Explanation Given information: The weight of fly ball is ( W ) is 1.2 lb. The radius of horizontal circle ( ρ ) is 6 in.. The mass of links AB , BC , AD , and DE are negligible. Calculation: Write the general equation of mass ( m ) of flyball. m = W g Here, g is the acceleration due to gravity. Sketch the free body diagram and kinetic diagram of fly ball as shown in Figure (1). Consider along y -axis as positive. Apply Newton’s law of equation along y -axis. Σ F y = 0 T D A cos 20 ° − T D E cos 30 ° − W = 0 0.93969 T D A − 0.86603 T D E = W (1) Here, T D A is the force in the link DA and T D E is the force in the link DE . Substitute 1.2 lb for W in Equation (1). 0.93969 T D A − 0.86603 T D E = 1.2 (2) Find the equation of force in the link DA using Equation (2): 0.93969 T D A = 0.86603 T D E + 1.2 T D A = 0.92160 T D E + 1.277 (3) Find the equation of force in the link DE using Equation (2): 0.86603 T D E = 0.93969 T D A − 1.2 T D E = 1.08506 T D A − 1.3856 (4) Consider along n -axis as positive. Apply Newton’s law of equation along n -axis. Σ F n = m a n T D A sin 20 ° + T D E sin 30 ° = m ( v 2 ρ ) Here, v is the velocity of fly ball. Substitute W g for m . T D A sin 20 ° + T D E sin 30 ° = W g ( v 2 ρ ) Substitute 1.2 lb for W , 32.2 ft/s 2 for g , and 6 in. for ρ . T D A sin 20 ° + T D E sin 30 ° = 1.2 32.2 ( v 2 6 in . × 1 ft 12 in . ) 0.34202 T D A + 0.5 T D E = 0.07453 v 2 v 2 = 4.589 T D A + 6.7087 T D E (5) The tension in the links not exceeds 17 lb

Vector Mechanics for Engineers: Statics and Dynamics

11th Edition

ISBN: 9780073398242

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Chapter 12.1, Problem 12.43P

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The range of the allowable values of v so that the magnitudes of the forces in the links do not exceed 17 lb.

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Chapter 12.1, Problem 12.42P

Chapter 12.1, Problem 12.44P

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