To determine The angular acceleration of the link OB . Explanation Draw the schematic diagram of the mechanism with instantaneous center C as shown in Figure (1). From Figure (1), consider the triangle Δ A B C and calculate the distance between the points B and C using the law of sine rule. sin 105 ° A B ¯ = sin 30 ° B C ¯ (I) Here, distance between the points B and C is B C ¯ and distance between the points A and B is A B ¯ . From Figure (1), consider the triangle Δ A C B and calculate the distance between the points A and C using the law of sine rule. sin 105 ° A B ¯ = sin 45 ° A C ¯ (II) Here, distance between the points A and C is A C ¯ . Write the expression for the angular velocity of the link AB . ω A B = v A A C ¯ (III) Here, velocity of point A is v A and angular velocity of the link AB is ω A B . Write the expression for the velocity of point B . v B = ( B C ¯ ) ω A B (IV) Here, velocity of point B is v B and distance between the points B and C is B C ¯ . Write the expression for the angular velocity of the link OB . ω O B = v B O B ¯ (V) Here, angular velocity of the link OB is ω O B and distance between the points O and B is O B ¯ . Write the expression for the acceleration of point B in vector form. a B = a A + a B / A ( a B ) n + ( a B ) t = a A + a B / A (VI) Here, acceleration of point B in vector form is a B , normal component of acceleration of point B in vector form is ( a B ) n , tangential component of acceleration of point B in vector form is ( a B ) t , acceleration of point A in vector form is a A and acceleration of point B with respect to point A in vector form is a B / A . Write the expression for the acceleration of point B with respect to point A in vector form. a B / A = α × r B / A − ω A B 2 r B / A (VII) Here, angular acceleration of the link AB in vector form is α and position vector of point B with respect to point A is r B / A . Conclusion: Substitute 180 mm for A B ¯ in Equation (I). sin 105 ° ( 180 mm ) = sin 30 ° B C ¯ sin 105 ° ( 0.180 m ) = sin 45 ° B C ¯ B C ¯ = 0.0932 m Substitute 180 mm for A B ¯ in Equation (II). sin 105 ° ( 180 mm ) = sin 45 ° A C ¯ sin 105 ° ( 0.180 m ) = sin 45 ° A C ¯ A C ¯ = 0.1318 m Substitute 2 m / s for v A and 180 mm for A B ¯ in Equation (III). ω A B = 2 m / s 0.1318 m = 15.18 rad / s Substitute 0 .0932 m for B C ¯ and 15.18 rad / s for ω A B in Equation (IV). v B = ( 0 .0932 m ) ( 15.18 rad / s ) = 1.414 m / s Substitute 1.414 m / s for v B and 120 mm for O B ¯ in Equation (V). ω O B = 1.414 m / s 120 mm = 1.414 m / s 0.120 m = 11.79 rad / s Write the expression for the angular acceleration of the link AB in vector form. α = α A B k Here, angular acceleration of the link AB is α A B and unit vector along z -direction is k . Since the point A is having the linear velocity and radius at infinity, the acceleration of point A in vector form is zero. a A = 0 Write the expression for the tangential component of acceleration of point B in vector form. ( a B ) t = ( O B ¯ ) α O B ( − sin 15 ° i − cos 15 ° j ) (VIII) Here, tangential component of acceleration of point B in vector form is ( a B ) t , angular acceleration of the link OB is α O B and unit vectors along x and y direction are i and j respectively. Substitute 120 mm for O B ¯ in Equation (VIII). ( a B ) t = ( 120 mm ) α O B ( − sin 15 ° i − cos 15 ° j ) = ( 0.120 m ) α O B ( − sin 15 ° i − cos 15 ° j ) = − 0

8th Edition

ISBN: 9781118885840

Author: James L. Meriam, L. G. Kraige, J. N. Bolton

Publisher: WILEY

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Chapter 5.6, Problem 145P

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The angular acceleration of the link OB.

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Chapter 5.6, Problem 144P

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Chapter 5 Solutions

Engineering Mechanics: Dynamics

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Car A is...Ch. 5.7 - For the instant represented, link CB is rotating...Ch. 5.7 - The disk rotates about a fixed axis through point...Ch. 5.7 - All conditions of the previous problem remain the...Ch. 5.7 - Prob. 182P Ch. 5.7 - Prob. 183P Ch. 5.7 - One wheel of an experimental vehicle F, which has...Ch. 5.8 - Prob. 185RP Ch. 5.8 - Prob. 186RP Ch. 5.8 - Prob. 187RP Ch. 5.8 - Prob. 188RP Ch. 5.8 - Prob. 189RP Ch. 5.8 - Roller B of the linkage has a velocity of 0.75 m/s...Ch. 5.8 - Prob. 191RP Ch. 5.8 - Prob. 192RP Ch. 5.8 - Prob. 193RP Ch. 5.8 - Prob. 194RP Ch. 5.8 - Prob. 195RP Ch. 5.8 - Prob. 196RP Ch. 5.8 - The isosceles triangular plate is guided by the...Ch. 5.8 - Prob. 198RP Ch. 5.8 - The hydraulic cylinder C imparts a velocity υ to...Ch. 5.8 - Prob. 200RP Ch. 5.8 - The figure illustrates a commonly used...Ch. 5.8 - Prob. 202RP Ch. 5.8 - Prob. 203RP Ch. 5.8 - Prob. 204RP Ch. 5.8 - Prob. 206RP Ch. 5.8 - For the slider-crank configuration shown, derive...Ch. 5.8 - Prob. 212RP

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