Explanation Given information : The forces acting on the pipe F 1 , F 2 , and F 3 are 300 N, 200 N, and 100 N. The anticlockwise moment acting about point A ( M A ) is 100 N ⋅ m . The clockwise moment acting about point B ( M B ) is 180 N ⋅ m . Determine the position vector for point A . r A = ( x A i + y A j + z A k ) (I) Here, the position vector for point A is r A , 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 , the distance of point A from the origin along the z direction is z A , the vector component along the x direction is i , vector component along the y direction is j, and vector component along the z direction is k. Determine the position vector for point B . r B = ( x B i + y B j + z B k ) (II) Here, the position vector for point B is r A , 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 . Force vector at point A : F 1 = F 1 k (III) Force vector at point B : F 2 = F 2 ( cos θ i − sin θ k ) (IV) Here, the angle for the force F 2 is θ . Force vector at point C : F 3 = F 3 j (V) Write the moment in the Cartesian vector form at point A: M 1 = M A k (VI) Write the moment in the Cartesian vector form at point B : M 2 = M B ( cos θ i − sin θ k ) (VII) Determine the equivalent force at point O . F R = F 1 + F 2 + F 3 (VIII) Determine the couple moment at point O . ( M R ) O = r A × F 1 + r B × F 2 + M 1 + M 2 (IX) Conclusion : Substitute 0 m for x A , 0.5 m for y A , and 0 m for z A in Equation (I). r A = { 0 i + 0.5 j + 0 k } m Substitute 0 m for x B , 1.1 m for y B , and 0 m for z B in Equation (II). r B = { 0 i + 0.5 j + 0 k } m Substitute 300 N for F 1 in Equation (III). F 1 = { 300 k } N Substitute 200 N for F 2 and 45 ° for θ in Equation (IV). F 2 = 200 ( cos 45 ° i − sin 45 ° k ) = { 141.42 i − 141.42 k } N Substitute 100 N for F 3 in Equation (V). F 3 = { 100 j } N Substitute 100 N ⋅ m for M A in Equation (VI). M 1 = { 100 k } N ⋅ m Substitute 180 N ⋅ m for M B and 45 ° for θ in Equation (VII). M 2 = 180 ( cos 45 ° i − sin 45 ° k ) = { 127.28 i − 127.28 k } N ⋅ m Substitute { 300 k } N for F 1 , { 141.42 i − 141.42 k } N for F 2 , and { 100 j } N for F 3 in Equation (VIII). F R = { 300 k } + { 141.42 i − 141.42 k } + { 100 j } = 141

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 4.8, Problem 134P

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The equivalent resultant force and the couple moment acting at point O.

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Chapter 4.8, Problem 133P

Chapter 4.8, Problem 135P

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Engineering Mechanics: Statics & Dynamics (14th Edition)

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The equivalent resultant force and the couple moment acting at point O .

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