Explanation Given information : The force ( F ) is 80 lb. Consider the x ′ , y ′ , and z axes. Determine the position vector for point AC . r A C = r C − r A = ( x ′ C i + y ′ C j + z C k ) − ( x ′ A i + y ′ A j + z A k ) = ( x ′ C − x ′ A ) i + ( y ′ C − y ′ A ) j + ( z C − z A ) k (I) Here, the position vector for point C is r C , the position vector for point A is r 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 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 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 Ay . r A y = r y − r A = ( x ′ y i + y ′ y j + z ′ y k ) − ( x ′ A i + y ′ A j + z ′ A k ) = ( x ′ y − x ′ A ) i + ( y ′ y − y ′ A ) j + ( z ′ y − z ′ A ) k (II) Here, the position vector for y ′ axis is r y , the distance of axis y from the origin along x ′ direction is x ′ y , the distance of axis y from the origin along y ′ direction is y ′ y , and the distance of axis y from the origin along z direction is z y . Determine the modulus of the position vector for point Ay . | r A y | = ( x ′ y − x ′ A ) 2 + ( y ′ y − y ′ A ) 2 + ( z y − z A ) 2 (III) Unit vector: u A y = r A y | r A y | (IV) Determine the moment of force F about y axis. M y = u A y ⋅ ( r A C × F ) (V) Consider x , y , z axis. Determine the position vector for point AC . r A C = r C − r A = ( x C i + y C j + z C k ) − ( x A i + y A j + z A k ) = ( x C − x A ) i + ( y C − y A ) j + ( z C − z A ) k (VI) Here, the position vector for point C is r C , the position vector for point A is r 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 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 , and the distance of point A from the origin along z direction is z A , 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 moment of force F about the y axis. M y = j ⋅ ( r A C × F ) (VII) Conclusion : Substitute − 6 cos 15 ° for x ′ C , 0 ft for x ′ A , 3 ft for y ′ C , 0 ft for y ′ A , 6 sin 15 ° for z C , and 0 ft for z O in Equation (I). r A C = { ( − 6 cos 15 ° − 0 ) i + ( 3 − 0 ) j + ( 6 sin 15 ° − 0 ) k } = ( − 5.80 + 3 j + 1.55 k ) ft Substitute − sin 30 ° for x ′ y , 0 ft for x ′ A , cos 30 ° for y ′ y , 0 ft for y ′ A , 0 ft for z y , and 0 ft for z A in Equation (II)

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ISBN: 9781323327067

Author: HIBBELER

Publisher: PEARSON C

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The area of science and mathematics concerned the motion of the objects, the relation between the force matter, and motion if specified. Forces exerted over the objects result in displacements or changes in the position of an object concerning its enviro…

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Chapter 4.5, Problem 60P

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The moment of the force F about the y axis.

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The moment of the force F about the y axis.

Solution Summary: The author explains the moment of the force F about the y axis and the position vector for point AC.

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