(a) Explanation Given info: The mass of the bat is 1.1 kg , the distance of the shoulder from the small end of the bat is 22.5 cm and the distance of the center of mass from the small end of the bat is 67.0 cm . The free body diagram shows the bat which is hold across his shoulder which is shown below. Figure (1) Apply Newton's second law of motion for torque about shoulder. ∑ τ = 0 F → h × 22.5 cm − m g × ( 67 cm − 22.5 cm ) Here, F → h is the force on the hand. m is the mass of the bat. g is the acceleration due to gravity. Substitute 1.1 kg for m and 9.81 m / s 2 for g in above equation to find force. [ ( F → h × 22.5 cm × 1 m 100 cm ) − ( 1 (b) To determine The magnitude and direction of the force exerted by the shoulder.

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ISBN: 8220103026918

Author: Walker

Publisher: PEARSON

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Chapter 11, Problem 29PCE

(a)

To determine

The magnitude and direction of the force exerted by the hand.

(b)

To determine

The magnitude and direction of the force exerted by the shoulder.

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Chapter 11, Problem 28PCE

Chapter 11, Problem 30PCE

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The magnitude and direction of the force exerted by the hand.

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The magnitude and direction of the force exerted by the hand.

The magnitude and direction of the force exerted by the shoulder.

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