Question

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Problem #7: The figure below shows a fourbar linkage and its dimensions in meters. The steel crank, coupler, and rocker have uniform cross sections of 50 mm diameter. In the instantaneous position shown, the crank OzA has o= 10 rad/sec and a = 10 rad/sec2. There is a horizontal force at P of F 300 N. Find all pin forces and the torque needed to drive the crank at this instant. L3= 1.85 L4= 0.86 AP=1.33 O4 -36° L1 2.22 L2=0.86

Answer 1

DETERMINE THE POSITION TO S6 PROGRAM FOUR BAR OF LINKS 3 AND 4. ACCELERATION VELOCITY, AND w3. 3. 285 Hadls : li - -109. 287 hodls 0346.028 04 106, i8q° j t, 417 hadUA 4 r -4 3, 4 2G hadls THE DISTANCE DETERMINE TO THE CG IN THE LRCS EACH OF ON THE THREE MOVING LINKS LINKS 2 AND 4: Rca 2 0.5 a o. 430m R ca2 = R ca4 o.5 c Ru4 o. 430m Rca 3 o.sb Reaa = O.925 m 3 LINK DETERMINE THE MASS AND MOMENT OF INERTIA OF EACH LNK A.dNk, a. s 4,o22 2 30. 164 A, d c. Ys 14,022 +a) 2 3duNK O. 866 12 2 3.d b m2 t m3 i2 dunk (4 Ia O.866 C m4 )2 1.녀9980 SET COORDINATE SYST EM AT THE LN CS UP AN CG PRAW ALL APPLI CABLE LINK, AND OF EACH -S서 STEM. VECTORS ON THE A CTING

OF EACH MOVING LINK BODY DIAGRAM DRAW A FREE 2 Resn Rea2 Rci 3 A LINKAG E wiTH GCS THE COMPLETE Fu3 FA3 FI2y R43 Fizz aas Re F324 R23 >F32z FBD OF LINK 3 OF LINK 2 FBD FaH ac4n Fiy LINK 4 OF FBD

COMPONENTS OF THE POSITiON THE AND CALCULATE VECTORS RCa2 Cos (02 +180*) R12x O. 348 m Riy RCa2 in (02 180" O. 253 m O. 348 m Rзъу O. 253 m R23z RCG3 Cos (03 +180°) O. 642 m RCG3 Aim 3+ 180) = - o. 666 m R43x (Rc63 - b) cos ( 0 3+180) o. E4 2 m (Recs-b)sin (03+180") O.666 m R342 Rca4 cos 4 O.12O m Rзнy RCh in G4 O.413 m R14 O.120 m t Rcés bin (o4 180°) = - 0.413 m o. 20 m (Rpa - Rca3 )os O2 (Rpa Rcmin 03 O. 24lm COMPONENTS OF THE ANDY CALCULATE THE MOVING OF ALL ACCELERATION оF THE CG& LINKS GLDBAL COORDINATE SYSTEM TH E aw2coa +in [- Rca sin Bz +jcos 0z - 67.048 mls aG22 Re (ag,) c 54, 028 m Im L0 in0

aa a ca3A Re Caa3 302 mls Im (a3) =-19.864 mla2 a G4x.Re (agy) = 49,187 m la 2 Im (a) -102, 440 mls THE CALCULAPE AND COMPONENTS OF THE EXTERNAL FORCE AT P "тHE CaS IN Fex F Fry E O N THIS GIVES- C e O R327 R12z m m m Ruзх о R23y -R23y m m C O O -R342 O m m2. ag2 N- IG22 N-' m- (mg a63x- Fpx)N (m ahy-Fpy)N- VI Rpr. Fay -Rey. Fp)Nm- M ag N m aayN (16ad-T4) N- m- O

R CF -12 46 N Ri N F12 940 N RN Fray F32 30 E N R3 N F3ay R N - 1a3 N 5.N R5 N 78 2 N REN 735 N F14 2219 N Ra N Rg. N m TI2 7, 14 N