Explanation Given: The magnitude of component vector force F 1 , F 1 = 90 N . The magnitude of component vector force F 2 , F 2 = 150 N . The magnitude of component vector force F 3 , F 3 = 200 N . The angle between F ′ and F 2 is 60 ° The angle between F ′ and y axis is 45 ° Illustrate the magnitude of components force F 1 , F 2 , and F 3 in x , y , and z directions as shown in Figure (1). Express the direction angles with magnitude of component force F 1 , F 2 , and F 3 as shown in Figure (2). Write the vector force F 1 in Cartesian vector form using Figure (1). F 1 = ( F 1 sin θ ) i + ( F 1 cos θ ) k (I) Use trigonometry to calculate the magnitude of the components from Figure (1). F z = ( F 2 sin 60 ° ) N (II) F ′ = ( F 2 cos 60 ° ) N (III) F x = ( F ′ sin 45 ° ) N (IV) F y = ( F ′ cos 45 ° ) N (V) Here, the component force in x , y , and z axes are F x , F y , and F z respectively, and the sum of forces F x and F y is F ′ . Calculate the resultant vector component force F 2 in terms of unit vectors ( i , j , and k ) that represents the direction cosines of F . F 2 = F x i + F y j + F z k (VI) Substitute 150 N for F 2 in Equation (II). F z = ( 150 sin 60 ° ) N = 150 × 0.866 = 129.9 N Substitute 150 N for F 2 in Equation (III). F ′ = ( 150 cos 60 ° ) N = 150 × 0.5 = 75 N Substitute 75 N for F ′ in Equation (IV). F x = ( 75 sin 45 ° ) N = 75 × 0 .7071 = 53.03 N Substitute 75 N for F ′ in Equation (V). F y = ( 75 cos 45 ° ) N = 75 × 0 .7071 = 53.03 N Substitute 53.03 N for F x , 53.03 N for F y , and 129.9 N for F z in Equation (VI). F 2 = [ 53.03 i + 53.03 j + 129.9 k ] (VII) Write the vector force F 3 in Cartesian vector form using Figure (1). F 3 = [ F 3 k ] N (VIII) Express the resultant force of vector force F 1 and F 2 . F R = F 1 + F 2 + F 3 (IX) Write the magnitude of the resultant force. F R = ( F R ) x 2 + ( F R ) y 2 + ( F R ) z 2 (X) Here, the magnitude of the resultant force measured along x , y , and z axis are ( F R ) x , ( F R ) y , and ( F R ) z

Engineering Mechanics: Statics & Dynamics, Student Value Edition Plus Modified Mastering Engineering Revision with Pearson eText -- Access Card Package

14th Edition

ISBN: 9780135841532

Author: HIBBELER, Russell , HIBBLER

Publisher: PEARSON

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Chapter 2.6, Problem 74P

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The magnitude (FR) and coordinate direction angles of the resultant force (α,β,γ).

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Chapter 2.6, Problem 73P

Chapter 2.6, Problem 75P

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Chapter 2 Solutions

Engineering Mechanics: Statics & Dynamics, Student Value Edition Plus Modified Mastering Engineering Revision with Pearson eText -- Access Card Package

Ch. 2.3 - In each case, construct the parallelogram law to...Ch. 2.3 - In each case, show how to resolve the force F into...Ch. 2.3 - Determine the magnitude of the resultant force...Ch. 2.3 - Two forces act on the hook. Determine the...Ch. 2.3 - Determine the magnitude of the resultant force and...Ch. 2.3 - Resolve the 30-lb force into components along the...Ch. 2.3 - The force F = 450 lb acts on the frame. Resolve...Ch. 2.3 - If force F is to have a component along the u axis...Ch. 2.3 - If = 60 and F = 450 N, determine the magnitude of...Ch. 2.3 - If the magnitude of the resultant force is to be...

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The magnitude ( F R ) and coordinate direction angles of the resultant force ( α , β , γ ) .

Solution Summary: The author illustrates the magnitude of the component vector force F_1, and the coordinate direction angles of it in x, y and z directions.

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The magnitude (FR) and coordinate direction angles of the resultant force (α,β,γ).

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