Explanation Given: The weight of the boy A is m A = 80 lb . The weight of the girl B is m B = 65 lb . The weight of the toboggan is m t = 20 lb . The distance between the A and B is 4 ft . Initially they starts from rest. Write the principle of conservation of linear momentum. ∑ m v 1 = ∑ m v 2 (I) Here, m is the mass of the system, v is the velocity of the system, the suffixes and 1 , 2 indicates the initial and final stages. Conclusion: According to law of conservation momentum. A goes to B : At initial: The boy, girl and toboggan are at rest, the initial momentum becomes zero. ∑ m v 1 = 0 At final: When A goes to B ’s position (moving to right that is positive direction), the girl B and toboggan moving in opposite direction (moving to left that is negative direction) to the A ’s motion. The final momentum is ∑ m v 2 = m A v A − ( m t + m B ) v B Substitute 0 for ∑ m v 1 and m A v A − ( m t + m B ) v B for ∑ m v 2 in Equation (1). 0 = m A v A − ( m t + m B ) v B (II) When, the displacement is the integral of velocity over the time. s = ∫ v d t Integrate the Equation (II) with respect to time. 0 = ∫ m A v A d t − ∫ ( m t + m B ) v B d t 0 = m A ∫ v A d t − ( m t + m B ) ∫ v B d t 0 = m A s A − ( m t + m B ) s B (III) Assume, that when A goes ( 4 − x ) distance to the right, the B and toboggan moves x distance to the left. Substitute ( 4 − x ) for s A and x for s B in Equation (III). 0 = m A ( 4 − x ) − ( m t + m B ) ( x ) 0 = 4 m A − m A x − m t x − m B x 0 = 4 m A − x ( m A + m B + m t ) x = 4 m A m A + m B + m t Substitute 80 lb for m A , 65 lb for m B and 20 lb for m t . x = 4 ( 80 lb ) 80 lb + 65 lb + 20 lb = 320 165 = 1.9393 ft A and B goes to A : At initial: The boy, girl and toboggan are at rest, the initial momentum becomes zero. ∑ m v 1 = 0 At final: When A and B goes to A ’s position (moving to left that is negative direction), the toboggan moving in opposite direction (moving to right that positive direction) to The A and B ’s motion

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 15.3, Problem 50P

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The final position of the toboggan just after the motion when A going to B ’s position and stops, both A , B going to A ’s original position.

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Chapter 15.3, Problem 49P

Chapter 15.3, Problem 51P

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The final position of the toboggan just after the motion when A going to B ’s position and stops, both A , B going to A ’s original position.

Solution Summary: The author calculates the final position of the toboggan just after the motion when A going to B's position and stops.

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The final position of the toboggan just after the motion when A going to B ’s position and stops, both A , B going to A ’s original position.

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Chapter 15 Solutions