Question 6.SP.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).
+↺ΣM_{E} = 0: − (2400 N)(3.6 m) + F(4.8 m) = 0
F = +1800 N F = 1800 N↑ ◂
+↑ΣF_{y} = 0: − 2400 N + 1800 N + E_{y}= 0
E_{y} = +600 N E_{y} = 600 N↑ ◂
\underrightarrow{+}\Sigma F_{x}=0\colon
\mathrm{E}_{x}=0 ◂
Now dismember the frame. Because 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.
+↺\begin{array}{r l r l}{{\Sigma M_{B}=0\colon\quad}}&{{\ {\displaystyle-(2400\,\mathrm{N})(3.6~m)+C_{y}(2.4~m)=0}}}&{{\qquad C_{y}=+3600\,\mathrm{N}}}\end{array} ◂
+↺\begin{array}{r l r l}{{\Sigma M_{C}=0\colon\;\;\;}}&{{~-(2400\,{ N})(1.2\,{ m})+B_{y}(2.4\,{ m})=0}}&{{~~~~B_{y}=+1200\,{ N}}}\end{array} ◂
\begin{array}{c c}{{\underrightarrow{+}\Sigma F_{x}=0\colon}}&{{\qquad-B_{x}+C_{x}=0}}\end{array}
Neither B_{x}~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}~and~C_{y} indicate that the force components B_{y}~and~C_{y} are directed as assumed.
Free Body: Member ABE.
+↺\Sigma M_{A}=0\colon\qquad\qquad\qquad\qquad B_{x}(2.7\,\mathrm{m})=0\qquad\qquad\qquad\qquad\qquad B_{x}=0 ◂
\underrightarrow{+}\Sigma F_{x} = 0\colon \qquad\qquad\qquad\qquad +B_{x} – A_{x} = 0 \qquad\qquad\qquad\qquad\qquad A_{x} = 0 ◂
+↑ΣF_{y} = 0\colon \qquad\qquad\qquad\qquad -A_{y} + B_{y} + 600~\mathrm{N} = 0
-A_{y} + 1200~\mathrm{N} + 600~\mathrm{N} = 0 \qquad\qquad A_{y} = + 1800~\mathrm{N} ◂
Free Body: Member BCD. Returning now to member BCD, you have
\underrightarrow{+}\Sigma F_{x} = 0\colon \qquad\qquad -B_{x} + C_{x} = 0 \qquad 0 + C_{x} = 0 \qquad\qquad C_{x} = 0 ◂
REFLECT and THINK: All unknown components have now been found. To check the results, you can verify that member ACF is in equilibrium.
+↺\Sigma M_{C}=(1800\:\mathrm{N})(2.4\,\mathrm{m})-A_{y}(2.4\,\mathrm{m})-A_{x}(2.7\,\mathrm{m})
\qquad\qquad= (1800 N)(2.4 m) − (1800 N)(2.4 m) − 0 = 0 (checks)
