Explanation Given information : The moment about hinged axis CD ( M C D ) is 500 lb ⋅ ft . Show the free body diagram of the panel as in Figure (1). Determine the position vector for point CA . r C A = r A − r C = ( x A i + y A j + z A k ) − ( x C i + y C j + z C k ) = ( x A − x C ) i + ( y A − y C ) j + ( z A − z C ) k (I) Here, the position vector for point A is r A , the position vector for point C is r C , Here, the position vector for point A is r A , the distance of point A from the origin along x direction is x A , the distance of point A from the origin along y direction is y A , the distance of point A from the origin along z direction is z A , the distance of point C from the origin along x direction is x C , the distance of point C from the origin along y direction is y C , the distance of point C from the origin along z direction is z C , the vector component along x direction is i, the vector component along y direction is j, and the vector component along z direction is k. Determine the position vector for point DB . r D B = r B − r D = ( x B i + y B j + z B k ) − ( x D i + y D j + z D k ) = ( x B − x D ) i + ( y B − y D ) j + ( z B − z D ) k (II) Here, the position vector for point B is r B , the position vector for point D is r D , the distance of point B from the origin along x direction is x B , the distance of point B from the origin along y direction is y B , the distance of point B from the origin along z direction is z B , the distance of point D from the origin along x direction is x D , the distance of point D from the origin along y direction is y D , and the distance of point D from the origin along z direction is z D . Determine the position vector for point AB . r A B = r B − r A = ( x B i + y B j + z B k ) − ( x A i + y A j + z A k ) = ( x B − x A ) i + ( y B − y A ) j + ( z B − z A ) k (III) Determine the modulus of position vector for point AB | r A B | = ( x B − x A ) 2 + ( y B − y A ) 2 + ( z B − z A ) 2 (IV) Force vector: F = F ( r A B | r A B | ) (V) Determine the position vector for point CD . r C D = r D − r C = ( x D i + y D j + z D k ) − ( x C i + y C j + z C k ) = ( x D − x C ) i + ( y D − y C ) j + ( z D − z C ) k (VI) Determine the modulus of position vector for point CD . | r C D | = ( x D − x C ) 2 + ( y D − y C ) 2 + ( z D − z C ) 2 (VII) Determine the unit vector for point CD . u C D = r C D | r C D | (VIII) Write the equation for moment of force F about CD axis. M C D = u C D ⋅ ( r C A × F ) (IX) Conclusion : Substitute 6 ft for x A , 0 ft for x C , 0 ft for y A , 0 ft for y C , 0 ft for z A , and 0 ft for z C in Equation (I). r C A = { ( 6 − 0 ) i + ( 0 − 0 ) j + ( 0 − 0 ) k } = ( 6 i + 0 j + 0 k ) ft Substitute 0 ft for x B , 0 ft for x D , 4 ft for y B , 8 ft for y D , 12 ft for z B , and 6 ft for z D in Equation (II). r D B = { ( 0 − 0 ) i + ( 4 − 8 ) j + ( 12 − 6 ) k } = ( 0 i − 4 j + 6 k ) ft Substitute 0 ft for x B , 6 ft for x A , 4 ft for y B , 0 ft for y A , 12 ft for z B , and 0 ft for z A in Equation (III)

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 4.5, Problem 59P

To determine

The magnitude of the force F in cable AB of the panel.

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Chapter 4.5, Problem 58P

Chapter 4.5, Problem 60P

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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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