Explanation Given information : The force F C is 250 N. 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 (I) Here, the position vector for point A is r A , the position vector for point B is r B , 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 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 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 CB . r C B = r B − r C = ( x B i + y B j + z B k ) − ( x C i + y C j + z C k ) = ( x B − x C ) i + ( y B − y C ) j + ( z B − z C ) k (II) Here, the position vector for point C is r C , 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 , and the distance of point C from the origin along z direction is z C Determine the modulus of the position vector for point CB . | r C B | = ( x B − x C ) 2 + ( y B − y C ) 2 + ( z B − z C ) 2 (III) Express the force F C as follows: F C = F C [ r C B | r C B | ] (IV) Determine the moment produced by the force F C about point O using the formula. M a − a = i ⋅ r A B × F C (V) Conclusion : Substitute –0.50 m for x B , –0.50 m for x A , 0 m for y B , –1.0 m for y A , 0 m for z B , and 0 m for z A in Equation (I). r A B = ( − 0.5 − ( − 0.5 ) ) i + ( 0 − ( − 1.0 ) ) j + ( 0 − 0 ) k = 0 i + 1.0 j + 0k = ( 1 .0j ) m Substitute –0.50 m for x B , –2.50 m for x C , 0 m for y B , − ( 1 + 1.5 cos 30 ° ) m for y C , –0.5 m for z B , and ( 1.5 sin 30 ° ) m for z C in Equation (II)

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

13th Edition

ISBN: 9780133027990

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 4.9, Problem 171RP

To determine

The moment of the force Fc

about the door hinged axis aa of the door.

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Chapter 4.9, Problem 170RP

Chapter 4.9, Problem 172RP

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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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The moment of the force F c about the door hinged axis aa of the door.

The moment of the force Fc about the door hinged axis aa of the door.

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Chapter 4 Solutions

The moment of the force Fc

about the door hinged axis aa of the door.