Lecture_4-Dot_Product-2.ppt

# Lecture_4-Dot_Product-2.ppt - the force along the pole AO...

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Force Vectors Lecture 4-Dot Product (section 2.9)

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PROBLEMS FOR RECITATION
Recitation Problem (RC-1)-1 Determine the force component along the rod AC. Assume B is located at the midpoint of the rod.

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Solution RC-1 Coordinates of points A, B, C and  D: ( 29 ( 29 ( 29 ( 29 0 , 6 , 4 : 0 , 4 , 3 : 2 , 2 , 5 . 1 : 4 , 0 , 0 : D C B A - - ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 k j i k j i k j i k j i - + = - + - + + = - + - = - + - + - - = 2 4 5 . 5 2 0 2 6 5 . 1 4 r 4 4 3 4 0 0 4 0 3 r BD AC Position vectors AC and BD: Force component along the rod: ( 29 ( 29 ( 29 N k j i k j i r r F F AC BD 14 . 99 2 4 5 . 5 2 4 5 . 5 4 4 3 4 4 3 600 r r u F 2 2 2 2 2 2 AC BD = - + + - + - + + - - + - = = =
Recitation Problem (RC-2)-2 Given: The force acting on the pole. Find: The angle between the force vector and the pole, and the magnitude of the projection of

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Unformatted text preview: the force along the pole AO. Plan : 1. Get r AO 2. θ = cos-1 {( F • r AO )/(F r AO )} 3. F OA = F • u AO or F cos θ r AO = {-3 i + 2 j – 6 k } ft. r AO = (3 2 + 2 2 + 6 2 ) 1/2 = 7 ft. F = {-20 i + 50 j – 10 k }lb F = (20 2 + 50 2 + 10 2 ) 1/2 = 54 . 77 lb Solution RC-2 θ = cos-1 {( F • r AO )/(F r AO )} θ = cos-1 {220/(54 . 77 × 7)} = 55.0° F • r AO = (-20)(-3) + (50)(2) + (-10)(-6) = 220 lb·ft u AO = r AO /r AO = {(-3/7) i + (2/7) j – (6/7) k } F AO = F • u AO = (-20)(-3/7) + (50)(2/7) + (-10)(-6/7) = 31 . 4 lb Or F AO = F cos θ = 54 . 77 cos(55 . 0°) = 31 . 4 lb...
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## This note was uploaded on 06/14/2009 for the course CE 221 taught by Professor Buch during the Spring '08 term at Michigan State University.

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Lecture_4-Dot_Product-2.ppt - the force along the pole AO...

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