Transcript AS - Mechanics - the SASPhysics.com

AS - Mechanics
Unit 2
Scalars and vectors
• All measurable physical quantities are either
scalars – they have a magnitude, or
vectors – they have a magnitude and direction
Examples:
Scalars
distance
Vectors
displacement
Vectors or Scalars?
Displacement
Weight
Distance
Energy
Lift
Speed
Drag
Mass
Force
Density
Temperature
Momentum
Time
Work
Power
Velocity
Acceleration
Area
Vectors or Scalars?
VECTORS
SCALARS
Lift
Time
Displacement
Weight
Drag
Force
Distance
Mass
Area
Density
Momentum
Acceleration
Velocity
Work
Temperature
Speed
Energy
Power
Addition and subtraction
• Scalars are generally positive numbers and can
be added/subtracted simply
• Vectors have to be added with directions taken
into account.
• Draw vectors as arrows with length = magnitude,
orientation = direction.
• To add, the vectors are placed nose-to-tail and the
hypotenuse of the resulting triangle represents the
“resultant” vector.
7m
5m
R
R2 = (52 + 72)
Resultant vectors
Right-angled triangles
trigonometric identities
• sinθ = o/h
• cosθ = a/h
• tanθ = o/a
Pythagoras’ theorem
• a2 = b 2 + c 2
If you are not comfortable
with the trigonometry,
vector problems can be
solved by careful scale
drawing (but this takes
longer....)
Scalene triangles
sine rule
• sinA/a = sinB/b = sinC/c
cosine rule
• a2 = b2 + c2 – 2bccosA
7m
95°
5m
R
R2 = 52 + 72 – 2×5×7×cos95°
Practice
• Now do Summary Questions 1, 2 on p.93
Components
• Any vectors can be described as the resultant of
two other vectors, therefore…
• …when it helps us, we break down a vector into
two “components” at right angles to each other
(e.g., one part north, one part east).
• This is the reverse operation of finding a resultant
in a right-angled triangle.
• For example
N
20 ms–1
40°
N = 20sin40°
E
E = 20cos40°
Practice
• Now do Mechanics examples 4
• And this:
A crate with a mass of 1500 kg is
suspended from a thin wire. The wire has
a breaking stress of 20 000 N. If the crate
is pulled sideways calculate the angle that
the wire must make with the vertical
before it breaks.
Equilibrium
• For forces to be in equilibrium, the
resultant force=0 (no nett force acts, a=0)
• For 2 forces in equilibrium:
– Forces must be equal, opposite and acting
along the same line
Can be in equilibrium
Can’t be in equilibrium
Equilibrium
• For 3 forces in equilibrium:
– The forces need not be acting along the same
line
– Solve problems by resolving into components
and equating them, or by completing vector
triangle
e.g. a ladder
leaning against a
wall:
Reaction
of wall
Reaction
of ground
weight
Resultant=0,
so vectors
form a closed
triangle
Inclined plane problems
• Sometimes, rather than vertical and
horizontal components, it is useful to
resolve a vector into components parallel
and perpendicular to a sloping surface
Reaction
(support)
Friction
No
motion, soan
forces are in
Consider
equilibrium.
object resting on
perpendicular:
Wcosq = R
a rough slope...
parallel: Wsinq = F
q
q
W
3 forces must make a triangle,
so tanq=F/R, W2=F2+R2
Tension problems
• Tightrope walker or bow and arrow q
(p.96, q. 3)
• How can they make the string perfectly
horizontal?
Moments
• A moment is the turning effect of a force
• Moment of a force about a point = force x
perpendicular distance
– Units: Nm
Principle of Moments
• If a body is acted on by more than one
turning force and remains in equilibrium,
then: Sum of clockwise
Sum of anticlockwise
moments
3m
500 N
=
moments
?
750 N
Density of metre rule
• Calculating weight of metre rule expt
• Calculating density of metre rule expt
• Combine your errors to find overall
uncertainty
• Find percentage difference from true value
• Are you within your experimental
‘tolerance’?
Centre of mass
• (aka centre of gravity, assuming
uniform gravitational field)
• Defined as:
– The point through which a single
force on a body has no turning effect
– Effectively the single
point at which the
whole weight
of the body
appears to act
Moments problems
• The pivot isn’t always in the middle!
– A shelf is supported as shown.
– (It is 38 cm deep.)
– Calculate the tension in the wire...
T
40o
Clockwise moment = anticlockwise moment
70 × 0.19 = Tsin40 × 0.38
P
So T = 54.5 N
70 N
Moments problems
• You can chose the point to take moments around...
Example problem
A truck is driven across a uniform bridge as shown below. The truck has a
mass of 4000 kg and the bridge is 20 m long and has a mass of 5000 kg.
(a) what is the total reaction at the supports?
(b) what is the reaction (R1 and R2) at each support when the truck is:
(i) 5 m from end A?
(ii) 12 m from end A?
R1
R2
Bob’s Trucks
B
A
20 m
Moments practice questions
• Textbook Summary Questions p.100
• TAP 203-5 practice questions
F
s
moment = Fs
F
s
couple = Fs
F
Couple and Torque
• Sometimes there are two
offset, equal, opposite
forces acting on a body to
turn it. This is called a
couple.
• The moment of the couple
(the “torque”) is defined as
the force × distance
F
between forces:
moment = F × d
F
d
Stability
• A stable equilibrium exists if a small
disturbance results in a body returning to
its original state
– eg marble in saucer
• An unstable equilibrium exists if a small
disturbance results in the body assuming a
new state
– eg pencil balanced on point
Toppling
• If an object is tilted and the line of force from the
centre of mass remains within the base, it will
not topple over.
• If it is tilted so far that the line of force from the
centre of mass moves outside the base, it will
topple over.
• Bus video
• For best stability, need
– low CoM
– wide base
Speed and velocity
• Distance is a scalar quantity.
• Speed is also a scalar quantity
– For motion at a constant speed:
speed = distance travelled / time taken
• Displacement is a vector
– Distance in a given direction
• Velocity is also a vector
– velocity = displacement / time taken
Changing velocity
• Acceleration is the change of velocity per
unit time
– Units: m/s2
– A vector
• +ve: increasing velocity
• -ve: decreasing velocity
• Acceleration can result in a change of
speed or direction
– eg motion in a circle at constant speed
Constant acceleration
• A special case
– Object moving in a straight line
– Constant rate of change of speed
•
change of velocity
a
time taken
vu
a
t
so v  u  at
speed
v
v–u
u
t
time
Constant acceleration
distance
covered
• average speed 
time taken
vu s
average speed 

