Question 6.SP.6: A 600-lb horizontal force is applied to pin A of the frame s......
A 600-lb horizontal force is applied to pin A of the frame shown. Determine the forces acting on the two vertical members of the frame.
STRATEGY: Begin as usual with a free-body diagram of the entire frame, but this time you will not be able to determine all of the reactions. You will have to analyze a separate member and then return to the entire frame analysis in order to determine the remaining reaction forces.
- A horizontal force of 600 lb is applied to pin A of the frame.
Final Answer
MODELING and ANALYSIS: Choosing the entire frame as a free body (Fig. 1), you can write equilibrium equations to determine the two force components \mathbf{E}_y \text { and } \mathbf{F}_y. However, these equations are not sufficient to determine \mathbf{E}_x \text { and } \mathbf{F}_x.
+↺ \Sigma M_E=0: \quad-(600 ~\mathrm{lb})(10 ~\mathrm{ft})+F_y(6 ~\mathrm{ft})=0
F_y=+1000 ~\mathrm{lb} \quad \mathbf{F}_y=1000 ~\mathrm{lb} \uparrow
+\uparrow \Sigma F_y=0: \quad E_y+F_y=0
E_y=-1000 ~\mathrm{lb} \quad \mathbf{E}_y=1000 ~\mathrm{lb} \downarrow
To proceed with the solution, now consider the free-body diagrams of the various members (Fig. 2). In dismembering the frame, assume that pin A is attached to the multi-force member ACE so that the 600-lb force is applied to that member. Note that AB and CD are two-force members.
Free Body: Member ACE
+\uparrow \Sigma F_y=0: \quad-\frac{5}{13} F_{A B}+\frac{5}{13} F_{C D}-1000 ~\mathrm{lb}=0
+↺ \Sigma M_E=0: \quad-(600 ~\mathrm{lb})(10~ \mathrm{ft})-\left(\frac{12}{13} F_{A B}\right)(10 ~\mathrm{ft})-\left(\frac{12}{13} F_{C D}\right)(2.5 ~\mathrm{ft})=0
Solving these equations simultaneously gives you
F_{A B}=-1040 ~\mathrm{lb} \quad F_{C D}=+1560 ~\mathrm{lb}
The signs indicate that the sense assumed for F_{C D} was correct and the sense for F_{A B} was incorrect. Now summing x components, you have
\stackrel{ \pm}{\rightarrow} \Sigma F_x=0: \quad 600 ~\mathrm{lb}+\frac{12}{13}(-1040 ~\mathrm{lb})+\frac{12}{13}(+1560 ~\mathrm{lb})+E_x=0
E_x=-1080 ~\mathrm{lb} \quad \mathbf{E}_x=1080 ~\mathrm{lb} \leftarrow
Free Body: Entire Frame. Now that \mathbf{E}_x is determined, you can return to the free-body diagram of the entire frame.
\stackrel{ \pm}{\rightarrow} \Sigma F_x=0: \quad 600 ~\mathrm{lb}-1080 ~\mathrm{lb}+F_x=0
F_x=+480 ~\mathrm{lb} \quad \mathbf{F}_x=480 ~\mathrm{lb} \rightarrow
REFLECT and THINK: Check your computations by verifying that the equation \Sigma M_B=0 is satisfied by the forces acting on member BDF.
+↺ \Sigma M_B=-\left(\frac{12}{13} F_{C D}\right)(2.5 ~\mathrm{ft})+\left(F_x\right)(7.5 ~\mathrm{ft})
=-\frac{12}{13}(1560 ~\mathrm{lb})(2.5 ~\mathrm{ft})+(480 ~\mathrm{lb})(7.5 ~\mathrm{ft})
=-3600 ~\mathrm{lb} \cdot \mathrm{ft}+3600 ~\mathrm{lb} \cdot \mathrm{ft}=0
