Copyright © John O’Connor St. Farnan’s PPS Prosperous For non-commercial purposes only….....
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Transcript of Copyright © John O’Connor St. Farnan’s PPS Prosperous For non-commercial purposes only….....
Copyright © John O’Connor
St. Farnan’s PPS
Prosperous
For non-commercial purposes only….. Enjoy!
Vectors and Scalars
Comments/suggestions please to the SLSS physics website forum
@ http://physics.slss.ie/forum
A scalar quantity is a quantity that has magnitude only and has no direction in space
Examples of Scalar Quantities:
Length Area Volume Time Mass
A vector quantity is a quantity that has both magnitude and a direction in space
Examples of Vector Quantities:
Displacement Velocity Acceleration Force
Vector diagrams are shown using an arrow
The length of the arrow represents its magnitude
The direction of the arrow shows its direction
Vectors in opposite directions:
6 m s-1 10 m s-1 = 4 m s-1
6 N 10 N = 4 N
Vectors in the same direction:
6 N 4 N = 10 N
6 m= 10 m
4 m
The resultant is the sum or the combined effect of two vector quantities
When two vectors are joined tail to tail
Complete the parallelogram The resultant is found by
drawing the diagonal
When two vectors are joined head to tail
Draw the resultant vector by completing the triangle
Solution: Complete the parallelogram (rectangle)
θ
The diagonal of the parallelogram ac represents the resultant force
2004 HL Section B Q5 (a)Two forces are applied to a body, as shown. What is the magnitude
and direction of the resultant force acting on the body?
5 N
12 N
5
12
a
b c
d
The magnitude of the resultant is found using Pythagoras’ Theorem on the triangle abc
N 13
512 Magnitude 22
ac
ac
675
12tan
5
12tan: ofDirection
1
ac
Resultant displacement is 13 N 67º with the 5 N force
13 N
45º5 N
90ºθ
Find the magnitude (correct to two decimal places) and direction of theresultant of the three forces shown below.
5 N
5
5
Solution: Find the resultant of the two 5 N forces first (do right angles first)
a b
cd
N 07.75055 22 ac
4515
5tan
7.07
N
10 N
135º
Now find the resultant of the 10 N and 7.07 N forces
The 2 forces are in a straight line (45º + 135º = 180º) and in opposite directions
So, Resultant = 10 N – 7.07 N = 2.93 N in the direction of the 10 N force
2.93
N
What is a scalar quantity? Give 2 examples What is a vector quantity? Give 2 examples How are vectors represented? What is the resultant of 2 vector quantities? What is the triangle law? What is the parallelogram law?
When resolving a vector into components we are doing the opposite to finding the resultant
We usually resolve a vector into components that are perpendicular to each other
y v
x
Here a vector v is resolved into an x component and a y component
Here we see a table being pulled by a force of 50 N at a 30º angle to the horizontal
When resolved we see that this is the same as pulling the table up with a force of 25 N and pulling it horizontally with a force of 43.3 N
50 Ny=25 N
x=43.3 N30º
We can see that it would be more efficient to pull the table with a horizontal force of 50 N
If a vector of magnitude v and makes an angle θ with the horizontal then the magnitude of the components are:
x = v Cos θ y = v Sin θ
vy=v Sin θ
x=v Cos θ
θ
y
Proof:
v
xCos
vCosx v
ySin
vSiny
x
60º
2002 HL Sample Paper Section B Q5 (a)A force of 15 N acts on a box as shown. What is the horizontalcomponent of the force?
Vert
ical
Com
ponent
Horizontal Component
Solution:
N 5.76015Component Horizontal Cosx
N 99.126015Component Vertical Siny
15 N
7.5 N
12.9
9 N
A person in a wheelchair is moving up a ramp at constant speed. Their total weight is 900 N. The ramp makes an angle of 10º with the horizontal. Calculate the force required to keep the wheelchair moving at constant speed up the ramp. (You may ignore the effects of friction).
Solution:
If the wheelchair is moving at constant speed (no acceleration), then the force that moves it up the ramp must be the same as the component of it’s weight parallel to the ramp.
10º
10º80º
900 N
Complete the parallelogram.Component of weight
parallel to ramp: N 28.15610900 Sin
Component of weight perpendicular to ramp:
N 33.88610900 Cos
156.28 N
886.33 N
2003 HL Section B Q6