Transcript Vectors & Scalars - The Grange School Blogs

Vectors and Scalars
Define Vector and Scalar quantities and give
examples of each.

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
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
Two forces are applied to a body, as shown. What is the magnitude
and direction of the resultant force acting on the body?
Solution:


Complete the parallelogram (rectangle)
The diagonal of the parallelogram ac
represents the resultant force
The magnitude of the resultant is found using
Pythagoras’ Theorem on the triangle abc
a
Magnitude  ac  12  5
ac  13 N
2
12
Direction of ac : tan  
5
12
   tan 1  67
5
2
b

12 N
d
θ
5N

5
12
c
Resultant displacement is 13 N 67º
with the 5 N force


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
Here a vector v is resolved into
an x component and a y
component
y

x


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
y=25 N
30º
x=43.3 N

We can see that it
would be more
efficient to pull the
table with a
horizontal force of
50 N
A force of 15 N acts on a box as shown. What are the horizontal and
vertical components of the force?
Horizontal Component  x  15Cos60  7.5 N
Vertical Component  y  15Sin 60  12.99 N
Vertical
12.99 N
Component
Solution:
60º
Horizontal
7.5 N
Component
Moments: turning forces
What is the equation for calculating moment of a force?
Keywords: All: calculate the moment of a force about a point (torque) (C)
Force
Most: apply principle of moments to simple balanced situations (B)
Moment
Some: employ principle of centre of mass in calculations (A)
Couple
Torque
COM
Weight
Turning
Key Equation:
Moment of a force = force x perpendicular distance
moment = fd
Keywords: All: calculate the moment of a force about a point (torque) (C)
Force
Most: apply principle of moments to simple balanced situations (B)
Moment
Some: employ principle of centre of mass in calculations (A)
Couple
Note definition
Torque
COM
Weight
Turning
The Principle of moments:
When an object is in equilibrium:
Sum of clockwise
moments about any
point
=
Sum of anticlockwise moments
about any point
clockwise moments – anticlockwise moments = zero (resultant force)
Centre of
gravity is
important
for
balancing
You have to keep the centre of gravity above the base if you want to balance.
My head is big
for my body
compared to
an adult’s
head. That’s
why I find it
harder to
balance.
Poor balance produces an overall
turning force.
This requires a turning force in the opposite
direction to counteract it.
Keywords: All: calculate the moment in single support situations (C)
Force
Most: calculate the moment in twin support situations (B)
Moment
Some: determine the moment of a couple (torque) (A)
Couple
• A uniform horizontal shelf of width 0.38 m is attached to a wall a
Torque
shown in the diagram. The total weight of the shelf and books is
COM
70 N. This weight acts from the middle of the shelf (0.19 m from
Weight
the wall).
Turning
1. Calculate the turning effect of the weight about point P.
2. Calculate the tension in the support wire.
T
40o
P
70 N
Keywords: All: calculate the moment in single support situations (C)
Force
Most: calculate the moment in twin support situations (B)
Moment
Some: determine the moment of a couple (torque) (A)
Couple
Torque
1. Moment = 70 N  0.19 m = 13.3 N m
COM
Weight
Turning
2. The component of T at 90 to the shelf
must
provide the moment to balance the moment of
the weight.
T
T sin 40°  0.38 m = 13.3 N m
T sin 40° = 13.3 / 0.38 = 35 N
T = 35 / sin 40°
T = 54.5 N
40o
P
70 N
Note that we take moments about point P. This is because there is a third force which acts on the shelf;
this is the contact force (or ‘reaction’) of the wall on the shelf. We do not know its magnitude or
direction but, since it acts through point P, it has no turning effect about P.
1) The plank is set up as shown and the balance
zeroed. When the student lies on the plank the
reading is 600 N. The balance is 2 m from the
student’s feet and the centre of gravity of the
student is 1.5 m from their feet. What is the
student’s weight?
1) The plank is set up as shown and the balance
zeroed. When the student lies on the plank
the reading is 600 N. The balance is 2 m from
the student’s feet and the centre of gravity of
the student is 1.5 m from their feet. What is
the student’s weight?
Taking moments about the feet.
600 x 2 = W x 1.5 so W = 800 N
Keywords: All: calculate the moment in single support situations (C)
Force
Most: calculate the moment in twin support situations (B)
Moment
Some: determine the moment of a couple (torque) (A)
Couple
Torque
COM
Weight
Turning
Couples
• A couple is a pair of equal and opposite forces NOT acting
along the same line
Keywords: All: calculate the moment in single support situations (C)
Force
Most: calculate the moment in twin support situations (B)
Moment
