Question 6.112: The piston C moves vertically between the two smooth walls. ...
The piston C moves vertically between the two smooth walls. If the spring has a stiffness of k = 15 lb/in., and is unstretched when θ = 0°, determine the couple M that must be applied to AB to hold the mechanism in equilibrium when θ = 30°.
Geometry:
\frac{\sin \psi}{8}=\frac{\sin 30^{\circ}}{12} \quad \psi=19.47^{\circ}
\phi=180^{\circ}-30^{\circ}-19.47=130.53^{\circ}
\frac{l^{\prime}_{AC}}{\sin 130.53^{\circ}}=\frac{12}{\sin 30^{\circ}} \quad l_{A C}^{\prime}=18.242 \mathrm{in} .
Free Body Diagram: The solution for this problem will be simplified if one realizes that member CB is a two force member. Since the spring stretches x=l_{A C}-l_{A C}^{\prime}=20-18.242=1.758 \mathrm{in}. the spring force is F_{\mathrm{sp}}=k x=15(1.758)=26.37 \mathrm{lb} .
Equations of Equilibrium: Using the method of joints [\mathrm{FBD}(\mathrm{a})] ,
+↑ \Sigma F_y=0 ; \quad F_{C B} \cos 19.47^{\circ}-26.37=0
F_{C B}=27.97 \mathrm{lb}
From FBD (b),
\hookrightarrow +\Sigma M_A=0 ; \quad 27.97 \cos 40.53^{\circ}(8)-M=0M=170.08 \mathrm{lb} \cdot \mathrm{in}=14.2 \mathrm{lb} \cdot \mathrm{ft}
