Transcript 02-Forces shorter

Forces
By Neil Bronks
Force causes a body to change
velocity…….. accelerate
The unit is called the
Newton (N)
Scalar vs. Vector
Scalar has only magnitude…..speed, mass
This car has a mass of
2000kg
Vector has magnitude and direction …….. Velocity, Force,
acceleration or displacement.
This car has a force of
500N
Distance and Displacement
Scalar- Distance
travelled 200m
VectorDisplacement
120m
80
60
0.5x10x20=100
Velocity
m/s
40
20
0.5x10x40=200
0
0.5x20x60
=600
40x20=800
10
20
30
40
Total Distance Traveled
=200+100+800+600=1700m
T/s
50
Motion Formula
v = u + at
A car starts from rest and accelerates for
12s at 2ms-2. Find the final velocity.
Using V = U + at = 0 + 2x12 = 24m/s
v2 = u2 + 2as
A car traveling at 30m/s takes 200m to
stop what is it’s deceleration?
Using V2 = U2 + 2as
0 = 900 + 2a (200)
a = -900/400=-2.25ms-2
Motion Formula
S = ut + 0.5at2
A train accelerates from rest at 10ms-2
for 12s find the distance it has traveled.
Using S = ut + 0.5at2 = 0x12 +0.5x10x144 =720m
Vector Addition
Speed in still air 120m/s

