Question

A M12 105° Р B

Crank and connecting rod mechanism shown in Figure 1, piston and cylinder friction between has been neglected. Static stability at the given crank position find the M12 torque required to make it happen. (O2A = 75 mm, AB = 350 mm, P = 0.9 kN)

Answer 1

- Date given O A = 75 mm = 0,075 m A 2 MI 4 105 Р AB = 350 mm = 0. 350 m P = 0.9 KN - 900 N Oz B - Foom Fig. wing sine rule we com sin (105) sin o say The АВ sim pa OBA sin cios) 0,075 sin (105) AB 0.350 = 11.9456 Mow FBD of link : Fuz sino 스 Moco F43 For static stability (EF) P = F43 cos o F42COS (1.9456) so F4z coso P N = Noomal Force. 900 F43 = Toote om body 4 by body 3 F43 = 919.90 Nm out omgle (-1.9456 ) Noco FBD OF link 3 Here F34 = Force by body 4 on body 3 F34 - F43 Copposite direction) F32 3 But 1534 1 = 1 F43 1 = gig. 90 N and F32 = F34 = 919.ge Nimm e 919.9e co cmgle (168.054) Fe = 919.ga cu cigle (-11.9456) F34

so Faz SO) F23 = 919-ge 2168.054) F32 = 919.ge ink 2. FBD OF Now F2z = gig.ge L 168.054 F23 F23 -899.gg + 190. 4139 kj mn 6 and r = Aoa ( cos (150) " + sin (105) ;) + 0.09244 ſ Mie -0.0194) mm (5 C C - Mow chie Moment due to Moment Force Faz to mie (-: Too sturicwly bulcmeed So, MIR x F .- j = -0.0194 0.07244 -899.99 190, 4319 O Mia = 61.500g Nom mm G - S, Too stwie stebiny Mie = 61.500g Num.