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Mechanics 1

Revision Notes

October 2016

1
...
3
Assumptions and approximations often used to simplify the mathematics involved:
...
Vectors in Mechanics
...
4
Parallel vectors
...
5
Resolving vectors in two perpendicular components
...
6
Vectors in mechanics
...
7
Relative displacement vectors
...
8
Closest distance between moving particles
...
9

3
...
10
Constant acceleration formulae
...
10
Speed-time graphs
...
Statics of a particle
...
13
Resultant of three or more forces
...
15
Types of force
...
17
Coefficient of friction
...
19

5
...
20
Newton’s laws of motion
...
22
Particles connected by pulleys:
...
26
Impulse and Momentum
...
28
Conservation of linear momentum, CLM
...
31

6
...
32
Moment of a Force
...
32
Moments and Equilibrium
...
33
Nearly tilting rods
...
35
Conservation of linear momentum, C
...
M
...
36

2

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1
...
have no mass,

f)

the tension in a light string which remains taut will be constant throughout its length
...


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3

2
...

A vector is a quantity which has both magnitude and direction
...
g
...

Vectors should be underlined, a letter without the underlining means the length of the vector
...

Example:

r = (32 + 52)

5

r = 3i + 5j =

m

180 - 



r=

x

3

tan  = 5/3



 = 59
...
0o with the x-axis
...

Example:
r is a vector of length 7 cm making an angle of 50o with the x-axis
...


Example: Find a vector p of length 15 in the direction of q =
Solution:


...


First find the length of q = q =
and as 15 = 15  10, p = 15 q  p = 15 

4

=

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Parallel vectors
Two vectors are parallel  one is a multiple of another:
 6i – 8j and 3i – 4j are parallel

e
...
6i – 8j = 2(3i – 4j)



Or


...


Example:
Add together,

3 miles on a bearing of 60 o and 4 miles on a bearing of 140o
...


Resolving vectors in two perpendicular components
y

F has components F cos  and F sin 
F sin 

as shown
...
, but

= r, usually!
A

To get from A to B
first go A to O using –a
a

then go O to B using b


= –a + b = b – a
...

Velocity is a vector so must be given either in component form or as magnitude and direction
...

Acceleration is a vector so must be given either in component form or as magnitude and
direction
...

If a particle moves from the point (2, 4) with a constant velocity v = 3i – 4j for 5 seconds
then its displacement vector will be velocity  time = (3i – 4j)  5 = 15i – 20j
and so its new position will be given by r = (2i + 4j) + (15i – 20j) = 17i – 16j
...
Find

Example: A particle is initially at the point (4, 11) and moves with velocity
its position vector after t seconds
...


rD – rC

rC

O

rD

D

Thus if a particle A is at rA = 3i – 4j and B is at rB = 7i + 2j
then the position of A relative to B is
= rA rel B = a – b = rA – rB = (3i – 4j) – (7i + 2j) = –4i – 6j
...

Particle B is intially at the point (6, 7) and travels with velocity 6i – 5j m s-1
...


(b)

Show that the particles collide and find the time and position of collision
...

y-coords = 4 – 2t = 7 – 5t  t = 1
...


Closest distance between moving particles
Example: Two particles, A and B, are moving so that their position vectors at time t are
rA =

and rB =


...


(b)

Find the distance between A and B at time t in terms of t
...


Solution:
= rB  rA =

(a)

(b)

The distance, d, between the particles is the length of
=

8



d 2 = (–1 +2t)2 + (1 + t)2 = 1 – 4t + 4t2 + 1 + 2t + t2
= 5t2 – 2t + 2



d =

5t 2  2t  2

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(c)

The minimum value of d will occur when the minimum value of d2 occurs so we
differentiate d2 with respect to t
...
F
...

Example: Particles A and B have velocities vA = (12t – 3) i + 4 j and
vB = (3t2 – 1) i + 2t j
...


Solution:


vA =

and

vB =

vA rel B = vA – vB =

When the relative velocity is parallel to the x-axis, the y-coordinate = 0
 4 – 2t = 0 which happens when t = 2


the relative velocity is parallel to the x-axis when t = 2
...


Kinematics of a particle moving in a straight line

Constant acceleration formulae
...


N
...
Units must be consistent - e
...
change km h1 to m s1 before using the formulae
...

Find the time(s) at which the particle is 12 m from O
...


Answer Particle is 12 m from O after 1 and 3 seconds
...

2) We assume that there is no air resistance, that the object is not spinning or turning and that
the object can be treated as a particle
...

4) Always state which direction (up or down) you are taking as positive
...

(a)

Find the greatest height reached
...


