Explanation Given: Draw the free body diagram of the sheet metal bracket as in Figure (1). Let us establish a position vector from O to A to O to C . The coordinates of the point O ( 0 , 0 , 0 ) , A ( 50 , 300 , 0 ) mm and point C ( 400 , 0 , 250 ) mm . Express the position vector r in the direction of OA using Figure (1). r O A = ( x A − x O ) i + ( y A − y O ) j + ( z A − z O ) k (I) Here, the coordinates of A and O are x A , x O , y A , y O , z A , and z O respectively. Express the position vector r in the direction of OC using Figure (1). r O C = ( x C − x O ) i + ( y C − y O ) j + ( z C − z O ) k (II) Here, the coordinates of A and C are x C , x O , y C , y O , z C , and z O respectively. Express the force F A C as a Cartesian vector form and the position vector r is directed from point A to point C . F A C = F A C u A C = F A C ( r A C r A C ) (III) Here, the unit vector in the direction of AC is u A C , the position vector in the direction of AC is r A C and the magnitude of the position vector in the direction of AC is r A C . Determine the unit vector of OA . u O A = ( r O A r O A ) = ( r O A ( x A − x O ) 2 + ( y A − y O ) 2 + ( z A − z O ) 2 ) (IV) Here, the position vector in the direction of OA is r O A and the magnitude of the position vector in the direction of OA is r O A . Determine the unit vector of OC . u O C = ( r O C r O C ) = ( r O C ( x C − x O ) 2 + ( y C − y O ) 2 + ( z C − z O ) 2 ) (V) Here, the position vector in the direction of OC is r O C and the magnitude of the position vector in the direction of OC is r O C . Apply the dot product as follows: u O A ⋅ u O C = A C cos θ (VI) Conclusion: Substitute 50 mm for x A , 300 mm for y A , 0 mm for z A , 0 for x O , y O , and z O in Equation (I). r O A = ( 50 − 0 ) i + ( 300 − 0 ) j + ( 0 − 0 ) k = 50 i + 300 j + 0 k Substitute 400 mm for x C , 0 mm for y C , 250 mm for z C , 0 for x O , y O , and z O in Equation (II)

MOD.MASTER.W/ETEXT ENG.MECHANICS CARD+BK

15th Edition

ISBN: 9780137519170

Author: HIBBELER

Publisher: PEARSON

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Chapter 2, Problem 7RP

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The angle θ between the edges of the sheet metal bracket.

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Chapter 2 Solutions

MOD.MASTER.W/ETEXT ENG.MECHANICS CARD+BK

Ch. 2 - Determine the magnitude of the resultant force...Ch. 2 - Two forces act on the hook. Determine the...Ch. 2 - Determine the magnitude of the resultant force and...Ch. 2 - Resolve the 30-lb force into components along the...Ch. 2 - The force F = 450 lb acts on the frame. Resolve...Ch. 2 - If force F is to have a component along the u axis...Ch. 2 - Determine the magnitude of the resultant force FR...Ch. 2 - Resolve the force F1 into components acting along...Ch. 2 - Resolve the force F2 into components acting along...Ch. 2 - Prob. 10P

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The angle θ between the edges of the sheet metal bracket.

Solution Summary: The author explains the angle between the edges of the sheet metal bracket and the position vector r in the direction of OA.

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