A n s also d a n s a n s v c 220 ft s v cb 245 rad s

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A n s . Also, d A n s . A n s . v C = 2.20 ft s ; v CB = 2.45 rad s ( + c ) 0 = - 5.196 + 2.12 v CB a : + b - v C = 3 - 2.12 v CB - v C i = (6 sin 30° i - 6 cos 30° j ) + ( v CB k ) * (3 cos 45° i + 3 sin 45° j ) v C = v B + v * r C > B v C = 2.20 ft > s ; v CB = 2.45 rad > s ( + c ) 0 = - 6 cos 30° + v CB (3) sin 45° ( : + ) - v C = 6 sin 30° - v CB (3) cos 45° B v C ; R = C 6 30° c S + D v CB (3) 45° b T v C = v B + v C > B 3 ft 2 ft AB = 3 rad/s ω = 45 ° θ = 30 ° φ C B A
16–67. Determine the velocity of point on the rim of the gear at the instant shown. A SOLUTION General Plane Mot i on : Applying the relative velocity e q uation to points B and C and referring to the kinematic diagram of the gear shown in Fig. a , E q uating the i components yields (1) (2) For points A and C , E q uating the i and j components yields Thus, the magnitude of v A is A n s . and its direction is A n s . u = tan - 1 C A v A B y A v A B x S = tan - 1 ¢ 3.2998 3.9665 = 39.8° v A = 2 A v A B x 2 + A v A B y 2 = 2 3.9665 2 + 3.2998 2 = 5.16 ft > s A v A B x = 3.9665 ft > s A v A B y = 3.2998 ft > s A v A B x i + A v A B y j = 3.9665 i + 3.2998 j A v A B x i + A v A B y j = - 4 i + A - 3.111 k B * A - 1.061 i + 2.561 j B v A = v C + v * r A > C v = 3.111 rad > s 3 = 2.25 v - 4 3 i = A 2.25 v - 4 B i 3 i = - 4 i + A - v k B * A 2.25 j B v B = v C + v * r B > C 3 ft / s 4 ft / s A O 0.75 ft 1.50 ft 45
C B A 0.5 m 60 1 m 45 v C 3 m / s 16–70. If the s lider block C i s mo v in g at determine the an g ular v elocity of B C and the crank AB at the in s tant s hown. v C = 3 m > s , SOLUTION Rotation About a Fixed Axis: Referrin g to Fi g . a , General Plane Motion: Applyin g the relati v e v elocity e q uation and referrin g to the kinematic dia g ram of link B C s hown in Fi g . b , E q uatin g the i and j component s yield s , Sol v in g , Ans. Ans. v AB = 4.39 rad > s v BC = 2.69 rad > s - 0.25 v AB = 0.7071 v BC - 3 0.4330 v AB = 0.7071 v BC 0.4330 v AB i - 0.25 v AB j = 0.7071 v BC i + (0.7071 v BC - 3) j 0.4330 v AB i - 0.25 v AB j = - 3 j + ( - v BC k ) * ( - 1 co s 45° i + 1 s in 45° j ) v B = v C + v BC * r B > C = 0.4330 v AB i - 0.25 v AB j = ( - v AB k ) * (0.5 co s 60° i + 0.5 s in 60° j ) v B = v AB * r B
v AB 6 rad / s 0.3 m 0.5 m C A B 60 30 16–74. If crank AB rotate s with a con s tant an g ular v elocity of determine the an g ular v elocity of rod B C and the v elocity of the s lider block at the in s tant s hown.The rod i s in a horizontal po s ition. v AB = 6 rad > s , SOLUTION Rotation About a Fixed Axis: Referrin g to Fi g . a , General Plane Motion: Applyin g the relati v e v elocity e q uation to the kinematic dia g ram of link B C s hown in Fi g . b , E q uatin g the i and j component s yield s (1) (2) Sol v in g E qs . (1) and (2) yield s Ans. Ans. v BC = 6.24 rad > s v C = 1.80 m > s 1.559 = 0.5 v BC - 0.8660 v C - 0.9 = - 0.5 v C - 0.9 i + 1.559 j = - 0.5 v C i + (0.5 v BC - 0.8660 v C ) j ( - 0.9 i + 1.559 j ) = ( - v C co s 60° i - v C s in 60° j ) + ( - v BC k ) * ( - 0.5 i ) v B = v C + v BC * r B > C = [ - 0.9 i + 1.559 j ] = (6 k ) * (0.3 co s 30° i + 0.3 s in 30° j ) v B = v AB * r B
4 5 3 250 mm 400 mm 300 mm 300 mm E B C D A 30 v A 4 m / s *16–76. If the s lider block A i s mo v in g downward at determine the v elocity of block C at the in s tant s hown. v A = 4 m > s , SOLUTION General Plane Motion: Applyin g the relati v e v elocity e q uation by referrin g to the kinematic dia g ram of link AB s hown in Fi g . a , E q uatin g j component, b U s in g the re s ult of , U s in g the re s ult of to con s ider the motion of link C DE , Fi g . b , E q uatin g j and i component s , b Ans. v C = 1.636 - 0.2(5.249) = 0.587 m > s : v CD = 5.249 rad > s 0 = 0.3464 v CD - 1.818 v C i = (1.636 - 0.2 v CD ) i + (0.3464 v CD - 1.818) j v C i = (1.636 i - 1.818 j ) + ( - v CD k ) * ( - 0.4 co s 30° i - 0.4 s in 30° j ) v C = v D + V

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