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/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. . q b r O A B C D E

. eq v = r er + r q eq er a = (r - r q 2 ) er + (r q + 2 r q ) eq b r O A B C D E Problem 12.128 Solving Problems on Your Own 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. 1. Kinematics: Examine the velocity and acceleration of the particle. In polar coordinates: v = r er + r q eq a = (r - r q 2 ) er + (r q + 2 r q ) eq . r = r er eq er q

q b r O A B C D E Problem 12.128 Solving Problems on Your Own 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. 2. Kinetics: Draw a free body diagram showing the applied forces and an equivalent force diagram showing the vector ma or its components.

q b r O A B C D E Problem 12.128 Solving Problems on Your Own 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. 3. Apply Newton’s second law: The relationship between the forces acting on the particle, its mass and acceleration is given by S F = m a . 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: S Fr = m ar = m ( r - r q 2 ) and S Fq = m aq = m ( r q + 2 r q ) . .

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

r = 2 b cos q, r = - 2 b sinq q, r = - 2 b sinq q - 2 b cosq q 2 A B q Problem 12.128 Solution . . . r = 2 b cos q, r = - 2 b sinq q, r = - 2 b sinq q - 2 b cosq q 2 . . 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 . . . . . .

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

= + SFr = mar: Fr = m ( r - r q2 ) Apply Newton’s second law. Problem 12.128 Solution r O A B q Fr Fq mar maq = + SFr = mar: Fr = m ( r - r q2 ) Fr = [- 989.79 - ( 3.13 )(152 )] = -13.16 lb Fr = 13.16 lb + SFq = maq: Fq = m ( r q + 2 r q ) Fq = [(3.13)(250) + 2 (-17.1)(15)] = 2.1 lb . . (4/16) 32.2 . . . (4/16) 32.2 Fq = 2.10 lb

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