Download PULLEYS - Mathematics with Mr Walters

MECHANICS M1
SLBS
DYNAMICS
NEWTON’S LAWS
1st LAW
A particle will remain at rest or will continue to move with
constant velocity in a straight line unless acted upon by a
resultant force.
2nd LAW
The Force F applied to a particle is proportional to the mass m
of the particle and the acceleration produced.
F = ma
3rd LAW
Every action has an equal and opposite reaction.
1 Newton is defined as the force that gives a mass of 1 kg an
acceleration of 1 ms-2
Example
A force of 10N acts on a particle of mass 2kg. Find the
acceleration.
F = ma  10 = 2a  a = 5ms
JMcC
–2
1
MECHANICS M1
SLBS
Example
A car of mass 500kg is accelerating at 3ms-2. The resistance
3ms to motion due to friction is 400N. Find the driving force of
the engine.
Always draw a force diagram
–2
3ms
500kg
–2
D
400N
Note: when using F = ma, F stands for Resultant Force
R(→) stands for “Resolving forces horizontally & use F = ma”
R(  ) D – 400 = 500  3
D = 1500 + 400
= 1900N
VERTICAL MOTION
Newton’s 2nd Law implies that the force on an object = mass × g
 Weight = mg
Example
A string is pulling up a particle of mass 4kg with an
acceleration of 1 ms-1. Find the tension in the string
T
4 kg
R(up) T – 4g = 4  1
T = 4 + (4  9·8)
4g
= 43·2 N
Ex 3A p 40
JMcC
2
MECHANICS M1
SLBS
TYPES OF FORCE
Weight
String in Tension
W = mg
Spring/Rod
in Thrust
Normal Reaction
Friction
direction of motion


W
R
F


W
Worksheet on Marking Forces
Ex 3B p 44
JMcC
3
F

MECHANICS M1
SLBS
RESOLVING FORCES INTO COMPONENTS
Q
F


O
F
cos 
 sin
P
But OQ = OP + PQ
OP = F cos 
and
PQ = F sin 
Therefore
F
F sin 
is equivalent
to

F cos 
Component of a force F in any direction
= F × Cosine of the angle between the force and the specified
direction
Note: cos(90 – ) = sin 
Example
8N
60
JMcC
8 sin 60 = 6.93
is equivalent to
8 cos 60 = 4
4
30
60
 60 = 11133
tan
sin
cos
60 =
cos 30
tan
sin
60
30
===
22223
MECHANICS M1
SLBS
TRIG RATIOS FOR 30°, 45° and 90°
1
sin 30 =
30
2
2
2
3
cos 30 =
tan
45
2 45 = 11 1 60
sincos
4545
 = =
2 2
1
30sin
8
cos
30
 30
1
tan 30 =
1
sin 45 =
45
2
2
1
cos 60 =
tan 60 =
3
2
1
2
3
3
1
2
cos 45 =
45
sin 60 =
1
2
tan 45 = 1
1
Example
8 sin 30
8N
is equivalent
to
30
8 cos 30 = 8 
8 sin 30 = 8 
3
2
1
2
8 cos 30
= 4 3
= 4
Ex 3C p 47
JMcC
5
MECHANICS M1
SLBS
FRICTION
 F = 2NFriction always opposes motion.
Consider an object being pulled along a rough horizontal
surface by a pulling force P
 F = 3N
F
When P=2N, object is at rest
P = 2N
 F = 2N
F
P = 3N
When P=3N, object is about to move
 F = 3N
 F = 3N
This is the maximum friction force
F
P = 4N
When P=4N, object is moving
 F = 3N only
It can be shown experimentally that:
F   R where  = coefficient of friction
max F =  R
JMcC
6
MECHANICS M1
SLBS
Example
R = 6g
A particle of mass 6kg is attempted to be pulled along a rough
horizontal surface. When the pulling is 35N, the particle is
about to move.
Calculate the value of μ.
R
F
R = 6g
35N
6g
R(  ) F = 35 and this is maximum friction
  =
F
R
=
35
6g
= 0·60 to 2dp
NOTE: When a force is applied to an object on a horizontal
surface that is inclined at an angle to the horizontal, the
Normal Reaction does not equal the Weight due to the effect
of this force.
R
P

F
R + Psin = W
F = Pcos
W
JMcC
7
MECHANICS M1
SLBS
Example
R
14 2 N
45
F
 =
1
7
Find the acceleration
8g
a
R + 14 2 sin 45 = 8g
1
R = 8g – 14 2 
F =
1
2
= 8g – 14 = 64·4 N
 64·4 = 9·2 N
7
R(  ) 14 2 cos 45 – F = 8a
14 2 
1
2
– 9·2 = 8a
14 – 9·2 = 8a
4·8 = 8a
a = 0·6 ms
–2
If a body is said to be in “limiting equilibrium”, this means that
it is “about to move” or is “on the point of moving.”
Ex 3D p 52
JMcC
8
MECHANICS M1
W cos
sin 