(c)

Find the velocity after 2 seconds
...
+
At the greatest height, h, the velocity will be 0 and so we have
+
Using


u = 14, v = 0, a = –98 and s = h (the greatest height)
...

Answer: Greatest height is 10 m
...


seconds to return to O
...


Answer: After 2 seconds the ball is travelling at 5
...


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11

Speed-time graphs
1) The area under a speed-time graph represents the distance travelled
...

Example: A particle is initially travelling at a speed of 2 m s1 and immediately accelerates at
3 m s2 for 10 seconds; it then travels at a constant speed before decelerating at a 2 m s2
until it stops
...

(b) Sketch a speed-time graph
...

Solution:
(a) For maximum speed: u = 2, a = 3, t = 10, v = u + at  v = 32 m s1 is maximum speed
...

(b) In the graph, T is the total time taken
...


12

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4
...


Resultant forces
As forces behave like vectors you can add two forces geometrically using a triangle or a
parallelogram
...

Example: F1 and F2 are two forces of magnitudes 9 N and 5 N and the angle between their
directions is 100o
...

Solution:
Using the cosine rule
R2 = 52 + 92  2  5  9  cos 80
R

 R = 950640…

5
80o

xo

100o

9

and, using the sine rule,

5
9  50640
...
0

Answer: The resultant force is 951 N at an angle of 312o with the 9 N force
...

Solution:
R = P + Q = (5i  7j) + (2i + 13j) = 3i + 6j
...


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13

Resultant of three or more forces
Reminder:
We can resolve vectors in two
perpendicular components as shown:
F has components F cos  and F sin 
...


For more than three forces continue with either of the above methods
...


60o
25o

9N

7N
8N

Solution:

First resolve the 7 N and 4 N
forces horizontally and vertically

4 sin 60 N

The resultant force 
is 4cos 60 + 9 – 7cos 25 = 465585… N

4 cos 60 + 9 N

7 cos 25 N

and the resultant force 
is (7sin 25 + 8) – 4sin 60 = 749423… N

7 sin 25 + 8 N

giving this diagram

14

4
...
49423

Answer: resultant is 882 N at an angle of 581o below the 9 N force
...

Solution: To sketch the vector polygon, draw the forces end to end
...

3N
3N
2N
48o

R1 N
4N

o

R2 N

4N

48o

2N

Combine the 3 N and 4 N forces to find the resultant R1 = 5 N with  = 36
...

It would probably be easier to resolve each force in two perpendicular directions as in the
previous example
...

If the sum of all the forces acting on a particle is zero (or if the resultant force is 0 N) then the
particle is said to be in equilibrium
...

Find the values of a and b
...

P+Q+R =0




Answer:

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a = –4 and b = –2

15

Example:
A particle is in equilibrium at O
under the forces shown in the diagram
...


60o

Solution:
Method (i)

P

12

First resolve Q in horizontal and vertical directions
Resolve   Q sin 60 = 12  Q = 13856
...


Q sin 60
Q cos 60

Answer P = 693 N and Q = 139 N

P

12

Method (ii)

P

Sketch a triangle of forces, showing the three forces adding up to 0
...
e
...

2) Non-contact forces: weight / gravity, magnetism, force of electric charges
...
B
...


16

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Friction
If we try to pull a box across the floor there is
a friction force between the box and the
floor
...


mg

When friction force is at its maximum and
the box is on the point of moving the box is
said to be in limiting equilibrium
...
B
...

Example: A particle of mass 2 kg rests in equilibrium on a plane which makes an angle of
25o with the horizontal
...

Solution: DRAW A DIAGRAM SHOWING ALL FORCES  the weight 2g N, the
friction F N and the normal reaction R N
...

Then draw a second diagram showing forces resolved along and perpendicular to the
slope
...


Friction force is 83 N and normal reaction is 18 N, to 2 S
...


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17

Coefficient of friction
...
The
ratio of this maximum friction force to the normal reaction between the surfaces is constant for
the two surfaces and is called the coefficient of friction
...

Example: A particle of mass 3 kg lies in equilibrium on a slope of angle 25o
...

Solution:

R
R

3g sin 25
25o

Res

F

F
3g cos 25

3g

 R = 3g cos 25 = 26645…

If we assume that the particle is in equilibrium
Res

 F = 3g sin 25 = 12424…

But the maximum friction force is Fmax = R = 06  26645… = 15987… N
Thus the friction needed to prevent sliding is 12
...
424… N
...
F
...

Example:
A particle of mass 6 kg on a slope of angle 30 o is being pushed by a horizontal
force of P N
...

Solution:

DRAW A DIAGRAM SHOWING ALL FORCES

As the particle is on the point of moving up the slope the friction force will be acting down
the slope, and as the particle is in limiting equilibrium the friction force will be at its
maximum or limiting value, F = R
...
to 2 S
...