Wind
50m/s
R2 = 1202 + 502 = 14400 + 2500
= 16900
Tan  = 50/120
R = 130m/s
 = 22.60
Friction is the force that opposes
motion
The unit is
called the
Newton (N)
Lubrication
reduces friction
Friction is the
force between
two bodies in
contact.
Momentum
V=? m/s
5m/s
2kg
3kg
3 m/s
In a closed system the linear momentum is always conserved
Momentum Before = Momentum After
Mass Moving x velocity before = Mass moving x velocity after
2kg x 5m/s = 2kg x (-3m/s) + 3kg x v
3v = 10 + 6
V = 5.333m/s
VERIFICATION OF THE PRINCIPLE OF
CONSERVATION OF MOMENTUM
t1
l
Dual
timer
Photogate
Light
beam
Card
Air track
Vehicle 1
Velcro pad
Vehicle 2
Recoil
m=2kg
Mass of canon=150kg
u=400m/s
Momentum of Recoil = Momentum of the Shoot
Mass Canon x Velocity Canon = Mass of Ball x Velocity of Ball
150 x v = 2 x 400
V= 800/150 =
5.3m/s
Newton’s Laws
• 1 /. Every body stays in it’s state of rest
or constant motion until an outside force
acts on it
• 2/. The rate of change of momentum is
proportional to the applied force and in
the direction of the applied force.
• F=ma
• 3/. To every action there is an equal and
opposite reaction
Newton 2
force Rate of change of Momentum
mv  mu
force
t
m(v  u )
force
t
(v  u )
but  a 
t
Forcem.a
Or Force=k.m.a where k=constant
As this is the basic constant so we say k=1 and
Force=m.a
TO SHOW THAT a F
Dual
timer
t1
Light
beam
Air
track
Photogate
Pulley
l
Card
s
Slotted
weights
Force and acceleration
If the forces acting on an object
are unbalanced then the object will
accelerate, like these wrestlers:
Force (in N) = Mass (in kg) x Acceleration (in m/s2)
F
M
A
Acceleration gives Net Force
Friction=?
900kg
Feng=5000N
2
a=3m/s
As net force causes acceleration F=m.a
Fnet = 900kg. 3m/s2
Fnet= 2700N
So Friction = Feng – 2700 = 2300N
Archimedes Principle
• A body in a fluid
experiences an
up-thrust equal
to the weight of
liquid displaced.
12N
20N
8N
Floatation
• A floating body
displaces its
own weight in
water.
MEASUREMENT OF g
Electromagnet
Switch
Electronic timer
h=½gt2
Ball bearing
h
Trapdoor
Newton's Law of Gravitation
• This force is always positive
• Called an inverse square law
F  m1m2
d2
Where
F = Gravitational Force
m1.m2 = Product of masses
d = Distance between their center
of gravity
Hookes Law Example
Force =Constant (k) x Extension
Example a/. A mass of 3kg causes an extension of
0.3m what is the spring constant?
3x9.8 = k x 0.3
K=98N/m
B/. What is the extension if 40N is put on the same spring?
Force = Spring Constant x Extension
40 = 98 x s
S = 40/98 = 0.41 m
Work done
When any object is moved around work will need
to be done on it to get it to move (obviously).
We can work out the amount of work done in
moving an object using the formula:
Work done = Force x Distance Moved
in J
in N
in m
W
F
D
Work Done = Energy Converted
Work Done raising an object = PE Stored
PE at top=KE at bottom
At the
bottom the
bob has no
PE only KE
KE = ½ mv2
At the top of the
oscillation the
pendulum bob
stops. All it’s
energy is PE
PE = mgh
h
Power
• The rate at which work
is done
• POWER = Work Done
time taken
Example
A jet takes 2mins to climb to 4000m. If the jet has
mass 200tonnes find the work done and the power?
Work Done = Force x Distance = 200x1000x9.81x4000
=7 x 109 Joules
Power = Work Done / Time = 7 x 109 Joules / 120
= 6.54 x 107 Watts
Pressure
Pressure depends on two things:
1) How much force is applied, and
2) How big (or small) the area on which this force is
applied is.
Pressure can be calculated using the equation:
F
Pressure (in N/m2) = Force (in N)
Area (in m2)
P
A
The Barometer
• The weight of the air
holds up the mercury.
• If we use water the
column is 10.4m high.
• 1 Atmosphere is
760mm of Hg.
VERIFICATION OF BOYLE’S LAW
1. .
Volume
scale
Bicycle pump
Tube with volume of
air trapped by oil
Reservoir of oil
Valve
Pressure
gauge
P
1/V
Plot a graph of P against 1/V.
A straight-line graph through the origin will
verify that, for a fixed mass of gas at constant
temperature, the pressure is inversely
proportional to the volume, i.e. Boyle’s law.
INVESTIGATION OF THE LAWS OF
EQUILIBRIUM FOR A SET OF CO-PLANAR FORCES
(2)
Support
Newton
balance
Newton
balance
Paperclips
w1
w2
w3
w4
?N
25N
10
15N
50
5N
60
70
?
10N
• First law coplanar forces
• Forces Up = Forces Down
25 + x = 15 + 5 +10 + 5
x = 10 N
90
5N
10N
25N
A
10
15N
50
5N
60
70
10N
?
90
5N
• Second law coplanar forces
• Take moments about A
Clockwise Moments = Anticlockwise Moments
10x15 + 50x5 + 70x50 + 90x5 = 60x25 + dx10
Circular Motion
• Angular Velocity
• =θ/t
• Units of Radians per
second
• Angle  time
t

A particle goes round a
circle in 4s what is it’s
angular velocity?
2 

 rads / second
4
2
Circular Motion
•
•
•
•
•
Linear Velocity(V)
m/s
V=  r
r=radius of motion
Always changing as
direction is always
changing this creates
acceleration
• If the radius is 6m


2
rads / second
v  r.  6.

2
 9.42m / s
Centripetal Acceleration
a = r 2
Always towards the centre
So the acceleration in the previous
example a= 6 (/2)2
=14.8m/s2
Satellites balance forces
• Balance of Gravity and Centripetal
• ((GMm)/d2)=mv2/d
Gravity F=-GmM/r2
Period of Orbit
((GMm)/d2)=mv2/d
(GM)/d=v2
(GM)/d=(2d/T)2
T2=42 d3 /GM
Simple Harmonic Motion
• Is a vibration where the
acceleration is
proportional to the
displacement
a  -s
•Further from centre =more acceleration
Hooke’s Law as SHM
Force  Extension
F  -s
m.a  -s
If mass is constant
a  -s
So motion under hookes
law is SHM
Pendulum
Split
cork
l
Bob
Timer
20:30
• If we displace
the bob by a
small angle it
vibrates with
SHM
T2
l
T  4
g
T 2 4 2


 slope
l
g
4 2
 g 
(slope)
2
l
2