Explanation Given: The magnitude of force, F 1 = 500 lb . The magnitude of force, F 2 = 800 lb . The angle between F ′ and the x axis is 30 ° The angle between F 2 and F ′ is 45 ° Show the magnitude of component forces F 1 and F 2 into x , y , and z direction as in Figure (1). Express the force F 1 as a Cartesian vector using Figure (1). F 1 = F 1 ( u F ) = F 1 x i + F 1 y j + F 1 z k (I) Here, the unit vectors are i , j , and k , the unit vector in the direction of F 1 is u F , the component force F 1 in x , y , and z directions are F 1 x , F 1 y , and F 1 z respectively. Express the force component F 1 x using Figure (1). F 1 x = 0 Express the force component F 1 y using Figure (1). F 1 y = F 1 cos θ (II) Express the force component F 1 z using Figure (1). F 1 z = F 1 sin θ (III) Use Figure (1) to calculate the cos θ and sin θ . cos θ = 4 5 sin θ = 3 5 Express the force F 2 as a Cartesian vector using Figure (1). F 2 = F 2 ( u F ) = F 2 x i + F 2 y j + F 2 z ( − k ) = F 2 x i + F 2 y j − F 2 z k (IV) Here, the unit vectors are i , j , and k , the unit vector in the direction of F 1 is u F , the component force F 2 in x , y , and z directions are F 2 x , F 2 y , and F 2 z respectively. Express the force component F 2 z using Figure (1). F 2 z = F 2 sin 45 ° (V) Express the sum of force component F ′ using Figure (1). F ′ = F 2 cos 45 ° (VI) Express the force component F 2 x using Figure (1). F 2 x = F ′ cos 30 ° (VII) Express the force component F 2 y using Figure (1). F 2 y = F ′ sin 30 ° (VIII) Express the resultant force acting on the hook. F R = F 1 + F 2 (IX) Conclusion: Substitute 500 lb for F 1 and 4 / 5 for cos θ in Equation (II)

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 2.6, Problem 18FP

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The resultant force as a Cartesian vector acting on the hook (FR).

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(b) A steel 'hot rolled structural hollow section' column of length 5.75 m, has the cross-section shown in Figure Q.5(b) and supports a load of 750 kN. During service, it is subjected to axial compression loading where one end of the column is effectively restrained in position and direction (fixed) and the other is effectively held in position but not in direction (pinned). i) Given that the steel has a design strength of 275 MN/m², determine the load factor for the structural member based upon the BS5950 design approach using Datasheet Q.5(b). [11] ii) Determine the axial load that can be supported by the column using the Rankine-Gordon formula, given that the yield strength of the material is 280 MN/m² and the constant *a* is 1/30000. [6] 300 600 2-300 mm wide x 5 mm thick plates. Figure Q.5(b) L=5.75m Pinned Fixed

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 resultant force as a Cartesian vector acting on the hook ( F R ) .

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