Some: determine the moment of a couple (torque) (A)
Couple
Torque
COM
Weight
• Show that the total moment = Fd
Turning
Couples
F
x
d
F
Keywords: All: calculate the moment in single support situations (C)
Force
Most: calculate the moment in twin support situations (B)
Moment
Some: determine the moment of a couple (torque) (A)
Couple
Torque
COM
Weight
• Show that the total moment = Fd
Turning
Couples
F
x
d
Clockwise moment = Fx
Clockwise moment = F(d-x)
Total = Fx + F(d-x)
Total = Fx + Fd – Fx
Total = Fd
F
Keywords:
All: make calculations using the equations of motion (C)
Velocity
Most: link algebraic and graphical methodologies (B)
Acceleration Some: derive the equations of motion from 1st principles (A)
Displacement
The equations of motion
Initial
Final
Time
Constant
 Only work when the acceleration is constant!
(Take acceleration due to gravity as = 9.8 ms-2 unless otherwise stated)
1.
A car accelerates from rest at 3 ms-2 for 4s. How far has it travelled?
2.
(a)
(b)
(c)
A stone is dropped from a cliff:
How far will it have fallen in 4s?
What will its velocity be at that point?
What is the average velocity during the 4s?
3.
Starting from rest a car travels for 2 minutes with a uniform acceleration
of 0.3 ms-2 after which its speed is kept constant until the car is brought to rest
with a uniform decceleration of 0.6ms-2 if the total distance travelled is 4500m
how long did the journey take?
4.
(a)
(b)
(c)
again?
A stone is thrown upward with a velocity of 12 ms-1
How far will it have risen in 1s?
What will its velocity be at that paint?
What is the maximum height that it will reach before coming down
Lesson 2
25
1.
A car accelerates from rest at 3 ms-2 for 4s. How far has it travelled?
Distance travelled = ½ at2 = ½x3x42 = 24 m
2.
A stone is dropped from a cliff:
(a)How far will it have fallen in 4s?
(a) s = ½ gt2 = ½ x 9.8 x 42 = 78.4 m
(b)What will its velocity be at that point?
(b) v = at = 9.8x4 = 39.2 ms-1
(c)What is the average velocity during the 4s?
(c) average velocity = 39.2/2 = 19.6 ms-1
Lesson 2
26
3.
Starting from rest a car travels for 2 minutes with a uniform acceleration
of 0.3 ms-2 after which its speed is kept constant until the car is brought to rest
with a uniform retardation of 0.6ms-2 if the total distance travelled is 4500m how
long did the journey take?
Initial acceleration time = 2 minutes = 120s (note conversion to seconds)
Distance travelled in that time = ½ 0.3x1202 = 2160 m
Speed after this initial acceleration = 0.3x120 = 36 ms-1
Distance travelled during final deceleration = 362/2x0.6 = 1080 m
Time of final deceleration = 1080/18 = 60 s
Distance of constant velocity section = 4500 – 2160 – 1080 = 1260
Time of constant velocity section = 1260/36 = 35 s
Therefore total time for the journey = 120 + 35 + 60 = 215 s = 3 min 35 s
4.
A stone is thrown upward with a velocity of 12 ms-1
(a)
How far will it have risen in 1 s?
(b)
What will its velocity be at that paint?
(c)
What is the maximum height that it will reach before coming down again?
(a) s = ut – ½ gt2 = 12 – ½ x9.8x1 = 7.1 m
(b) v = u –gt = 12 – 9.8x1 = 2.2 ms-1
(c) maximum height (h) when vertical velocity = 0
v2 = 0 = 122 – 2x9.8xh
h = 7.5m
Lesson 2
27
Projectile : an object acted upon only
by the force of gravity
Putting our SUVAT equations to good use
Starting points....
1.We need to consider the vertical and horizontal
components of motion separately.
2.The acceleration involved is always g (downwards) &
only affects the vertical component
3.Any horizontal velocity is constant and unaffected by g
4.We’ll consider up as positive and down as negative
Note : Vertically, the motion of the dropped
ball and the thrown ball is identical
Note : Vertically, the distance travelled
between each photo increase (acceleration)
Note : Horizontally there is no change in velocity, (1
square per photo all the way across)
A worked example....
An object is projected at a horizontal
velocity of 15 ms-1 from the top of a 35m
tower. Calculate the following
a)How long it takes to reach the ground
b)How far it travels horizontally
c)It’s speed at impact
A worked example....
An object is projected at a horizontal
velocity of 15 ms-1 from the top of a 35m
tower. Calculate the following
a)How long it takes to reach the ground
(Using S=ut + ½ at2 = 2.67s)
a)How far it travels horizontally
(Using S = ½ (u + v) t or S = velocity x time = 40m)
a)It’s vertical speed at impact
(Using V2 = U2 + 2aS = 26.2 ms-1)
An example....
A ball is projected horizontally at 0.52 ms-1 across
the top of an inclined board which is 60cm wide
(horizontally) and 120cm high (vertically in the
direction of the incline). The ball reaches the
bottom of the board in 0.9s. Calculate the
following:
a)The distance travelled across the board [0.468m]
b)Its acceleration on the board [2.96ms-2]
c)Its speed at the bottom of the board [2.8ms-1]