CHAPTER 2 – PART 3: POSITION & FORCE VECTORS

Today’s Objectives:

Students will be able to :

a) Represent a position vector in Cartesian coordinate form, from given geometry.

b) Represent a force vector directed along a line.

Class contents:

• Daily quiz

• Applications

• Position vectors

• Force vectors

• Examples

• Group problem solving

Wing strut

APPLICATIONS

How can we represent the force along the wing strut in a 3-D Cartesian vector form?

POSITION VECTOR

A position vector is defined as a fixed vector that locates a point in space relative to another point.

Consider two points, A & B, in 3-D space. Let their coordinates be (XA, YA, ZA) and ( XB, YB, ZB ), respectively.

The position vector directed from A to B, r AB , is defined as

r AB = {( XB – XA ) i + ( YB – YA ) j + ( ZB – ZA ) k }m

Please note that B is the ending point and A is the starting point. So ALWAYS subtract the “tail” coordinates from the “tip” coordinates!

FORCE VECTOR DIRECTED ALONG A LINE (Section 2.8)

If a force is directed along a line, then we can represent the force vector in Cartesian Coordinates by using a unit vector and the force magnitude. So we need to:

a) Find the position vector, r AB , along two points on that line.

b) Find the unit vector describing the line’s direction, uAB = (rAB/rAB).

c) Multiply the unit vector by the magnitude of the force, F = F uAB .

Example 2.12

An elastic rubber band is attached to points A and B. Determine its length and its direction measured from A towards B.

SolutionPosition vector

r = [-2m – 1m]i + [2m – 0]j + [3m – (-3m)]k = {-3i + 2j + 6k}m

Magnitude = length of the rubber band

Unit vector in the director of ru = r /r = -3/7i + 2/7j + 6/7k

mr 7623 222

Example 2.12

Solutionα = cos-1(-3/7) = 115°

β = cos-1(2/7) = 73.4°

γ = cos-1(6/7) = 31.0°

Example 2.12

The man pulls on the cord with a force of 350N. Represent this force acting on the support A, as a Cartesian vector and determine its direction.

Example 2.13

SolutionEnd points of the cord are A (0m, 0m, 7.5m) and B (3m, -2m, 1.5m)r = (3m – 0m)i + (-2m – 0m)j + (1.5m – 7.5m)k

= {3i – 2j – 6k}mMagnitude = length of cord AB

Unit vector, u = r /r = 3/7i - 2/7j - 6/7k

mmmmr 7623 222

Example 2.13

SolutionForce F has a magnitude of 350N, direction specified by u

F = Fu = 350N(3/7i - 2/7j - 6/7k) = {150i - 100j - 300k} N

α = cos-1(3/7) = 64.6° β = cos-1(-2/7) = 107° γ = cos-1(-6/7) = 149°

The roof is supported by cables. If the cables exert FAB = 100N and FAC = 120N

on the wall hook at A, determine the magnitude of the resultant force acting at A.

GROUP PROBLEM SOLVING

GROUP PROBLEM SOLVING (continued)

GROUP PROBLEM SOLVING (continued)

GROUP PROBLEM SOLVING (continued)