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Problem 12.128

Pin B weighs 4 oz and is free to slide in a horizontal plane along the rotating arm OC and along the circular slot DE of radius b = 20 in. Neglecting friction and assuming that q = 15 rad/s and q = 250 rad/s 2 for the position q = 20 o , determine for that position (a) the

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Problem 12.128

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  1. Pin B weighs 4 oz and is free to slide in a horizontal plane along the rotating arm OC and along the circular slot DE of radius b = 20 in. Neglecting friction and assuming that q = 15 rad/s and q = 250 rad/s2 for the position q = 20o, determine for that position (a) the radial and transverse components of the resultant force exerted on pin B, (b) the forces P and Q exerted on pin B, respectively, by rod OC and the wall of slot DE. r C B D . q O . . b A b E Problem 12.128

  2. Problem 12.128 r C B D q Pin B weighs 4 oz and is free to slide in a horizontal plane along the rotating arm OC and along the circular slot DE of radius b = 20 in. Neglecting friction and assuming that q = 15 rad/s and O b A b . E . . q = 250 rad/s2 for the position q = 20o, determine for that position (a) the radial and transverse components of the resultant force exerted on pin B, (b) the forces P and Q exerted on pin B, respectively, by rod OC and the wall of slot DE. 1. Kinematics: Examine the velocity and acceleration of the particle. In polar coordinates: v = rer + rqeq a = (r - r q 2 ) er + (r q + 2 r q ) eq eq . er . r = rer . . . . . . . q Solving Problems on Your Own

  3. Problem 12.128 r C B D q Pin B weighs 4 oz and is free to slide in a horizontal plane along the rotating arm OC and along the circular slot DE of radius b = 20 in. Neglecting friction and assuming that q = 15 rad/s and O b A b . E . . q = 250 rad/s2 for the position q = 20o, determine for that position (a) the radial and transverse components of the resultant force exerted on pin B, (b) the forces P and Q exerted on pin B, respectively, by rod OC and the wall of slot DE. Solving Problems on Your Own 2. Kinetics: Draw a free body diagram showing the applied forces and an equivalent force diagram showing the vector ma or its components.

  4. Problem 12.128 r C B D q Pin B weighs 4 oz and is free to slide in a horizontal plane along the rotating arm OC and along the circular slot DE of radius b = 20 in. Neglecting friction and assuming that q = 15 rad/s and O b A b . E . . q = 250 rad/s2 for the position q = 20o, determine for that position (a) the radial and transverse components of the resultant force exerted on pin B, (b) the forces P and Q exerted on pin B, respectively, by rod OC and the wall of slot DE. 3. Apply Newton’s second law: The relationship between the forces acting on the particle, its mass and acceleration is given by SF = ma . The vectors F and a can be expressed in terms of either their rectangular components or their radial and transverse components. With radial and transverse components: SFr = mar = m ( r - r q 2 ) and SFq = maq = m ( r q + 2 r q ) . . . . . . . Solving Problems on Your Own .

  5. Problem 12.128 Solution r C B D q O b A b B E r b O 2q q A b q = 20o q = 15 rad/s q = 250 rad/s2 . . . . . . . Kinematics. r = 2 b cos q r = - 2 b sin qq r = - 2 b sin qq - 2 b cos qq2 . . .

  6. B Problem 12.128 Solution r b O 2q q A b . . . . . . . . . . . . . r = 2 b cos q, r = - 2 b sinqq, r = - 2 b sinqq - 2 b cosqq2 For: b = 20/12 ft, q = 20o, q = 15 rad/s q = 250 rad/s2 r = 2 (20/12 ft) cos 20o = 3.13 ft r = - 2 (20/12 ft) sin 20o (15 rad/s) = - 17.1 ft/s r = -2(20/12 ft) sin 20o (250 rad/s2 ) - 2(20/12 ft) cos 20o (15 rad/s)2 r = - 989.79 ft/s2 . . .

  7. Problem 12.128 Solution r C B D q O b A b E maq Fq Fr mar = r r B B q q A A O O (a) Radial and transverse components of the resultant force exerted on pin B. Kinetics; draw a free body diagram.

  8. Problem 12.128 Solution maq Fq Fr mar = r r B B q q A A O O . . (4/16) 32.2 . . (4/16) 32.2 Apply Newton’s second law. + SFr = mar: Fr = m ( r - rq2 ) Fr = [- 989.79 - ( 3.13 )(152)] = -13.16 lb Fr = 13.16 lb + SFq = maq: Fq = m ( r q + 2 rq ) Fq = [(3.13)(250) + 2 (-17.1)(15)] = 2.1 lb . . . Fq = 2.10 lb

  9. Problem 12.128 Solution r C B D q O b A b E q q 40o Fq Q Fr q = r r B B P q q A A O O Fq = - Q sin q + P 2.10 = - 14.0 sin 20o + P P = 6.89 lb Fr = - Q cos q -13.16 = - Q cos 20o Q = 14.00 lb 20o

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