Explanation Given: The magnitude of force in the direction of AB , F A B = 600 N . The magnitude of force in the direction of AC , F A C = 450 N . Show the magnitude of forces F A B and F A C as in Figure (1). Establish a position vector from A to C . The coordinates of the tail A ( 0.5 , − 1.5 , 0 ) m are subtracted from the coordinates of the head C ( − 1.5 , 0.5 , 3.5 ) m . Express the position vector r in the direction of AC using Figure 1. r A C = ( x C − x A ) i + ( y C − y A ) j + ( z C − z A ) k (I) Here, the coordinates of A and C are x A , x C , y A , y C , z A , and z C , respectively. Express the force F A C as a Cartesian vector form, and the position vector r is directed from point A to point C . F A C = F A C u A C = F A C ( r A C r A C ) (II) Here, the unit vector in the direction of AC is u A C , the position vector in the direction of AC is r A C , and the magnitude of the position vector in the direction of AC is r A C . Write the magnitude of the position vector in the direction of AC . r A C = ( x C − x A ) 2 + ( y C − y A ) 2 + ( z C − z A ) 2 (III) Establish a position vector from A to B . The coordinates of the tail A ( 0.5 , − 1.5 , 0 ) m are subtracted from the coordinates of the head B ( − 1.5 , − 2.5 , 2 ) m . Express the position vector r in the direction of AB using Figure 1. r A B = ( x B − x A ) i + ( y B − y A ) j + ( z B − z A ) k (IV) Here, the coordinates of A and B are x A , x B , y A , y B , z A , and z B , respectively. Express the force F A B as a Cartesian vector form, and the position vector r is directed from point A to point B . F A B = F A B u A B = F A B ( r A B r A B ) (V) Here, the unit vector in the direction of AB is u A B , the position vector in the direction of AB is r A B , and the magnitude of the position vector in the direction of AB is r A B . Write the magnitude of the position vector in the direction of AB . r A B = ( x B − x A ) 2 + ( y B − y A ) 2 + ( z B − z A ) 2 (VI) Express the resultant force as a Cartesian vector. F R = F A C + F A B (VII) Express the magnitude of F R . F R = ( F R ) x 2 + ( F R ) y 2 + ( F R ) z 2 (VIII) Write the cosines of the coordinate direction angles. cos α = ( F R ) x F R (IX) cos β = ( F R ) y F R (X) cos γ = ( F R ) z F R (XI) Conclusion: Substitute –1.5 m for x C , 0.5 m for x A , 0.5 m for y C , 1.5 m for y A , 3.5 m for z C , and 0 m for z A in Equation (I). r A C = { ( − 1.5 − 0.5 ) i + ( 0.5 − ( − 1.5 ) ) j + ( 3.5 − 0 ) k } m = { − 2 i + 2 j + 3.5 k } m Substitute –1.5 m for x C , 0.5 m for x A , 0.5 m for y C , 1.5 m for y A , 3.5 m for z C , and 0 m for z A in Equation (III). r A C = { ( − 1

Engineering Mechanics: Statics and Study Pack (13th Edition)

13th Edition

ISBN: 9780133027990

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 2.8, Problem 91P

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

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Chapter 2.8, Problem 90P

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Chapter 2 Solutions

Engineering Mechanics: Statics and Study Pack (13th Edition)

Ch. 2.3 - Determine the magnitude of the resultant force...Ch. 2.3 - Determine the magnitude of the resultant force....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 - Resolve this force into components acting along...Ch. 2.3 - along the v axis. Prob. F2-6 Ch. 2.3 - Determine the magnitude of the resultant force FR...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...Ch. 2.3 - Determine the magnitude of the resultant force FR...

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

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

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Chapter 2 Solutions