Vectors and Forces

8/13/2019 Vectors and Forces

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ALL YOU NEED TO KNOW ABOUTVECTORS AND FORCES

A vector is a mathematical entity used to describe a physical quantity

having both magnitude and direction. You have probably already takencourses involving vector addition and multiplication etc. Here we summarizethe important features of vectors that you will need for this course and alsotry to organize them so that you dont have so many things to remember.

Elementary treatments of vectors are often based on trigonometric meth-ods, such as triangles and polygons of forces. These can be useful for visual-ization in two dimensions, but three-dimensional trigonometry is conceptu-ally difficult and almost impossible to represent on the page with intelligablegures. Also trigonometric methods require you to be familiar with the var-ious trigonometric identities, such as formul for compound angles, the sineand cosine rules, etc. We can avoid all this by performing all the necessaryoperations using algebraic operations rather than trigonometry.

Cartesian vectors

For this purpose, the rst step in almost any vector problem will be to writethe vector in terms of its Cartesian components. For example, the vector F in Cartesian coordinates x, y,z can be written

F = {F x , F y , F z } . (1)

This notation is easy to write and has the added advantage that it is similar

to the notation used for vectors in linear algebra classes.The sum of two vectors is obtained by summing the respective Cartesian

components. For example

F + G = {F x , F y , F z } + {G x , G y , G z }= {(F x + Gx ), (F y + Gy), (F z + Gz )} . (2)

The magnitude of a vector |F | is dened by

|F | = F 2x + F 2y + F 2z . (3)

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Unit vectors

A unit vector is a vector with unit magnitude. Here we shall denote it by thesymbol

e = {ex , ey , ez } . (4)

It follows from the denition that|e |2 = e2x + e

2

y + e2

z = 1 . (5)

However, we use the special notation i , j , k for unit vectors aligned with thex , y, or z coordinate axes respectively. Notice that with this notation

i = {1, 0, 0} ; j = {0, 1, 0} ; k = {0, 0, 1} . (6)

A general vector F can be dened in terms of magnitude and direction by the product

F = e |F | , (7)

where e is a unit vector in the same direction as F . From this equation, wesee that

e = F |F |

= F x|F |

, F y|F |

, F z|F |

. (8)

Another way to represent the vector F is

F = i F x + j F y + k F z . (9)

Position vectors

A point in space P with Cartesian coordinates ( x,y,z ) can be dened by a

position vector r P = {x,y,z } . (10)

This is a vector with dimensions of length representing a line drawn fromthe origin O to P . If another point Q has coordinates ( x , y , z ), we canconstruct the relative position vector r P Q representing the line drawn fromP to Q, as shown in Figure 1. From this gure, it is clear that

r Q = r P + r P Q (11)

and hence

r P Q = r Q r P = {x , y , z } { x,y,z } = {(x x), (y y), (z z )} . (12)

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P

Q

O x

y

z

r

r

rP

PQ

Q

Figure 1

Dot product

The dot product of two vectors F , G is dened as

F G = F x G x + F y G y + F z G z . (13)

This denes a scalar quantity and hence is also known as the scalar product .An alternative form is

F G = |F || G | cos() , (14)

where is the angle between the two vectors.A special case of this equation is the result

e i = cos(x ) , (15)

where x is the angle between the unit vector e and the x-axis. Also, since

e i = {ex , ey , ez } {1, 0, 0} = ex , (16)

using (13,4,6), we haveex = cos(x ) . (17)

Similar results can be established for the y and z -axes, so

e = {cos(x ), cos(y ), cos(z )} . (18)

For this reason, the components ex , ey , ez of the unit vector are sometimesreferred to as direction cosines . Notice that from (5),

cos2 (x ) + cos2 (y ) + cos

2 (z ) = 1 , (19)

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so only two of the three direction cosines can be chosen independently.Equations (13, 14) are useful for nding the angle between two intersect-

ing lines. For example, if the two lines are represented as (known) relativeposition vectors r , s , we have

r s = |r || s | cos() = r x s x + r ys y + r z s z . (20)The magnitudes of the two vectors are easily found using (3) after whichequation (20) can be solved for .

Another important use of the dot product is to nd the component of avector F in a specied direction. If the direction is specied by a known unitvector e , then the component of F along e is simply

F e . (21)

Examples

1. A straight bar AB extends from the point A(3, 7, 10) to B ( 2, 5, 3),where the coordinates are in meters. Find the coordinates of thepoint C on the bar if C is 2 meters from A.

Solution

The coordinates of C can be combined in a position vector r C . We can alsowrite

r C = r A + r AC ,

where

r A = {3, 7, 10} .We know that the magnitude of r AC is 2 meters and that its direction is

the same as that of r AB . In other words, eAC = e AB . For the latter, we have

r AB = r B r A = { 2, 5, 3} { 3, 7, 10} = { 5, 2, 13} ,

so the magnitude

|r AB | = ( 5)2 + ( 3)2 + 13 2 = 14 .25 mand the unit vector in direction AB is

e AB = r AB|r AB | = { 5, 2, 13}14.25 = { 0.351, 0.140, 0.912} = eAC .

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Thus

r AC = 2e AC = 2{ 0.351, 0.140, 0.912} = { 0.702, 0.281, 1.825} m

and

r C = r A + r AC = {3, 7, 10} + { 0.702, 0.281, 1.825}= {2.298, 6.719, 8.175} m.

2. A rope is attached to the wall at the point P (3, 10, 0), wherecoordinates are in meters and the y-direction is vertically upwards.A man standing at the point Q(5, 0, 8) holds the other end of therope. If he pulls on the rope with a force of 100 N, nd a vectordescription of the force F applied by the rope to the wall. Also,if the man weighs 600 N, nd the component of his weight in thedirection P Q .

Solution

Here, we know the magnitude of F . All we have to do is to nd its direction.As in the rst example, we can nd a position vector describing the wholelength P Q of the taut rope and hence nd a unit vector in the same direction.We have

r P Q = r Q r P = {5, 0, 8} { 3, 10, 0} = {2, 10, 8} .

The magnitude of this vector is

|r P Q | = 22 + ( 10)2 + 8 2 = 12 .96 mand hence the unit vector

e P Q = r P Q|r P Q |

= {2, 10, 8}

12.96 = {0.154, 0.772, 0.617} .

ThusF = 100e P Q = {15.4, 77.2, 61.7} N.

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The mans weight is dened by the vector

W = {0, 600, 0} N,

so its component in the direction P Q is obtained from the dot product

W e P Q = {0, 600, 0}{0.154, 0.772, 0.617} = ( 600) ( 0.772) = 463.2 N.

The other terms in the dot product are zero.

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