quad -(2400 \text{ N})(3.6 \text{ m})+C_y(2.4 \text{ m})=0 \quad C_y=+3600 \text{ N} \\ +\circlearrowleft \sum{M_C}=0: \quad \quad -(2400 \text{ N})(1.2 \text{ m})+B_y(2.4 \text{ m})=0 \quad B_y=+1200 \text{ N}\\ \underrightarrow{+}\sum{F_x}=0: \quad \quad -B_x+C_x=0Neither B_x \text{ nor }C_x can be obtained by considering only member BCD; you need to look at member ABE. The positive values obtained for B_y \text{ and }C_y indicate that the force components \pmb{\text{B}}_{y} \text{ and }\pmb{\text{C}}_yare directed as assumed.
Free Body: Member ABE.
\begin{matrix} +\circlearrowleft \sum{M_A}=0: & B_x(2.7 \text{ m})&=&0 && B_x=0 \\ \underrightarrow{+}\sum{F_x}=0: & -B_x-A_x&=&0 && A_x=0 \\ +\uparrow \sum{F_y}=0: & -A_y+B_y+600 \text{ N}&=&0 \\ & -A_y+1200 \text{ N}+600 \text{ N}&=&0 && A_y=+1800 \text{ N} \end{matrix}Free Body: Member BCD. Returning now to member BCD, you have
\underrightarrow{+}\sum{F_x}=0: \quad \quad \quad -B_x+C_x=0 \quad \quad 0+C_x=0 \quad \quad \quad \quad C_x=0REFLECT and THINK: All unknown components have now been found.
To check the results, you can verify that member ACF is in equilibrium.
