E_-_Position___Force_Vectors_Along_Line

# E_-_Position___Force_Vectors_Along_Line - FORCE VECTORS...

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Unformatted text preview: FORCE VECTORS Position and Force Vectors Directed Along a Line LEARNING OBJECTIVES Be able to represent a position vector in Cartesian coordinate form, from given geometry. Be able to represent a force vector directed along a line. PRE-REQUISITE KNOWLEDGE Units of measurements. Trigonometry concepts. Vector concepts. Directional vector concepts. POSITION VECTOR The position vector (labeled r) is defined as a vector which locates a point in space relative to another point. If r extends from the origin O (0, 0, 0) to point P (x, y, z), then the position vector rP can be expressed as: Z rP = (x-0)i + (y-0)j + (z-0)k r P (x, y, z) zk O xi X Y yj POSITION VECTOR - continued If r extends from point A (xA, yA, zA) to point B (xB, yB, zB), it is designated rAB and is expressed as: rAB = i(xB xA) + j(yB yA) + k(zB zA) POSITION VECTOR - EXAMPLES 1. Position vector extends from point A (3, 0, 9) to point B (1, 4, 9) can be expressed as: r = - 2i + 4j + 0k FORCE VECTOR DIRECTED ALONG A LINE The direction of a force F is often specified by two points through which its line of action passes. To determine the properties (components and directional angles) of F, one can use both the position and the unit vectors as follows: r F= F u = F r EXAMPLE 1 Express the 340 pound force (F) in the cord as a cartesian vector Solution 1 The coordinates of points A and B are The position vector from point A to point B can then be expressed as: r = ( xB - x A ) i + ( y B - y A ) j + ( z B - z A ) k A : ( 8,9,0 ) B : ( 0,0,12 ) = ( 0 - 8) i + ( 0 - 9 ) j + (12 - 0 ) k r = -8i - 9 j + 12k r = [(-8)2 + (-9)2 + 122]0.5 = 17 ft The Cartesian force vector F = Fu = F(r/r) is given as r - 8i - 9 j + 12k F = F = 340 = -160i - 180 j + 240k lb 17 r The directional angles of F are: = cos-1(-8/17) =118o, = cos-1(-9/17) =122o, & = cos-1(12/17)=45o EXAMPLE 2 Each of the four forces has a magnitude of 28 kN. Determine the weight of the vehicle. SOLUTION 2 The coordinates for points A, B, C, D and E are: A = ( 6,-4,0 ) ; B = ( 6,4,0 ) ; C = ( - 6,4,0 ) D = ( - 6,-4,0 ); E = ( 0,0,12 ) The position vectors of the forces are rEA = 6 i - 4 j - 12 k rEB = 6 i + 4 j - 12 k rEC = - 6 i + 4 j - 12 k rED = - 6 i - 4 j - 12 k The magnitudes of the position vectors of the forces rEA = rEB = rEC = rED = 6 + 4 + 12 = 14 m 2 2 2 SOLUTION 2 The resultant of the forces F is given by: F = FEA + FEB + FEC + FED 6i - 4 j - 12k 6i + 4 j - 12k F = 28 + 28 14 14 - 6i + 4 j - 12k - 6i - 4 j - 12k + 28 + 28 14 14 - 48k F = 28 ; F = -96k kN 14 Vehicle weight is 96 kN ...
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E_-_Position___Force_Vectors_Along_Line - FORCE VECTORS...

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