Question

Two forces F13=50 N Z230° and F14= 100 N Z200° are acting on links 3 The free-body diagrams of the moving links are shown in figure Example 5.3. F 50° FA В For the mechanism shown in Fig. 5.30 AA= 22=80, AB=az =100, y C 3 D А) T12 Figure 5.30 20° 013 4 012 0, 14 Aol Bo Х BoB= 24=120, A B = a;= 140, AC=b3 mm. When 612=60°, from kinematic analysis 013=29,980, = 70, BC=80 and BoD= ba=90 014 = 96.40 and 4 respectively. 5.31.

Answer 1

hey, what do you want to ask ?

okay .. lemme explain the procedure how we can solve just kind of problem.

ya, it is a four bar mechanism . so for analyse we have to take individual bar.

for each link . u should have to consider that they are always in equilibrium at every instant.

second step is the each link should also be in rotational equilibrium.

if u find that number of unknowns are more than number of equilibrium equation then, we have to find the compatibility equation for solving the unknown value.

this is the complete procedure.

hope , it will helps u.

thanks!