Explanation Given information: The unstretched length of the cord ( l 0 ) is 1 ft . The weight of the beanbag and holder at C ( W ) is 4 lb . The velocity of the beanbag ( v 2 ) is 31 ft / s . The member OBC is inclined at angle ( α ) is 30 ° . The length of the member OB ( l O B ) is 2 ft . The length of the member BC ( l B C ) is 1ft . The horizontal distance of OA ( x ) is 2 ft . The bag is released at angle ( θ ) is 45 ° . The acceleration due to gravity ( g ) is 32.2 ft / s 2 . Calculation: Show the free body diagram of the given system as in Figure (1). Calculate the mass of the cord ( m ) using the formula: W = m g m = W g Substitute 4 lb for W and 32.2 ft / s 2 for g . m = 4 32.2 = 0.1242 lb ⋅ s 2 / ft Consider the position 1. Calculate the length of AB ( l 1 ) using the law of cosines: l 1 = ( l O B ) 2 + ( l O A ) 2 + ( 2 × l O B × l O A × cos α ) Substitute 2 ft for l O B , 2 ft for l O A , and 30 ° for α . l 1 = ( 2 ) 2 + ( 2 ) 2 + ( 2 × 2 × 2 × cos 30 ° ) = 3.864 ft Calculate the stretch in cord ( s 1 ) using the relation: s 1 = l 1 − l 0 Substitute 3.864 ft for l 1 and 1 ft for l 0 . s 1 = 3.864 − 1 = 2.864 ft Calculate the height of bean bag ( h 1 ) using the relation: h 1 = − ( l O B + l B C ) sin α Substitute 2 ft for l O B , 1ft for l B C and 30 ° for α . h 1 = − ( 2 + 1 ) sin 30 ° = − 1.5 ft Calculate the potential energy stored in the cord due to elongation at position 1 ( V 1 ) e using the formula: ( V 1 ) e = 1 2 k s 1 2 Substitute 2.864 ft for s 1 . ( V 1 ) e = 1 2 × k × ( 2.864 ) 2 = 4.10125 k Calculate the potential energy stored in the cord due to gravitation at position 1 ( V 1 ) g using the formula: ( V 1 ) g = m g h 1 Substitute 0.1242 lb ⋅ s 2 / ft for m , 32.2 ft / s 2 for g and − 1.5 ft for h 1 . ( V 1 ) g = 0.1242 lb ⋅ s 2 / ft × 32.2 × ( − 1.5 ) = − 5.99886 lb ⋅ ft Calculate the potential energy stored in rod in position 1 ( V 1 ) using the relation: V 1 = ( V 1 ) e + ( V 1 ) g Substitute 4.10125 k for ( V 1 ) e and − 5.99886 lb ⋅ ft for ( V 1 ) g . V 1 = 4.10125 k − 5.99886 Here, the kinetic energy in the position 1 ( T 1 ) is zero. Consider the position 2. Here, the length ( l O B ' ) and ( l O B ) are same of 2 ft . Calculate the length of AB’ ( l 2 ) using the law of cosines: l 2 = ( l O B ' ) 2 + ( l O A ) 2 − ( 2 × l O B ' × l O A × cos θ ) Substitute 2 ft for l O B ' , 2 ft for l O A and 45 ° for θ

11th Edition

ISBN: 9781259633133

Author: BEER

Publisher: MCG

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Solution Summary: The author shows the free body diagram of the given system as in Figure (1).

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