Explanation Given: The magnitude of force at point B , F B = 630 N . The magnitude of force at point C , F C = 520 N . The magnitude of force at point D , F D = 750 N . Show the magnitude of forces F B , F C , and F D with coordinates of B , C , and D as in Figure (1). Establish a position vector from A to C . The coordinates of the tail A ( 0 , 6 , 2.5 ) m are subtracted from the coordinates of the head C ( 2 , 0 , 4 ) 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 C as a Cartesian vector form and the position vector r is directed from point A to point C . F C = F C u A C = F 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 , 6 , 2.5 ) m are subtracted from the coordinates of the head B ( − 3 , 0 , 4.5 ) 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 B as a Cartesian vector form and the position vector r is directed from point A to point B . F B = F B u A B = F 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) Establish a position vector from A to D . The coordinates of the tail A ( 0 , 6 , 2.5 ) m are subtracted from the coordinates of the head D ( 3 , 0 , − 3.5 ) m. Express the position vector r in the direction of AD using Figure 1. r A D = ( x D − x A ) i + ( y D − y A ) j + ( z D − z A ) k (VII) Here, the coordinates of A and D are x A , x D , y A , y D , z A , and z D , respectively. Express the force F D as a Cartesian vector form and the position vector r is directed from point A to point D . F D = F D u A D = F D ( r A D r A D ) (VIII) Here, the unit vector in the direction of AD is u A D , the position vector in the direction of AD is r A D , and the magnitude of the position vector in the direction of AD is r A D . Write the magnitude of the position vector in the direction of AD . r A D = ( x D − x A ) 2 + ( y D − y A ) 2 + ( z D − z A ) 2 (IX) Express the resultant force as a Cartesian vector. F R = F B + F C + F D (X) Express the magnitude of F R . F R = ( F R ) x 2 + ( F R ) y 2 + ( F R ) z 2 (XI) Write the cosines of the coordinate direction angles. cos α = ( F R ) x F R (XII) cos β = ( F R ) y F R (XIII) cos γ = ( F R ) z F R (XIV) Conclusion: Substitute 2 m for x C , 0 for x A , 0 m for y C , 6 m for y A , 4 m for z C , and 2.5 m for z A in Equation (I). r A C = { ( 2 − 0 ) i + ( 0 − 6 ) j + ( 4 − 2.5 ) k } m = { 2 i − 6 j + 1.5 k } m Substitute 2 m for ( x C − x A ) , 6 m for ( y C − y A ) , and 1.5 m for ( z C − z A ) in Equation (III). r A C = ( 2 ) 2 + ( − 6 ) 2 + ( 1.5 ) 2 = 4 + 36 + 3 = 43 m Substitute 3 m for x B , 0 for x A , 0 m for y B , 6 m for y A , 4.5 m for z B , and 2.5 m for z A in Equation (IV). r A B = ( − 3 − 0 ) i + ( 0 − 6 ) j + ( 4

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 108P

To determine

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

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Chapter 2.8, Problem 107P

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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 ) .

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

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Chapter 2 Solutions