Explanation Given information : The force vector F 1 is { 600 j } lb . The force vector F 2 is { − 450 j } lb . The force vector F 3 is { 300 k } lb . The anti-clockwise moment M is 600 lb ⋅ ft . Show the free body diagram of the forces acting on the rectangular block by a wrench as in Figure (1). Show the resultant force as in Figure (2). Write the resultant force vector. F R = F 1 + F 2 + F 3 (I) Determine the resultant force F R . F R = ( F 1 ) 2 + ( F 2 ) 2 + ( F 3 ) 2 (II) Determine the unit vector for the force. u F R = F R F R (III) Determine the equivalent couple moment on the wrench. M W = M W u F R (IV) Determine the position vector for point PA . r P A = r A − r P = ( x A i + y A j + z A k ) − ( x P i + y P j + z P k ) = ( x A − x P ) i + ( y A − y P ) j + ( z A − z P ) k (V) Here, the position vector for point A is r A , the position vector for point P is r P , the distance of point A from the origin along the x -direction is x A , the distance of point A from the origin along the y -direction is y A , and the distance of point A from the origin along the z -direction is z A , the distance of point P from the origin along the x -direction is x P , the distance of point P from the origin along the y -direction is y P , the distance of point P from the origin along the z -direction is z P , the vector component along the x -direction is i, the vector component along the y -direction is j, and the vector component along the z -direction is k. Determine the position vector for point PB . r P B = r B − r P = ( x B i + y B j + z B k ) − ( x P i + y P j + z P k ) = ( x B − x P ) i + ( y B − y P ) j + ( z B − z P ) k (VI) Here, the position vector for point B is r B , the distance of point B from the origin along the x -direction is x B , the distance of point B from the origin along the y -direction is y B , and the distance of point B from the origin along the z -direction is z B . Determine the position vector for point PC . r P C = r C − r P = ( x C i + y C j + z C k ) − ( x P i + y P j + z P k ) = ( x C − x P ) i + ( y C − y P ) j + ( z C − z P ) k (VII) Here, the position vector for point C is r C , the distance of point C from the origin along the x -direction is x C , the distance of point C from the origin along the y -direction is y C , and the distance of point C from the origin along the z -direction is z C . Determine the value of ( M P ) A . ( M P ) A = r P A × F 1 (VIII) Determine the value of ( M P ) B . ( M P ) B = r P B × F 3 (IX) Determine the value of ( M P ) C . ( M P ) C = r P C × F 2 (X) Determine the couple moment acting on the rectangular block. M W = ( M P ) A + ( M P ) B + ( M P ) C + M (XI) Conclusion : Substitute { 600 j } lb for F 1 , { − 450 i } lb for F 2 , and { 300 k } lb for F 3 in Equation (I). F R = { 600 j − 450 i + 300 k } = { − 450 i + 600 j + 300 k } lb Substitute { 600 j } lb for F 1 , { − 450 i } lb for F 2 , and { 300 k } lb for F 3 in Equation (II). F R = ( 600 ) 2 + ( − 450 ) 2 + ( 300 ) 2 = 808 lb Thus, the magnitude of resultant force F R is 808 lb _ . Substitute { − 450 i + 600 j + 300 k } lb for F R and 808 lb for F R in Equation (III). u F R = { − 450 i + 600 j + 300 k } 808 = − 0.556 i + 0.743 j + 0.371 k The value of M W is collinear with F R . Substitute { − 0.556 i + 0.743 j + 0.371 k } for u F R in Equation (IV). M W = M W { − 0.556 i + 0.743 j + 0.371 k } = { − 0.556 M W i + 0.743 M W j + 0.371 M W k } lb ⋅ ft Substitute 0 ft for x A , x for x P , 4 ft for y A , y for y P , 2 ft for z A , and 0 ft for z P in Equation (V). r P A = ( 0 − x ) i + ( 4 − y ) j + ( 2 − 0 ) k = { − x i + ( 4 − y ) j + 2 k } ft Substitute 3 ft for x B , x for x P , 4 ft for y B , y for y P , 0 ft for z B , and 0 ft for z P in Equation (VI). r P B = ( 3 − x ) i + ( 4 − y ) j + ( 0 − 0 ) k = { ( 3 − x ) i + ( 4 − y ) j + 0 k } ft Substitute 3 ft for x C , x for x P , 4 ft for y C , y for y P , 2 ft for z C , and 0 ft for z P in Equation (VII)

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

13th Edition

ISBN: 9780133027990

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Question

Chapter 4.8, Problem 136P

To determine

The magnitude of the resultant force, the couple moment for the wrench, and the line of action of the wrench intersecting with the x–y plane.

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Chapter 4.8, Problem 135P

Chapter 4.8, Problem 137P

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Chapter 4 Solutions

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

Ch. 4.4 - F41. Determine the moment of the force about point...Ch. 4.4 - F42. Determine the moment of the force about point...Ch. 4.4 - F43. Determine the moment of the force about point...Ch. 4.4 - Neglect the thickness of the member.Ch. 4.4 - F45. Determine the moment of the force about point...Ch. 4.4 - F46. Determine the moment of the force about point...Ch. 4.4 - F47. Determine the resultant moment produced by...Ch. 4.4 - F48. Determine the resultant moment produced by...Ch. 4.4 - F49. Determine the resultant moment produced by...Ch. 4.4 - Express the result as a Cartesian vector.

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