SLBS
COMPONENTS OF THE WEIGHT OF AN OBJECT PLACED
ON A SLOPE
is equivalent
to
W sin 



W cos 
W
Component of Weight down the slope = W sin 
Component of weight into the slope
= W cos 
You must remember these!
CALCULATING THE COEFFICIENT OF FRICTION
R
F
Limiting Equilibrium

W
R = W cos 
F = W sin 
F = R   =
F
R
=
W sin 
W sin 
= tan 
If tan  >   object will accelerate down the slope
If tan     object will remain at rest
JMcC
9
MECHANICS M1
SLBS
–2
25

Example
2ms
Force P is pulling a particle of mass 8kg up an inclined smooth
slope of 25° at an acceleration of 2ms-2. Find P
2ms
R
–2
P
8 kg
25
8g
R(up slope) P – 8g sin 25 = 8  2
P = 16 + 8g sin 25
= 49·1 N
Sometimes, the angle of the slope may be given in the form
3
4 x
4angle whose sine is x
4
arcsin
i.e.
the
sin  = and tan  =
= arccos
y
55 y
3
Do not use Trigonometry and calculate the angle, rather use
Pythagoras:
 = arccos
3
5
5
4
3
 sin  =
4
5
and tan  =
4
3
If you find θ = 53.13…° and
use 53.1, you will lose accuracy
JMcC
10
MECHANICS M1
SLBS
Example
30
A particle of mass 10kg rests in limiting equilibrium on a slope
inclined at 30° to the horizontal. Calculate the coefficient of
friction between the object and the slope.
R
F
10kg
30
10g
There is no motion, so the forces up and down the slope, must
balance each other out.
R(up slope)
F = 10g sin 30
R( slope)
R = 10g cos 30
Max F  F =  R   =
F
R
=
10g sin 30
10g cos 30
= tan 30
=
1
3
Ex 3E p 55
JMcC
11
MECHANICS M1
SLBS
LIFTS
Example
2ms
A lift if mass 500kg contains a parcel of mass 50kg. Find the
tension in the lift cable and the reaction between the parcel
and the lift floor if the lift is accelerating:
a) upwards at 2 ms-2
b) downwards at 2 ms-2
–2
T
a) Lift + Parcel
2ms
R(up) T – 550g = 550  2
–2
(1)
T = 1100 + 550g
T = 6490 N
2ms
–2
550g
Lift Only
2ms
T
R(up) T – 500g – R = 500  2
(2)
1590 – R = 1000
–2
R = 1590 – 1000
= 590N
2 ms
–2
R
500g
Parcel Only
R(up) R – 50g = 50  2
2 ms
–2
R
(3)
R = 100 + 50g
= 590N
50g
Note: Equation (2) + Equation (3) gives Equation (1) Therefore options!
JMcC
12
MECHANICS M1
2 ms
–2
SLBS
Always state which objects you are considering and mark up an
appropriate diagram
b) Lift & Parcel
T
R(down) 550g – T = 550  2
550g – 1100 = T
2 ms
2 ms
T = 4290N
–2
–2
550g
Parcel Only
R
R(down)
50g – R = 50  2
50g – 100 = R
R = 390N
2 ms
–2
50g
McEx 5A
JMcC
13
MECHANICS M1
SLBS
CONNECTED PARTICLES
The tension in the tow-bar of a car towing a caravan is
represented by:
T
T
Example
0.5ms
–2
A car of mass 1200 kg tows a caravan of mass 700 kg along a
horizontal road. The car is accelerating at 0.5 ms-2. Ignoring
any resistances, find the driving force of the engine and the
tension in the tow-bar.
0.5ms
700kg
T
R(  ) Car & Caravan
T
–2
1200kg
D
D = 1900  0·5
= 950 N
R(  ) Caravan only
T = 700  0·5
= 350 N
JMcC
14
MECHANICS M1
SLBS
In practice, there will be resistance to motion. Often, it is
proportional to the mass of the objects. Looking again at the
previous example where there is a total resistance to motion of
