Explanation Draw the free body diagram of force F B and F C from point A as in Figure (1). Establish a position vector from A to B to A to C . The coordinates of the point A ( 0 , 0 , 6 ) m , B ( 3 , − 2 , 0 ) m and point C ( 2 , 3 , 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 B as a Cartesian vector form and position vector r directed from point A to point B . F B = F B u A B = F B ( r A B r A B ) = F 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 C as a Cartesian vector form and position vector r directed from point A to point C . F C = F C u A C = F C ( r A C r A C ) = F 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 B + F C (V) Determine the magnitude of the resultant force at point A . F R = ( F R x ) 2 + ( F R y ) 2 + ( F R z ) 2 (VI) Here, the magnitude of resultant force in x , y , and z coordinate axes are F R x , F R y , and F R z , respectively. Conclusion: Substitute 3 m for x B , 2 m for y B , 0 m for z B , 0 for x A , 0 for y A , and 6 m for z A in Equation (I). r A B = ( 3 − 0 ) i + ( − 2 − 0 ) j + ( 0 − 6 ) k = { 3 i − 2 j − 6 k } m Substitute 2 m for x C , 3 m for y C , 0 m for z C , 0 for x A , 0 for y A , and 6 m for z A in Equation (II)

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 2.8, Problem 23FP

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The magnitude of the resultant force at A.

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Can you solve it analytically using laplace transforms and with Matlab code as well please. Thank You

Chapter 2 Solutions

Engineering Mechanics: Statics & Dynamics (14th Edition)

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 of the resultant force at A .

Solution Summary: The author explains how to determine the magnitude of the resultant force at A by drawing a free body diagram of force from A to B.

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The magnitude of the resultant force at A.

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