Explanation Draw the free body diagram of component of force F A acting parallel and perpendicular to the axis of the pole as in Figure (1). Draw the force F y and F x as in Figure (2). Establish a position vector from O to A . The coordinates of the point A ( − 4 , 4 , 2 ) ft and O ( 0 , 0 , 0 ) ft . Express the position vector r in the direction of O A using Figure (1). r O A = ( x A − x O ) i + ( y A − y O ) j + ( z A − z O ) k (I) Here, the unit vectors are i , j , and k , the coordinates of A and O are x A , x O , y A , y O , z A , and z O , respectively. Express the magnitude of the position vector r in the direction of O A . r O A = ( x A − x O ) 2 + ( y A − y O ) 2 + ( z A − z O ) 2 (II) Determine the unit vector in the direction of OA . u O A = ( r O A r O A ) (III) Use Figure (1) to find the value of F ′ and F z . F ′ = F cos 60 ° (IV) Determine the component of force in z coordinate axes. F z = F sin 60 ° (V) Use Figure (2) to find the value of F y and F x . F y = F ′ cos 30 ° (VI) F x = − F ′ sin 30 ° (VII) Determine the force vector in Cartesian form. F = F x i + F y j + F z k (VIII) Determine the force parallel to the pole OA . ( F A ) par = F ⋅ u O A (IX) Determine the force perpendicular to the pole OA using Pythagoras theorem. ( F A ) per 2 + ( F A ) par 2 = F 2 ( F A ) per = F 2 − ( F A ) par 2 (X) Conclusion: Substitute 0 ft for x O , 0 ft for y O , 0 ft for z O , 4 ft for x A , 4 ft for y A , and 2 ft for z A in Equation (I). r O A = ( − 4 − 0 ) i + ( 4 − 0 ) j + ( 2 − 0 ) k = { − 4 i + 4 j + 2 k } ft Substitute 0 ft for x O , 0 ft for y O , 0 ft for z O , 4 ft for x A , 4 ft for y A , and 2 ft for z A in Equation (II). r O A = ( − 4 − 0 ) 2 + ( 4 − 0 ) 2 + ( 2 − 0 ) 2 = 16 + 16 + 4 = 6 ft Substitute { − 4 i + 4 j + 2 k } ft for r O A and 6 ft for r O A in Equation (III)

Engineering Mechanics: Statics & Dynamics, Student Value Edition Plus Modified Mastering Engineering Revision with Pearson eText -- Access Card Package

14th Edition

ISBN: 9780135841532

Author: HIBBELER, Russell , HIBBLER

Publisher: PEARSON

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Chapter 2.9, Problem 30FP

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The component of force acting parallel (FA)par and perpendicular (FA)per to the axis of the pole.

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Chapter 2.9, Problem 29FP

Chapter 2.9, Problem 31FP

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Engineering Mechanics: Statics & Dynamics, Student Value Edition Plus Modified Mastering Engineering Revision with Pearson eText -- Access Card Package

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The component of force acting parallel ( F A ) par and perpendicular ( F A ) per to the axis of the pole.

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The component of force acting parallel (FA)par and perpendicular (FA)per to the axis of the pole.

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