2
t
but remember v  u  at
u  at  u s
so

2
t
1 2
or s  ut  at
2
Constant acceleration

v u
u  v t
a
and s 
t
2

v  u  u  v t
as 
.
t
2
2
2
v u
as 
2
2
2
v  u  2as


“suvat” equations
• Describe motion with
constant acceleration
• 5 variables, 4 equations
• Pick the right one and
solve any problem!
• Now do some practice…
v  u  at
(u  v)t
s
2
1 2
s  ut  at
2
2
2
v  u  2as
Travel graphs
•
•
•
•
Distance-time graphs
Displacement-time graphs
Speed-time graphs
Velocity-time graphs
• http://phet.colorado.edu/simulations/sims.p
hp?sim=The_Moving_Man
Distance–time graphs
• Distance is simply the length of ground
covered, regardless of direction
• e.g., walking around a square of side 3 m,
distance travelled is 12 m
– Displacement = 0
s
t
• Gradient = speed (always +ve)
3m
Displacement–time graph
• We must define a starting position and
direction
– At starting position the distance = 0
– One direction of travel is positive, the opposite
direction is negative
– Only plot the component of the distance in the
direction of interest
Displacement–time graphs
• Try to describe the motion shown in the graph
– What does the slope of the line represent?
– What does the slope of the dotted line tell you?
Displacement–time graphs
Constant speed forward
Constant speed backwards
Slope=average
velocity of
return journey
Slope = velocity
Speed=5/0.42=11.9 km/h
stationary
After 160 minutes, we
are back where we
started
Calculating velocity
• The slope of the graph gives the velocity
distance travelled
slope 
time taken
(a) Slope = 60/10 = 6.0 m/s
(b) Slope = 0/5 = 0.0 m/s
(b)
(a)
(c)
(c) Slope = -100/25 = -4.0 m/s
(d)
(d) Slope = 40/15 = 2.7 m/s
• The steeper the line, the higher the speed
Displacement-time graphs
How would
you
represent
something
getting
slower?
x
t
Note: displacement can also become negative, if object travels in the opposite
direction
Velocity is gradient of the
distance/time graph
• If velocity is
changing, the
instantaneous
speed is given by
the gradient of the
tangent to the
curve.
Speed and Velocity
• The velocity of an object gives its
instantaneous speed and direction
• As with distance, the sign of the velocity
indicates the direction
– a negative velocity means speed in the
opposite direction
Velocity
• Going from A
to B: + velocity
• Going from C
to F: - velocity
Velocity-time graphs
• Try to describe the motion shown in the graph
– What does the slope of the line represent?
– Where is the object not moving?
10
8
Velocity (m/s)
6
4
2
Time (s)
0
0
-2
-4
-6
10
20
30
40
Velocity-time
graphs
Constant speed
Constant acceleration
Gradual
slowing
forwards
10
More rapid slowing
8
Velocity (m/s)
6
4
stationary
2
Time (s)
0
0
10
20
30
40
-2
-4
-6
Reversing direction
and speeding up
Constant speed
backwards
Slowing to a
stop
Acceleration
• Acceleration is rate of change of velocity
– Given by the slope of a velocity-time graph
velocity
constant
acceleration
increasing
acceleration
Constant
negative
acceleration
time
Average acceleration=
velocity change/time taken
Displacement
• Displacement=velocity × time
• Found as the area under the graph
ds
–  ds   dt   v.dt
dt
Example – bouncing ball
• A tennis ball is dropped from a height of 2
m above a hard level floor, and falls to the
floor in 0.63 s. It rebounds to a height of
1.5 m, rising to a maximum height 1.18 s
after it was released. Draw a velocity–time
graph indicating velocity and time at key
points of the motion.
• The ball falls to the floor in
0.63 s. Its average speed
during the fall is
average speed 
distance
2m

