Determine the components of the forces acting on each member of the frame shown.

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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).

[latex]\begin{array}{llr}+\curvearrowleft \Sigma M_{E}=0:. & -(2400 \mathrm{~N})(3.6 \mathrm{~m})+F(4.8 \mathrm{~m})=0 &\ & \mathbf{F}= +1800 \mathrm{~N} & \mathbf{F}=1800 \mathrm{~N} \uparrow \+\uparrow \Sigma F_{y}=0: & -2400 \mathrm{~N}+1800 \mathrm{~N}+E_{y}=0 \& E_{y}=+600 \mathrm{~N} & \mathbf{E}_{y}=600 \mathrm{~N} \uparrow \\stackrel{+}{ ightarrow} \Sigma F_{x}=0: & & \mathbf{E}_{x}=0 \end{array}[/latex]

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.

[latex]\begin{aligned}&+\curvearrowleft \Sigma M_{B}=0: \quad-(2400 \mathrm{~N})(3.6 \mathrm{~m})+C_{y}(2.4 \mathrm{~m})=0 \quad C_{y}=+3600 \mathrm{~N}\&+\curvearrowleft \Sigma M_{C}=0: \quad-(2400 \mathrm{~N})(1.2 \mathrm{~m})+B_{y}(2.4 \mathrm{~m})=0 \quad B_{y}=+1200 \mathrm{~N}.\&\stackrel{+}{ ightarrow} \Sigma F_{x}=0: \quad-B_{x}+C_{x}=0\end{aligned}[/latex]

Neither [latex]B_{x}[/latex] nor [latex]C_{x}[/latex] can be obtained by considering only member BCD; you need to look at member ABE. The positive values obtained for By and [latex]C_{y}[/latex] indicate that the force components [latex]B_{y}[/latex] and [latex]C_{y}[/latex] are directed as assumed.

Free Body: Member ABE.

[latex]\begin{aligned}+\curvearrowleft \Sigma M_{A}=0: & B_{x}(2.7 \mathrm{~m}) =0 & B_{x}=0 \\stackrel{+}{ ightarrow \Sigma F_{x}=0:} & A_{x}-A_{x} =0 & A_{x}=0 \+\uparrow \Sigma F_{y}=0: &-A_{y}+B_{y}+600 \mathrm{~N} =0 & \&-A_{y}+1200 \mathrm{~N}+600 \mathrm{~N} =0 & A_{y}=+1800 \mathrm{~N}\end{aligned}[/latex]

Free Body: Member BCD. Returning now to member BCD, you have

[latex]+\Sigma F_{x}=0: \quad-B_{x}+C_{x}=0 \quad 0+C_{x}=0 \quad C_{x}=0[/latex]

REFLECT and THINK: All unknown components have now been found.To check the results, you can verify that member ACF is in equilibrium.

[latex]\begin{aligned}+\curvearrowleft \sum M_{C}.&=(1800 \mathrm{~N})(2.4 \mathrm{~m})-A_{y}(2.4 \mathrm{~m})-A_{x}(2.7 \mathrm{~m}) \&=(1800 \mathrm{~N})(2.4 \mathrm{~m})-(1800 \mathrm{~N})(2.4 \mathrm{~m})-0=0\end{aligned}[/latex]

Question 6.SP.5: Determine the components of the forces acting on each member...

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