(a) Explanation Given info: The magnitude of the hanging mass is 4.25 kg . The angle made by F → total above the horizontal is 20.0 ° . Draw the free body diagram of the hanging mass. Figure (1) Formula to calculate the weight of the hanging mass is, W = m g Here, m is the magnitude of the attached mass. g is the gravitational acceleration. Under equilibrium condition, the tension force exerted by the strings balances the weight of the hanging mass. Write the equilibrium equation forces in vertical direction from the free body diagram. T = W Here, T is the tension force exerted by the string. W is the weight of the hanging mass. Substitute m g for W to find the value of T . T = m g Substitute 4.25 kg for m and 9.81 m / s 2 for g to find T . T = ( 4.25 kg ) × ( 9.81 m / s 2 ) = 41.6925 N ≈ 41.7 N Since, the pulleys are ideal. So, the tension force in the entire string remains same which is equal to the weight of the hanging mass. The different pulleys simply changes the direction of the tension force, its magnitude does not change at all. Therefore, | F → 1 | = | F → 2 | = | F → 3 | = T = 41.7 N Consider the given diagram. Write the vector representation of force F → 1 . F → 1 = ( | F → 1 | cos θ ) i ^ + ( | F → 1 | sin θ ) j ^ Here, θ is the angle made by force F → 1 with the horizontal. Substitute 41.7 N for | F → 1 | to find F → 1 . F → 1 = { ( 41.7 N ) cos θ } i ^ + { ( 41.7 N ) sin θ } j ^ Write the vector representation of force F → 2 . F → 2 = ( | F → 2 | cos 20.0 ° ) i ^ + ( | F → 2 | sin 20.0 ° ) j ^ Substitute 41.7 N for | F → 2 | to find F → 2 . F → 2 = { ( 41.7 N ) cos 20.0 ° } i ^ + { ( 41.7 N ) sin 20.0 ° } j ^ = ( 39.2 N ) i ^ + ( 14.3 N ) j ^ Write the vector representation of force F → 3 . F → 3 = ( | F → 3 | cos 20.0 ° ) i ^ + ( − | F → 3 | sin 20.0 ° ) j ^ Substitute 41.7 N for | F → 3 | to find F → 3 . F → 3 = { ( 41.7 N ) cos 20.0 ° } i ^ + { − ( 41.7 N ) sin 20.0 ° } j ^ = ( 39.2 N ) i ^ − ( 14.3 N ) j ^ Formula to calculate the total force experienced by the knee is, F → total = F → 1 + F → 2 + F → 3 Substitute { ( 41.7 N ) cos θ } i ^ + { ( 41.7 N ) sin θ } j ^ for F → 1 , ( 39.2 N ) i ^ + ( 14.3 N ) j ^ for F → 2 and ( 39.2 N ) i ^ − ( 14 (b) To determine The magnitude of the force F → total that is applied to the femur.

5th Edition

ISBN: 9780321976444

Author: James S. Walker

Publisher: PEARSON

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Chapter 6, Problem 34PCE

(a)

To determine

The angle θ required to produce the alignment of F→total at an angle of 20.0° above the horizontal.

(b)

To determine

The magnitude of the force F→total that is applied to the femur.

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Chapter 6, Problem 33PCE

Chapter 6, Problem 35PCE

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Physics (5th Edition)

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The angle θ required to produce the alignment of F → total at an angle of 20.0 ° above the horizontal.

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The angle θ required to produce the alignment of F→total at an angle of 20.0° above the horizontal.

The magnitude of the force F→total that is applied to the femur.

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Chapter 6 Solutions