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...