Explanation Given: The magnitude of force in the direction of AB , F A B = 150 N . The magnitude of force in the direction of AC , F A C = 200 N . The angle between x axis and point B is 60 ° . Show the coordinate direction angles with magnitude of the resultant force as in Figure (1). Establish a position vector from A to C . The coordinates of the tail A ( 0 , − 1.5 , 4 ) m are subtracted from the coordinates of the head C ( 3 , − 2 , 0 ) 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 , − 1.5 , 4 ) m are subtracted from the coordinates of the head B ( 3 cos 60 ° , 3 sin 60 ° , 0 ) . 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 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 3 m for x C , 0 for x A , –2 m for y C , –1.5 m for y A , 0 for z C , and 4 m for z A in Equation (I). r A C = { ( 3 − 0 ) i + ( − 2 − ( − 1.5 ) ) j + ( 0 − 4 ) k } m = { 3 i − 0.5 j − 4 k } m Substitute 3 m for ( x C − x A ) , 0.5 m for ( y C − y A ) , and –4 m for ( z C − z A ) in Equation (III). r A C = ( 3 ) 2 + ( − 0.5 ) 2 + ( − 4 ) 2 = 9 + 0.25 + 16 = 5

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ISBN: 9780132915540

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 2.8, Problem 99P

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

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Chapter 2.8, Problem 98P

Chapter 2.8, Problem 100P

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Chapter 2 Solutions

Engineering Mechanics: Statics

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

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

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Chapter 2 Solutions