380 N
Resistance on car
= 1200k where k = constant of proportion
Resistance on caravan = 700k
0.5ms
–2
Total resistance
= 1900k = 380N  k = 0·2
 Resistance on car
= 1200  0·2 = 240 N
Resistance on caravan = 700  0·2 = 140 N
0.5ms
700kg
T
T
–2
1200kg
D
240 N
140 N
R(  ) Car & Caravan
D – 140 – 240 = 1900  0·5
D = 950 + 380
= 1330 N
R(  )
Caravan only
T – 140 = 700  0·5
T = 350 + 140
= 490 N
JMcC
15
MECHANICS M1
SLBS
If the car brakes, the tow-bar will go into THRUST
T
T
Example
In the previous example, the car reduces the driving force to
zero and applies the brakes. Given that the resistances to
motion are unchanged and the thrust in the tow-bar is 300N,
find the value of the braking force
a
700kg
140 N
300
300
B
1200kg
240 N
Caravan only 300 + 140 = 700a
440 = 700a
a = 0·62857...
Car & Caravan B + 380 = 1900  0·62857...
B = 1194·285... – 380
= 814N
McEx 5B
JMcC
16
MECHANICS M1
SLBS
PULLEYS
Example
Particles of mass 3kg and 5kg are attached to the ends of a
light inextensible string which passes over a smooth fixed
pulley. The system is released from rest. Find the
acceleration of the system and the tension in the string.
Find T and a
T
T
a
T
T
3 kg
5 kg
a
3g
R(↓) 5 kg
R(↑) 3 kg
5g
5g
T
−
−
(1) + (2)
Subst in (2)
T
−
T
3g
=
=
5a
3a
(1)
(2)
2g
a
=
=
8a
g/4 m/s2 (= 2.45 m/s2)
3g
T
T
=
=
=
3g/4
3g/4 + 3g
15g/4 N (= 36.75 N)
Note: Smooth pulley =>Tension in both portions of the string is the same.
Inextensible string =>Acceleration of both particles is the same
(and they have the same speeds at any given instant)
Exam questions often ask where in your solutions you have used these
assumptions.
McEx 5C
JMcC
17
MECHANICS M1
SLBS
Example (Horizontal Table)
Two particles A and B of masses 4kg and 10 kg respectively are
connected by a light inextensible string. Particle A rests on a
rough horizontal table where the coefficient of friction
between A and the table is ½. The string passes over a smooth
pulley fixed at the edge of the table and B is hanging freely.
Find a)
the acceleration of the system
b)
the tension in the string
R
a
A
T
T
Find T and a
T
F
4g
T
a
10g
R(↑)
R
R(→) A
R(↓) B
=
4g
T −
10g −
(1) + (2)
Subst. in(1) T
JMcC
−
=>
F
=
½ × 4g
2g =
T =
4a (1)
10a (2)
8g =
a =
14a
8g/14
2g =
T =
T =
16g/7
16g/7 + 2g
30g/7 =
42 N
=
=
=
2g
4g/7 ms−2
5.6 ms−2
18
MECHANICS M1
SLBS
Note: Force on the pulley
T sin 45
T
is equivalent to
T cos 45
T
T sin 45
T cos 45
T sin 45 components cancel each other out!
Force on Pulley =
=
=
=
2T cos 45
2 × 30g/7 × 1/√2
30g√2/7
59.4 N
McEx 5D
JMcC
19
MECHANICS M1
SLBS
Example (Pulley on a Slope)
A package A of mass 26 kg rests on a smooth inclined plane of
5
angle arcsin
. A light inextensible rope is attached to A
13
and passes over a smooth fixed pulley at the top of the plane.
Another package B of mass 30 kg hangs freely from the other
end of the rope. The system is released from rest. Find the
acceleration of the system and the tension in the rope.
a
T
R
T

