Explanation Consider the free body diagram given below. The force R → , exerted on the foot by the tibia, can be resolved into vertical and horizontal components. Consider the system is in equilibrium. Write the expression for the force along the x direction. ∑ F x = 0 R sin 15.0 ° − T sin θ = 0 (I) Rearrange equation (I) to obtain an expression for R . R = T sin θ sin 15.0 ° (II) Write the expression for the force along y direction. ∑ F y = 0 700 N − R cos 15.0 ° + T cos θ = 0 (III) Write the expression for the torque acting at the point O . ∑ τ O = 0 − ( 700 N ) [ ( 18.0 cm ) cos θ ] + T ( 25.0 cm − 18 .0 cm ) = 0 (IV) Rearrange equation (IV) to obtain the expression for T . T = ( 1800 N ) cos θ (V) Substitute, equation (V) in (II). R = ( 1800 N sin 15.0 ° ) sin θ cos θ (VI) Substitute, equation (VI) in (IV). 700 N= ( 1800 N sin 15.0 ° ) sin θ cos θ cos 15.0 ° + T cos θ sin θ cos θ = tan 15.0 ° cos 2 θ + 0.1042 (VII) Square equation (VII) and substitute, 1 − cos 2 θ for sin 2 θ . ( 1 − cos 2 θ ) cos 2 θ = tan 2 15.0 ° cos 4 θ + ( 0

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN: 9781133104261

Author: Raymond A. Serway, John W. Jewett

Publisher: Cengage Learning

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Torque

Torque is the rotational equivalent of linear force in physics and mechanics. Based on the course of study, it is also known as the moment of force, rotational force, or turning effect.

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Chapter 10, Problem 69P

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The value of T, R, and θ.

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Chapter 10, Problem 68P

Chapter 10, Problem 70P

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The value of T , R , and θ .

Solution Summary: The author analyzes the value of T, R and theta . The force exerted on the foot by the tibia can be resolved into vertical and horizontal components.

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