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19

5
...


Newton’s laws of motion
...

2) For a particle with constant mass, m kg, the resultant force F N acting on the particle and
its acceleration a m s2 satisfy the equation F = m a
...


Example: A box of mass 30 kg is being pulled along the ground by a horizontal force of
95 N
...

Solution:

DRAW A DIAGRAM SHOWING ALL FORCES

No need to resolve as the forces
are already at 90o to each other
...
F
...
The tension in the string is 30 N
...

Solution:

DRAW A DIAGRAM SHOWING ALL FORCES

Resolve upwards  30 – 2g = 2a


30

a = 52
a

Answer Acceleration is 52 m s2 upwards
...


20

2g

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Example:
A particle of mass 25 kg is being pulled up a slope at angle of 25o above the
horizontal by a rope which makes an angle of 15o with the slope
...

Solution:
300

R

R + 300 sin 15
15o

a

a
F

300 cos 15

25o

F + 25g sin 25

25o

25g cos 25
25g

 R + 300 sin 15 = 25g cos 25

Res

 R = 14439969…

and, since moving, friction is maximum  F = R =

Res

,

 14439969 = 360999…

F = ma
 300 cos 15  (F + 25g sin 25) = 25a


a = 600545…

Answer the acceleration is 6
...
F
...
Then put in all forces on each particle: don’t forget Newton’s third law there
will be some ‘equal and opposite’ pairs of forces
...
If the
tension in the lift cables is 7000 N find the acceleration of the lift and the force between
the floor and the man’s feet
...

Draw a second diagram showing the combined system, that is, the man and the lift as
‘one particle’
...
B
...


Hint: Draw a LARGE lift and put the man in the middle (not touching the floor)
...
F
...
F
...
The driving force
exerted by the truck is 1500 N and there is no resistance to motion
...

Solution: Draw a clear diagram, separating the truck and the trailer to show the forces on
each one
...

If the force in the tow bar is T N then this force will be pulling the trailer and pulling
back on the truck, since the truck is accelerating
...

For the truck and trailer combined
Res , F = ma 

1500 = 2000a  a = 075 m s2

For the trailer only
Res , F = ma 

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T = 700a  T = 700 × 075 = 525 = 530 N

to 2 S
...


23

Particles connected by pulleys:
The string will always be inextensible and light and the pulley (or peg) will always be smooth
or light
...

2) As the string is light, the tension in the string will be constant along its length
...

Example: Particles of mass 3 kg and 5 kg are attached to the ends of a light inextensible
string which passes over a fixed smooth pulley
...

The system is released from rest; find the greatest height of the lighter mass above its initial
position in the subsequent motion
...

Solution:
The two particles will move together until the heavier one hits
the floor
...

Thus, our problem has two distinct parts
...

Since the string is light and the pulley is smooth the tensions on both sides will be equal in
magnitude
...

Both particles will have travelled 2 m and so, considering the 3 kg particle,
+ u = 0, a = 025g, s = 2, v = ? so using
 v2 = 2  025g  2 = g 

24

v2 = u2 + 2as

v =

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2) The remaining motion takes place freely under gravity as the string will have become
slack when the heavier mass hit the floor!
For the 3 kg mass
+ u =

, a = g, v = 0, s = ?

 0 = g + 2  g  s



so using v2 = u2 + 2as

s = 05

The 3 kg mass travelled 2 m before the 5 kg mass hit the floor and then moved up a further
05 m after the string became slack
...
5 m above its initial position
...
The block is attached to an inextensible, light string which passes over a light,
smooth pulley
...
The
coefficient of friction between the block and the slope is
...

(a) What can you assume because the string is light and inextensible?
(b) What can you assume because the pulley is light and smooth?
(c) Find the magnitude of the force exerted on the pulley by the string
...

(b) Because the pulley is light and smooth, the tensions in the string on either side of the
pulley will be equal
...
So we first find the tensions
...
F
...

(a) We know that the velocity v m s1 of a body of mass m kg moving with a constant
acceleration a m s2 for time t seconds is given by
v = u + at, where u is the initial velocity
...

Newton's Second Law states that F = ma


Ft = mat



Ft = mv  mu
...

(b) We define the impulse, I, of a constant force F N acting for a time t seconds to be
I = Ft Newton-seconds (Ns)
...

(d) The equation I = Ft = mv  mu of paragraph (a) can now be thought of as
Impulse = Change in Momentum
...
B
...

Example: A ball of mass 2 kg travelling in a straight line at 4 m s1 is acted on by a constant
force of 3 N acting in the direction of motion
...

