Explanation Draw the free body diagram of force F A B and F A C as in Figure (1). Establish a position vector from A to B to A to C . The coordinates of the point A ( 0 , 0 , 3 ) m , B ( 2 , 4 , 0 ) m and point C ( − 3 , − 4 , 0 ) 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 (I) Here, the coordinates of A and B are x A , x B , y A , y B , z A , and z B , respectively. 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 (II) Here, the coordinates of A and C are x C , x A , y C , y A , z C , and z A , respectively. Express the force F A B as a Cartesian vector form and position vector r 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 ) = F A B ( r A B ( x B − x A ) 2 + ( y B − y A ) 2 + ( z B − z A ) 2 ) (III) Here, the unit vector in the direction of AB is u A B , position vector in the direction of AB is r A B , and magnitude of the position vector in the direction of AB is r A B . Express the force F A C as a Cartesian vector form and position vector r 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 ) = F A C ( r A C ( x C − x A ) 2 + ( y C − y A ) 2 + ( z C − z A ) 2 ) (IV) Here, the unit vector in the direction of AC is u A C , position vector in the direction of AC is r A C , and magnitude of the position vector in the direction of AC is r A C . Determine the resultant force at point A . F R = F A B + F A C (V) Determine the magnitude of resultant force at point A . F R = ( F R ) x 2 + ( F R ) y 2 + ( F R ) z 2 (VI) Here, the magnitude of the resultant force in x , y , and z coordinate axes are ( F R ) x , ( F R ) y , and ( F R ) z , respectively. Write the cosines of the coordinate direction angles. cos α = ( F R ) x F R (VII) cos β = ( F R ) y F R (VIII) cos γ = ( F R ) z F R (IX) Conclusion: Substitute 2 m for x B , 4 m for y B , 0 for z B , 0 for x A , 0 for y A , and 3 m for z A in Equation (I). r A B = ( 2 − 0 ) i + ( 4 − 0 ) j + ( 0 − 3 ) k = { 2 i + 4 j − 3 k } m Substitute 3 m for x C , 4 m for y C , 0 for z C , 0 for x A , 0 for y A , and 3 m for z A in Equation (II). r A C = ( − 3 − 0 ) i + ( − 4 − 0 ) j + ( 0 − 3 ) k = { − 3 i − 4 j − 3 k } m Substitute 2 m for x B , 4 m for y B , 0 for z B , 0 for x A , 0 for y A , and 3 m for z A , 200 N for F A B , and { 2 i + 4 j − 3 k } m for r A B in Equation (III)

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 2.8, Problem 100P

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The magnitude of the resultant force acting at point A.

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

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Chapter 2.8, Problem 99P

Chapter 2.8, Problem 101P

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

Solution Summary: The author explains the magnitude of the resultant force acting at point A and the coordinate direction angles of it.

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The magnitude of the resultant force acting at point A.

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

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Chapter 2 Solutions

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

Solution Summary: The author explains the magnitude of the resultant force acting at point A and the coordinate direction angles of it.

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

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Chapter 2 Solutions

The magnitude of the resultant force acting at point A.

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