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.

52.62′ (using the cosine theorem for the triangle ABC). 5.2. Static Force Analysis of Machinery 341 F 13 Fazy a Ezay В F *34x B F 23x 20° A 014 G12x G14x Во AO Gizy G141 only when F34x and F34y are determined. The three equilibrium equations for link 3 must also be written (note that F34 and F43 are of equal -F34 24 sin014 + F34a4 cos 14 – F14 b4 sin(200-014)=0 (EMB=0) (6) 50° F14 F. 32y F32x… T12 012 Figure 5.31. The three equilibrium equations for link 4 are: F34x + G14x – F14 cos20º = 0 (2Fx=0) (1) F34y + Gl4y – F14 sin20º = 0 (Fy=0) (2) -F34x 44 sin014 + F34y84 cos014 – F 14 b4 sin(20°-014)=0 (EMBo=0) (3) There are four unknowns in three equations, therefore the equations obtained from one free-body diagram is not enough to solve for the unknowns. Equations 1 and 2 can be used to solve for G14x and Giay , F23x – F34x – F1zcos(509= 0 F23y – F34y – F13sin(509) = 0 magnitude) (EFx=0) (EFy=0) Where O=