Explanation Given: The magnitude of force F is 300 N . Explanation: Show the free body diagram of the projected component as in Figure (1). Calculations: Write the x -component of point A from Figure 1. A x = − [ 300 mm + 300 mm ( sin30 ° ) ] = − [ ( 300 mm ) 1 m 1 , 000 mm + { ( 300 mm ) 1 m 1 , 000 mm } ( sin30 ° ) ] = − [ 0 .3 m + 0 .3 m ( 0.5 ) ] = − 0.45 m Write the y -component of point A from Figure 1. A y = 300 mm = ( 300 mm ) 1 m 1000 mm = 0 .3 m Write the z -component of point A from Figure 1. A z = 300 mm ( cos30 ° ) = [ ( 300 mm ) 1 m 1000 mm ] ( cos30 ° ) = 0 .3 m ( 0.8660 ) = 0.2598 m Establish a position vector from O to A. Write the coordinates of the points as: O ( 0 , 0 , 0 ) mm A ( − 0.45 , 0.3 , 0.2598 ) m Calculate the vector force F from Figure 1. F = ( − F 30 ° sin 30 ° ) i + ( F cos 30 ° ) j + ( F i n 30 ° cos 30 ° ) k (I) Express the unit vector u in the direction of OA . u O A = r O A r O A = ( x A − x O ) i + ( y A − y O ) j + ( z A − z O ) k ( x A − x O ) 2 + ( y A − y O ) 2 + ( z A − z O ) 2 (II) Here, the position vector r in the direction of OA is r O A , the length of the cable O A is r O A , and the coordinates of O and A in the x , y and z directions are x O , x A , y O , y A , z O , and z A , respectively

13th Edition

ISBN: 9780132915540

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 2.9, Problem 139P

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The magnitude of the projected component of force that is acting along the line OA.

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Chapter 2.9, Problem 138P

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The magnitude of the projected component of force that is acting along the line O A .

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