AS - Mechanics
Unit 2
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
Vectors or Scalars?
Density
AccelerationVelocity
Time
Speed
Temperature
Energy
Mass
Distance
Area
Force
Momentum
Work
Power
Weight
Drag
Lift
Displacement
Vectors or Scalars?VECTORS SCALARS
Lift Time
Displacement Distance
Weight Mass
Drag Area
Force Density
Momentum Work
Acceleration Temperature
Velocity 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.
5 m
7 m
R2 = (52 + 72)
R
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....)
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 = 20sin40°E = 20cos40°
20 ms–1
40°
E
N
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:
weight
Reaction of wall
Reaction of ground 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
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
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
momentsSum of anticlockwise
moments=
500 N 750 N
3 m ?
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 thewhole weightof the bodyappears 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
70 N
P
Clockwise moment = anticlockwise moment
70 × 0.19 = Tsin40 × 0.38
So T = 54.5 N
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...
Moments practice questions
• Textbook Summary Questions p.100
• TAP 203-5 practice questions
moment = Fs
s
F
couple = Fs
s
F
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 between forces:moment = F × d
F
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
• speed
time
u
v
t
v–u
atuvt
uva
a
so
takentime
velocityof change
Constant acceleration
•
2
2
1or
2 so
remember but 2
speed average
takentime
covered distance speed average
atuts
t
suatu
atuvt
suv
Constant acceleration
asuv
uvas
tvu
t
uvas
tvus
t
uva
2
2
2.
2 and
22
22
“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
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
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
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 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
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)
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
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?
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
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
Displacement
• Displacement=velocity × time
• Found as the area under the graph–
dtvdtdt
dsds .
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
• 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
• 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
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
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 vertical components v
vcos
vsin
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
NOT EXAMIN
ABLE
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 FireNOT E
XAMINABLE
Cannon Fire
• The cannon balls will hit each other.
NOT EXAMIN
ABLE
Projectile motion
• Now check that you can do the problems...
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