Solution:
 +

4 m s1

v m s1

t=0

t=5

3N

The impulse of the force is 3  5 = 15 Ns in the direction of motion
...

Using I = mv – mu we have 15 = 2v – 2  4


v = 115

Answer: speed after 5 seconds is 115 m s1
...
Before the impulse the ball is travelling at 16 m s1 and the impulse of the bat on
the ball is 50 Ns
...

Solution:
Take the direction of motion of the ball as positive and let the speed after impact
be x m s1
...


Answer:

velocity after impact is 17

m s1 away from the bat
...

(a) If a hockey ball is hit by a hockey stick then the impulse on the ball is an external impulse
on the ball
...

If we were considering just one ball then the impulse of collision would be external to that
ball
...

If we were considering the satellite alone then the impulse of the explosion would be
external to the satellite
...

If there are no external impulses acting on a system then the total momentum of that system is
conserved (i
...
remains the same at different times)
...

Note that if there is an external impulse acting on the system then the momentum
perpendicular to that impulse is conserved
...

CLM  m1u1 + m2 u2 = m1v1 + m2v2
Example: A railway truck of mass 1500 kg is travelling in a straight line at 3 m s1
...
They collide (without
breaking up) and couple together
...

ALWAYS DRAW A DIAGRAM  before and after (and sometimes during)
You must always choose which direction is positive, then take note of the directions of the
arrows in your diagram
...


Answer Speed is 02 m s1 in the direction of the 1000 kg truck's initial velocity
...

After impact A is now travelling in the opposite direction at 3 m s1, and B continues to
travel in its original direction but with speed 2 m s1
...
Find the mass of B
...


ALWAYS DRAW A DIAGRAM  before and after (and sometimes during)
before

after

5

6

2

m

A

B

A

3

2

2

m

B

+ 
Taking left as positive
No external impulse  momentum conserved
CLM  total momentum before = total momentum after
 2  (5) + m  6 = 2  3 + m  2

Answer

30

4m = 16

 m=4

Mass of ball B is 4 kg
...

In this case the impulses on the two particles will be equal in magnitude but opposite in
direction
...

The assumptions involved are that the string is light (mass can be ignored) and inextensible
(does not stretch)
...
They are moving away from each other with speeds uP = 3 m s1 and
uQ = 4 m s1
...

(a) Find this common velocity
...

Solution:

ALWAYS DRAW A DIAGRAM  before and after and this time during

Let common speed after the string has become taut be v
before

P

during

3

4

2

5

Q

I
P

2

I

after

5

Q

P

v

v

2

5

Q

Taking direction to the right as positive
 +
(a) No external impulse 

CLM  total momentum conserved



2  (3) + 5  4 = 2  v + 5  v



v=2

(b) To find impulse consider only one particle, P
...

For particle P using I = mv – mu


I = 2  v  2  (3)



I = 10

Answer:

but v = 2

Common speed is 2 m s1 and Impulse = 10 Ns

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31

6
...

Moments are measured in newton-metres, Nm and the sense - clockwise or anti-clockwise
should always be given
...

y

4N

3

3

7N

2
P
x
5

Taking moments about P clockwise
moment = 7  2  4  3 = 2 Nm clockwise
...

(ii) The sum of the moments of all the forces about any point must be zero
...

4N

FN
B

A

D

C

3m

2m

xm
7N

13 N

Solution:
(i)

Resolve 

 4 + F = 13 + 7 

(ii)

The sum of moments about any point must be zero
...


Non-uniform rods
A uniform rod has its centre of mass at its mid-point
...
g
...
Its centre of mass is
3 metres from A
...

A mass of 20 kg is placed at P so that the system is in equilibrium
...


R

Moments about the pivot

A

 1  25g = x  20g


x = 125

G
3m



M
1m

xm

AP = 525 m
...


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33

Nearly tilting rods
If a rod is supported at two points A and B then when the rod is about to tilt about B the normal
reaction at A will be 0
...

PQ = 2 m, PA = 06 m and AB =07 m
...

The rod is on the point of tilting about B: find the value of M
...

RA

RB
X

A

P
06 m

B

G
04 m

03 m

Q
05 m

8g

02 m

Mg

If the rod is on the point of tilting about B then the reaction at A will be 0


RA = 0
...

We could find the value of RB, but if we take moments about B the moment of RB is 0,
whatever the value of RB
...
8 kg
...
L
...

Two balls, masses m1 and m2, are moving with speeds u1 and u2
...

There are no external impulses acting on the system
Let the internal impulse between the balls be I, acting in opposite directions on each ball
...
L
...

Note that if there is an external impulse acting on the system then the momentum
perpendicular to that impulse is conserved

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