Explanation Free body diagram represents the magnitude and direction of force acting on a system. There are three forces, one is the force exerted by both hands of athlete, second is the force exerted by both hands of the athlete, and finally the weight of the athlete. Total torque in this case will be zero. Consider the foot of the athlete is fixed as a pivot and remaining part of the body is rotating in counter clockwise direction. Thus the motion of hand and body except foot responsible for torque Write the expression for total torque ∑ τ = ( 2 F h ) ( 1.70 m ) − w ( 1.15 m ) = 0 (I) Here, F h is the force acting on hands, w is the weight of the athlete. Rearrange equation (II) to obtain an expression for force acting on hands F h = w ( 1.15 m ) 2 ( 1.70 m ) = 0.338 × w (II) Write the expression for weight of athlete w = m g (III) Here, m is the mass of athlete, g is acceleration due to gravity The force on hands and feet are acting along positive y direction, and gravitational force due to weight is acting in downward direction. Write the expression for total force ∑ F y = 2 F h + 2 F f − w = 0 (IV) Rearrange equation (III) to obtain an expression for F f F f = 1 2 w − F h (V) Conclusion: Substitute 82 kg for m , and 9.8 m / s 2 for g in equation (III) w = 82 kg × 9

10th Edition

ISBN: 9780321902788

Author: Hugh D. Young, Philip W. Adams, Raymond Joseph Chastain

Publisher: PEARSON

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Chapter 10, Problem 38P

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The force exerts on each feet and hands of the athlete.

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Chapter 10, Problem 37P

Chapter 10, Problem 39P

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