 3.17 m s 1.
time
0.63 s
• Its maximum speed (the
speed with which it hits the
floor) is then 2 × 3.17 ms–1
= 6.35 ms–1. On the
rebound, the average
speed is
average speed 
5.45
velocty
/ms-1
0.63
distance 1.5 m

 2.73 m s 1.
time
0.55 s
6.35
• The time in this equation is
calculated from 1.18 s –
0.63 s = 0.55 s. The
maximum speed on the
rebound must then be 2 ×
2.73 ms–1 = 5.45 ms–1.
1.18
Time / s
• Check that you agree
with these graphs for
the bouncing ball
Thrust SSC
In 1997 Thrust SSC was
driven to a supersonic
world record speed of 771
mph (peak) and 767 mph
(mile average) (about 334
m s-1and 332 m s-1).
In their research the Thrust
SSC Development Team
predicted that the car’s
velocity would initially
increase as shown in the
graph below.
Thrust
SSC
(a) Describe in words only
(no numerical values) the
predicted acceleration
(i) during the first 4 seconds,
(ii) from 4 s to 30 s.
(b) Use the graph to predict
the size of the
acceleration at 12 s.
(c) Use the same graph to
predict the car’s
displacement after 10 s.
Thrust
SSC
Non-uniform acceleration
• For uniform acceleration can use “suvat”
equations
• Non-uniform acceleration problems can be
tackled graphically
0
1
2
3
4
5
6
7
8
9
10
Velocity (ms-1)
0
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16
Time (s)
e.g.100 m sprint
2-step “suvat” problems
• For problems where the acceleration
changes from one uniform value to
another (eg acceleration followed by
retardation) we can tackle each step
separately.
• See example p.125
Freefall and terminal velocity
• When the only force acting is gravity, a
body is in freefall and acceleration is
constant (g)
• When air resistance is considered, an
object accelerates until drag force=gravity
– No net force acts, so no acceleration
– Object falls at terminal velocity
– (drag force increases with speed)
Motion detectives
• Now make sure you can do the
problems...
Projectile motion
• We consider the horizontal and vertical
velocities of a projectile independently
– No acceleration in horizontal direction
(neglecting air resistance)
– Acceleration due to gravity in the vertical
direction
• Can use “suvat” equations, having
resolved velocity into horizontal and
vsinq
vertical components
q
v
vcosq
Projectile motion
• Horizontally:
x = ut
• Vertically:
y = ut – ½gt2
• so object follows a
parabolic path
Trajectory
Misconceptions
• Moving faster in the horizontal direction
does not change any movement in the
vertical direction.
Package Drop
• The package follows a parabolic path and
remains directly below the plane at all times.
• The vertical velocity changes due to gravity.
• The horizontal velocity is constant.
Trajectory & Range
•
•
•
Maximum range is at 45°
Note: the AQA specification requires
you to be able to solve problems
involving horizontal or vertical launch
only.
Phet simulation
Classic Problem
• A zookeeper finds an escaped monkey
hanging from a light pole. The zookeeper
aims a banana launcher at the monkey. At
the moment the zookeeper shoots the
banana, monkey lets go. Does the
monkey catch the banana?
Classic Problem
• Does the monkey catch the banana?
Cannon Fire
Cannon Fire
• The cannon balls will hit each
other.
Projectile motion
• Now check that you can do the problems...