In the frame shown, members ACE and BCD are connected by a pin at C and by the link DE. For the loading shown, determine the force in link DE and the components of the force exerted at C on member BCD.

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STRATEGY: Follow the general procedure discussed in this section. First treat the entire frame as a free body, which will enable you to find the reac-tions at A and B. Then dismember the frame and treat each member as a free body, which will give you the equations needed to find the force at C.
MODELING and ANALYSIS: Since the external reactions involve only three unknowns, compute the reactions by considering the free-body dia-gram of the entire frame (Fig. 1).

[latex]\begin{array}{rlrl}+\uparrow \Sigma F_{y}=0: & A_{y}-480 \mathrm{~N}=0 & A_{y}=+480 \mathrm{~N} & \mathbf{A}_{y}=480 \mathrm{~N} \uparrow \+\curvearrowleft\Sigma M_{A}=0:. & -(480 \mathrm{~N})(100 \mathrm{~mm})+B(160 \mathrm{~mm})=0 & & \& & B=+300 \mathrm{~N} & \mathbf{B}=300 \mathrm{~N} ightarrow \\stackrel{+}{ ightarrow} \Sigma F_{x}=0: & B+A_{x}=0 & & \& 300 \mathrm{~N}+A_{x}=0 & A_{x}=-300 \mathrm{~N} & \mathbf{A}_{x}=300 \mathrm{~N} \leftarrow\end{array}[/latex]

Now dismember the frame (Figs. 2 and 3). Since only two members are connected at C, the components of the unknown forces acting on ACE and BCD are, respectively, equal and opposite. Assume that link DE is in tension (Fig. 3) and exerts equal and opposite forces at D and E, directed as shown.

Free Body: Member BCD. Using the free body BCD (Fig. 2), you can write and solve three equilibrium equations:

[latex]\begin{aligned}&+\curvearrowright \Sigma M_{C}=0:\&\left(F_{D E} \sin \alpha ight)(250 \mathrm{~mm})+(300 \mathrm{~N})(80 \mathrm{~mm})+(480 \mathrm{~N})(100 \mathrm{~mm})=0\&F_{D E}=-561 \mathrm{~N} \quad F_{D E}=561 \mathrm{~N} \mathrm{} C\&\stackrel{+}{ ightarrow} \Sigma F_{x}=0: \quad C_{x}-F_{D E} \cos \alpha+300 \mathrm{~N}=0\&C_{x}-(-561 \mathrm{~N}) \cos 28.07^{\circ}+300 \mathrm{~N}=0 \quad C_{x}=-795 \mathrm{~N}\&+\uparrow \Sigma F_{y}=0: \quad C_{y}-F_{D E} \sin \alpha-480 \mathrm{~N}=0\&C_{y}-(-561 \mathrm{~N}) \sin 28.07^{\circ}-480 \mathrm{~N}=0 \quad C_{y}=+216 \mathrm{~N}\end{aligned}[/latex]

From the signs obtained for [latex]C_{x}[/latex] and [latex]C_{y}[/latex], the force components [latex]C_{x}[/latex] and [latex]C_{x}[/latex] exerted on member BCD are directed to the left and up, respectively. Thus, you have

[latex]\mathbf{C}_{x}=795 \mathrm{~N} \leftarrow \mathbf{C}_{y}=216 \mathrm{~N} \uparrow[/latex]

REFLECT and THINK: Check the computations by considering the free body ACE (Fig. 3). For example,

[latex]\begin{aligned}+\curvearrowleft\Sigma M_{A}.&=\left(F_{D E} \cos \alpha ight)(300 \mathrm{~mm})+\left(F_{D E} \sin \alpha ight)(100 \mathrm{~mm})-C_{x}(220 \mathrm{~mm}) \&=(-561 \cos \alpha)(300)+(-561 \sin \alpha)(100)-(-795)(220)=0\end{aligned}[/latex]

Question 6.SP.4: In the frame shown, members ACE and BCD are connected by a p...

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