26g
a
30g
R(down) B
R(up slope) A
30g – T = 30a
(1)
T – 26gsin = 26a
T – 26g 
5
13
= 26a
T – 10g = 26a
(1) + (2)
20g = 56a
a =
Subst. in (2)
20g
56
=
T – 10g = 26 
T =
JMcC
(2)
270g
14
=
135g
7
5g
14
5g
14
= 3·5 ms
=
–2
130g
14
= 189 N
20
MECHANICS M1
SLBS
PROBLEMS WHERE FORCES CHANGE
Example
Two particles A and B of masses 4kg and 5kg respectively, are
connected by a light inextensible string which passes over a
smooth fixed pulley. They are originally held so that both
particles are 2m above the ground. The system is released
from rest. B hits the ground and does not rebound.
Find A’s greatest height above the ground.
a
T
5kg
4g
(1) + (2)
5g −
T −
B
A
4kg
R(↓) B
R(↑) A
T
T =
4g =
a
5a (1)
4a (2)
g = 9a
a =
JMcC
5g
2m
g
9
21
MECHANICS M1
SLBS
We need to find the speed that A and B have when B hits the
ground
Use constant acceleration equations
2
u = 0
a =
2
v – u = 2as
g
2
g
2
v – 0 = 2 
9
s = 2
v =
9
 2
4g
g
0
h
v
A
Now A acts under gravity only,
so its acceleration = -g
4m
B
u =
4g
9
a = – g
2
2
v – u = 2as
0 –
v = 0
4g
9
= – 2gh
h =
2
9
m
s=h
Greatest Height reached by A
=
4 + h = 4
2
9
m
Old M1 Book Ex 5D Q 13-16
Ex 3F p64
JMcC
22
MECHANICS M1
SLBS
MOMENTUM and IMPULSE
v = u + at  a =
F = ma =
v – u
t
m(v – u)
t
Ft = mv – mu
(*)
MOMENTUM
=
mass × velocity =
mv
IMPULSE
=
Force × time
Ft
=
Both quantities are measured in Newton-seconds (Ns)
(*) =>
Impulse = Change in Momentum
Example
A particle of mass 3 kg increases its speed from 2 ms-1 to 5 ms-1
Find its gain in momentum
Initial momentum
Final momentum
Gain in momentum
=
=
=
3×2
3×5
15 − 6
=
=
=
6 NS
15 Ns
9 Ns
Example
A ball of mass 100g hits a wall at 10m/s and rebounds at 8 m/s.
Find the impulse exerted by the wall on the ball.
10 m/s
0.1 kg
8 m/s
Ex 3G p 68
JMcC
Initial Momentum = 0.1  10 = 1 Ns
Final Momentum = 0.1  ( – 0.8) = – 0.8 Ns
Impulse = Change in Momentum
= 1 – ( – 0.8) = 1 + 0.8 = 1.8 Ns
Note: Momentum may be positive or negative
23
MECHANICS M1
SLBS
The Principle of Conservation of Momentum
When two particles A and B collide, they exert equal and
opposite forces, and hence impulses, on each other.
The impulse that A exerts on B (equal to B’s change in
momentum) is therefore equal and opposite to the impulse that
B exerts on A (equal to A’s change in momentum).
If these changes are equal and opposite, there is no change in
the total amount of momentum of both particles taken
together.
Hence, we have the Principle of Conservation of Momentum:
Unless acted upon by external forces, the total momentum
of a system remains constant.
So, the total momentum before impact equals the total
momentum after impact.
Example
A particle A of mass 3kg travelling in a straight line at 5 m/s
catches up with another particle B of mass 2kg moving in the
same direction at 2 m/s. After the collision, A has a speed of 1
m/s in its original direction. Find the speed of particle B
Before
3kg
After
C of M
5 m/s
2 m/s
2kg
1 m/s
v
2v + (3  1) = (2  2) + (3  5)
2v + 3 = 4 + 15
2v = 16
v = 8 m/s
JMcC
24
MECHANICS M1
SLBS
If two objects stick together, they are said to COALESCE
Example
A particle of mass 4kg travelling at 3 m/s catches up with a
particle of mass 2kg travelling in the same direction at 2 m/s.
On impact, the particles coalesce. Find the speed of the
combined particle.
Before
3 m/s
4kg
2kg
After
6kg
C of M
2 m/s
v
6v = (4  3) + (2  2)
6v = 12 + 4
6v = 16
v =
JMcC
8
3
m/s
25
MECHANICS M1
SLBS
Remember that momentum can be negative
Example
Two particles of masses 8kg and 5kg are approaching each
other with speeds of 7 m/s and 3 m/s respectively. After the
collision, the lighter particle reverses its direction and has a
speed of 6 m/s. Find the speed of the heavier particle after
the collision.
Before
8kg
After
3 m/s
7 m/s
5kg
v
6 m/s
Decide on which direction to take as positive.
In this case, we shall take from left to right (→) as positive.
C of M
8v + (5  6) = (8  7) – (5  3)
8v + 30 = 56 – 15
8v = 11
v =
11
8
m/s
Ex 3H p72
JMcC
26
MECHANICS M1
SLBS
JERKS
The principle of conservation of momentum also applies when a
string is jerked taut.
Example
Two particles of masses 4kg and 5kg are joined together by a
light inextensible string. Initially they are at rest with the
string slack. The 5kg particle is projected away from the
other particle with a speed of 2 m/s. Find their common speed
when the string jerks taut.
Before
5kg
4kg
After
2 m/s
v
v
After the jerk, both particles move with the same speed.
C of M
4v + 5v = 5  2
9v = 10
v =
10
9
m/s
Old M1 Book Ex 5F p 123 Q 13-18
Mixed Ex 3I p73
JMcC
27