For the linkage shown, determine the force Q required for equilibrium when l = 18 in., M = 600 lb in., and θ = 70°.

Given,

l = 18 in.,
M = 600 lb in.,
θ = 70°

Let us consider the small virtual displacement of the mechanism as shown in figure:

From the principle of virtual work,

δU = 0

or, MδΦ - QδC = 0 ---(1)

From the figure, we get the following triangles;

From ΔABB’:

\( \tan \delta \phi = \frac{\delta X_b}{l/2} \)

\( \delta \phi = \frac{2\delta X_b}{l} \)

\( \delta X_B = \frac{l\delta \phi}{2} \)

Since the horizontal displacement is the same, \( \delta X_c = \delta X_B = \frac{l\delta \Phi}{2} \)

Now, From ΔCC’X:

\( \cos \Theta = \frac{\delta X_c}{\delta C} \)

\( \delta C = \frac{\delta X_c}{\cos \Theta} \)

\( \delta C = \frac{l\delta \Phi}{2\cos \Theta} \)

\( \delta C = \frac{18\delta \Phi}{2\cos 70} \)

\( \delta C = 26.31 \delta \phi \) ----- (2)

Then, using eqn (2) in eqn (1):

\( 600\delta \phi - Q \times 26.31\delta \phi = 0 \)

∴ \( Q = 22.80 \text{ lb} \)

Engineering Mechanics