Explanation Given: The magnitude of component force, F 1 = 15 kN The magnitude of component force, F 2 = 26 kN The magnitude of component force, F 3 = 36 kN The angle between the magnitude of component force F 1 and y -axis is 40 ° . The angle between the magnitude of component force F 3 and x -axis is 30 ° . Show the magnitude of the component force F 1 into x and y -axis components as in Figure (1). Show the magnitude of the component force F 2 into x and y -axis components as in Figure (2). Show the magnitude of the component force F 3 into x and y -axis components as in Figure (3). Show the magnitude of the resultant force F R into x and y -axis components as in Figure (4). Use Figure (2) to find out cos θ and sin θ . cos θ = 12 13 sin θ = 5 13 Resolve the component vector force F 1 into x and y components. F 1 = { ( F 1 sin 40 ° i ) + ( F 1 cos 40 ° j ) } kN (I) Resolve the component vector force F 2 into x and y components. F 2 = { ( − F 2 cos θ i ) + ( F 2 sin θ j ) } kN (II) Resolve the component vector force F 3 into x and y components. F 3 = { ( F 3 cos 30 ° i ) + ( − F 3 sin 30 ° j ) } kN (III) Express the vector resultant force F R . F R = F 1 + F 2 + F 3 (IV) Calculate the magnitude of resultant force F R . F R = ( F R ) x 2 + ( F R ) y 2 (V) Calculate the orientation θ of the resultant force. tan θ = ( F R ) y ( F R ) x (VI) Conclusion: Substitute 15 kN for F 1 in Equation (I). F 1 = { ( 15 sin 40 ° i ) + ( 15 cos 40 ° j ) } kN = { [ ( 15 × 0.642 ) i ] + [ ( 15 × 0.766 ) j ] } kN = { 9.64 i + 11.5 j } kN Substitute 26 kN for F 2 , 12 13 for cos θ , and 5 13 for sin θ in Equation (II). F 2 = { [ − ( 26 ) 12 13 i ] + [ ( 26 ) 5 13 j ] } kN = { − 24 i + 10 j } kN Substitute 36 kN for F 3 in Equation (III)

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 2.4, Problem 51P

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The magnitude of the resultant force FR and its orientation θ measured counterclockwise from the positive x axis.

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Chapter 2.4, Problem 50P

Chapter 2.4, Problem 52P

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Engineering Mechanics: Statics & Dynamics (14th Edition)

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The magnitude of the resultant force F R and its orientation θ measured counterclockwise from the positive x axis.

Solution Summary: The author explains the magnitude of the component force F_1 into x and y -axis components.

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The magnitude of the resultant force FR and its orientation θ measured counterclockwise from the positive x axis.

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