Lecture_4-Dot_Product-2.ppt

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

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Force Vectors Lecture 4-Dot Product (section 2.9)
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
PROBLEMS FOR RECITATION
Background image of page 2
Recitation Problem (RC-1)-1 Determine the force component along the rod AC. Assume B is located at the midpoint of the rod.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 = - + + - + - + + - - + - = = =
Background image of page 4
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
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: the force along the pole AO. Plan : 1. Get r AO 2. = cos-1 {( F r AO )/(F r AO )} 3. F OA = F u AO or F cos 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 = cos-1 {( F r AO )/(F r AO )} = cos-1 {220/(54 . 77 7)} = 55.0 F r AO = (-20)(-3) + (50)(2) + (-10)(-6) = 220 lbft 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 = 54 . 77 cos(55 . 0) = 31 . 4 lb...
View Full Document

Page1 / 6

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

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online