as - mechanics unit 2. scalars and vectors all measurable physical quantities are either scalars –...
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AS - Mechanics
Unit 2
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Scalars and vectors
• All measurable physical quantities are either scalars – they have a magnitude, orvectors – they have a magnitude and direction
Examples:
Scalars Vectors
distance displacement
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Vectors or Scalars?
Density
AccelerationVelocity
Time
Speed
Temperature
Energy
Mass
Distance
Area
Force
Momentum
Work
Power
Weight
Drag
Lift
Displacement
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Vectors or Scalars?VECTORS SCALARS
Lift Time
Displacement Distance
Weight Mass
Drag Area
Force Density
Momentum Work
Acceleration Temperature
Velocity Speed
Energy
Power
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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.
5 m
7 m
R2 = (52 + 72)
R
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Resultant vectorsRight-angled triangles
trigonometric identities• sinθ = o/h• cosθ = a/h• tanθ = o/a
Pythagoras’ theorem• a2 = b2 + c2
5 m
7 m
R2 = 52 + 72 – 2×5×7×cos95°
Scalene triangles
sine rule• sinA/a = sinB/b = sinC/c
cosine rule• a2 = b2 + c2 – 2bccosA
95°
R
If you are not comfortable with the trigonometry, vector problems can be solved by careful scale drawing (but this takes longer....)
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Practice• Now do Summary Questions 1, 2 on p.93
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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 = 20sin40°E = 20cos40°
20 ms–1
40°
E
N
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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.
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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
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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:
weight
Reaction of wall
Reaction of ground Resultant=0,
so vectors form a closed triangle
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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
W
Reaction (support)
Friction
Consider an object resting on a rough slope...
No motion, so forces are in equilibrium.
perpendicular: Wcos= R
parallel: Wsin= F
3 forces must make a triangle, so tan=F/R, W2=F2+R2
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Tension problems
• Tightrope walker or bow and arrow q (p.96, q. 3)
• How can they make the string perfectly horizontal?
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Moments
• A moment is the turning effect of a force
• Moment of a force about a point = force x perpendicular distance– Units: Nm
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Principle of Moments
• If a body is acted on by more than one turning force and remains in equilibrium, then: Sum of clockwise
momentsSum of anticlockwise
moments=
500 N 750 N
3 m ?
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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’?
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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 thewhole weightof the bodyappears to act
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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
70 N
P
Clockwise moment = anticlockwise moment
70 × 0.19 = Tsin40 × 0.38
So T = 54.5 N
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Moments problems
Example problemA 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?
20 m
R2
R1
A
Bob’s TrucksB
• You can chose the point to take moments around...
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Moments practice questions
• Textbook Summary Questions p.100
• TAP 203-5 practice questions
moment = Fs
s
F
couple = Fs
s
F
F
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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 between forces:moment = F × d
F
F
d
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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
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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
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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
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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
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Constant acceleration
• A special case– Object moving in a straight line– Constant rate of change of speed
• speed
time
u
v
t
v–u
atuvt
uva
a
so
takentime
velocityof change
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Constant acceleration
•
2
2
1or
2 so
remember but 2
speed average
takentime
covered distance speed average
atuts
t
suatu
atuvt
suv
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Constant acceleration
asuv
uvas
tvu
t
uvas
tvus
t
uva
2
2
2.
2 and
22
22
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“suvat” equations
• Describe motion with constant acceleration
• 5 variables, 4 equations
• Pick the right one and solve any problem!
• Now do some practice…
asuv
atuts
tvus
atuv
2
2
12
)(
22
2
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Travel graphs
• Distance-time graphs
• Displacement-time graphs
• Speed-time graphs
• Velocity-time graphs
• http://phet.colorado.edu/simulations/sims.php?sim=The_Moving_Man
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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
• Gradient = speed (always +ve) 3m
t
s
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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
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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?
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Displacement–time graphsConstant speed forward
stationary
Constant speed backwards
After 160 minutes, we are back where we started
Slope=average velocity of return journey
Slope = velocity
Speed=5/0.42=11.9 km/h
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Calculating velocity• The slope of the graph gives the velocity
• The steeper the line, the higher the speed
takentime travelleddistance
slope
Slope = 60/10 = 6.0 m/s
(a)
(b)
(c)
(a)
(c)
(d)
Slope = -100/25 = -4.0 m/s
Slope = 40/15 = 2.7 m/s
(d)
Slope = 0/5 = 0.0 m/s(b)
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Displacement-time graphs
Note: displacement can also become negative, if object travels in the opposite direction
How would you represent something getting slower?
t
x
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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.
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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
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Velocity
• Going from A to B: + velocity
• Going from C to F: - velocity
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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?
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Velocity-time graphsConstant acceleration Constant speed
forwards Gradual slowing
More rapid slowing
stationary
Reversing direction and speeding up
Constant speed backwards
Slowing to a stop
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Acceleration
• Acceleration is rate of change of velocity– Given by the slope of a velocity-time graph
time
velocity
increasingacceleration
constantacceleration
Constantnegative
acceleration
Average acceleration= velocity change/time taken
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Displacement
• Displacement=velocity × time
• Found as the area under the graph–
dtvdtdt
dsds .
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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.
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• The ball falls to the floor in 0.63 s. Its average speed during the fall is
• 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
• 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.
6.35
5.45
velocty /ms-1
Time / s0.631.18
.sm17.3s63.0
m2
time
distancespeedaverage 1
.sm73.2s0.55
m1.5
time
distancespeedaverage 1
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• Check that you agree with these graphs for the bouncing ball
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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
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(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
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Non-uniform acceleration
• For uniform acceleration can use “suvat” equations
• Non-uniform acceleration problems can be tackled graphically
e.g.100 m sprintTime (s)
Velocity (ms-1)
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
0
1
2
3
4
5
6
7
8
9
10
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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
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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)
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Motion detectives
• Now make sure you can do the problems...
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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 vertical components v
vcos
vsin
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Projectile motion
• Horizontally:x = ut
• Vertically:
y = ut – ½gt2
• so object follows a parabolic path
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Trajectory
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Misconceptions
• Moving faster in the horizontal direction does not change any movement in the vertical direction.
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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.
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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
NOT EXAMIN
ABLE
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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?
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Classic Problem
• Does the monkey catch the banana?
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Cannon FireNOT E
XAMINABLE
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Cannon Fire
• The cannon balls will hit each other.
NOT EXAMIN
ABLE
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Projectile motion
• Now check that you can do the problems...