This preview shows pages 1–5. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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. PREREQUISITE 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 = (x0)i + (y0)j + (z0)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: = cos1(8/17) =118o, = cos1(9/17) =122o, & = cos1(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 ...
View
Full
Document
 Spring '08
 Buch

Click to edit the document details