EVERY OBJECT REMAINS AT REST OR MOVES WITH CONSTANT
VELOCITY UNLESS ACTED UPON BY AN EXTERNAL FORCE
In other words, if you want an object to accelerate, you have to apply a force.

WHEN AN APPLIED FORCE CAUSES AN OBJECT TO ACCELERATE
• THE FORCE AND ACCELERATION ARE IN THE SAME DIRECTION
• THE MAGNITUDE OF THE FORCE IS PROPORTIONAL TO THE
MAGNITUDE OF THE ACCELERATION AND TO THE MASS OF THE
OBJECT
This law fits well with common sense:
• For a given object, a larger acceleration needs a larger force
• The more massive an object, the greater the force needed to produce a given acceleration.

Symbolically, Newton’s second law is
OR
The SI unit of force is the newton, which is defined as the force needed to accelerate 1kg of mass at 1ms -2 .
With this definition, the value of k is 1 and Newton’s second law becomes:
F = ma
This is the equation of motion of an object. In other words, resultant force equals mass times acceleration. It is very important to remember that F does stand for resultant force.

Example 1
The engine of a car of mass 900kg produces a driving force of 2000N. There are resistive forces of 650N. Find the acceleration of the car on a level road.
R
2000N 650N
900g
F = ma
2000 – 650 = 900a
1350 = 900a a = 1.5ms
-2

Example 2
A horizontal force of 50N is applied to a sledge of mass 20kg resting on level snow. The sledge accelerates at 2.2ms
-2 . Find the coefficient of friction between the sledge and the snow.
R a = 2.2ms
-2
μR 50N
Resolving vertically
R = 20g
20g
Resolving horizontally
F = ma
50 – 20gμ = 20 x 2.2
6 = 20gμ
μ = 0.0306

Example 3
An object of mass 10kg is acted upon by forces (3i + 6j)N, (2i – 3j)N and (i + 2j)N relative to some coordinate system. Find the magnitude and direction of the acceleration of the object.
F = ma
3i + 6j + 2i – 3j + i + 2j = 10a
6i + 5j = 10a
a = (0.6i + 0.5j)ms -2
0.5
|a|
θ
0.6
|a| = 0.6
2 + 0.5
2
|a| = 0.781ms -2
Tanθ =
0.5
0.6
θ = 39.8°

Example 4
A crane lifts a 120kg object on the end of its cable, which has negligible mass. At first the object accelerates at 2ms -2 . It then travels at a uniform speed and finally slows to rest with an acceleration of -1.2ms
-2 . Find the tension in the cable during each stage of the motion.
Constant
Accelerating Speed Decelerating
T T T
2ms -2 0ms -2 -1.2ms
-2
120g
F = ma
T – 120g = 120 x 2
T = 240 + 120g
T = 1416N
120g
T = 120g
T = 1176N
120g
F = ma
T – 120g = 120 x -1.2
T = 120g – 144
T = 1032N

Example 5
An object of mass 8kg is being towed by a light string up a slope inclined at 20° to the horizontal. The string is inclined at 30° to the slope. There is a frictional resistance of 40N.
The object accelerates up the slope at 0.8ms
-2 . Find:
(a) The tension in the string
(b) The normal reaction between the object and the slope.
T
0.8ms
40N
-2 R
20°
20°
8g
30°
(a) Resolve parallel to plane
F = ma
Tcos30 – 40 – 8gsin20 = 8 x 0.8
Tcos30 = 6.4 + 40 + 8gsin20
T = 84.5N
(b) Resolve perpendicular to plane
R + Tsin30 = 8gcos20
R = 8gcos20 – Tsin30
R = 31.4N

Example 6
A block of mass 5kg moves on a rough horizontal plane with coefficient of friction 0.2 under the action of a horizontal force of 30N. If the block starts from rest, find the distance it travels in the first 3 seconds of motion.
R a = ?ms
-2
Resolving vertically
R = 5g
0.2R
30N
Resolving horizontally
F = ma
5g
30 – 0.2 x 5g = 5a a = 4.04ms
2 s = ?
u = 0ms -1 v = s = ut + ½ at 2 s = 0 x 3 + ½ x 4.04 x 3 2 a = 4.04ms
-2 t = 3s s = 18.18m

Example 7
A particle of mass 6kg is initially moving with a speed of 8ms -1 on a rough horizontal surface with a coefficient of friction 0.25. Find the distance it moves across the rough surface before coming to rest.
R a = ?ms
-2 u = 8ms -1
Resolving vertically
R = 6g
0.25R
Resolving horizontally
F = ma
6g
– 0.25 x 6g = 6a a = -2.45ms
2 s = ?
u = 8ms -1 v = v 2 = u 2 + 2as
0ms -1
0 = 64 + 2(-2.45)s s = 13.06m
a = -2.45ms
-2 t =

FOR EVERY ACTION THERE IS AN EQUAL AND OPPOSITE REACTION
In other words, if an object A exerts a force on a second object B (by direct contact or at a distance by magnetic attraction, gravitation etc.) then B will exert a force on A. The two forces will be of equal magnitude and in opposite directions.
If A and B are parts of the same system, the force of A on B and the force of B on A cancel out. They are forces internal to the system and do not affect the acceleration of the system. They are only taken into account if you want to calculate the acceleration of object
A (or B) alone.

Example 8
A man of mass 90kg stands in a lift of mass 300kg. The cable of the lift has a tension of
4056N. Find the reaction between the man and the floor of the lift.
4056N System as a whole
4056N
Forces on the man
R
R
90g
300g a ms -2 a ms -2
390g
F = ma
4056 – 390g = 390a a = 0.6ms
2
90g
F = ma
R – 90g = 90 x 0.6
R = 936N
0.6ms
-2

Example 9
An engine of mass 10 tonnes is pulling a truck of mass 3 tonnes. The resistance on the engine and the truck are 4000N and 1500N respectively. The driving force of the engine is
14000N. Find the acceleration of the system and the tension in the coupling between the engine and the truck.
1500N
R
T
4000N
R
E a
(a) System as a whole
R a
14000N
T T
5500N 14000N
Truck
1500N
3000g 10000g
13000g
F = ma
R
T
3000g
0.654ms
-2
T
F = ma
T - 1500 = 3000 x 0.654
14000 – 5500 = 13000a a = 0.654ms
-2
T = 3462N

Example 9 involved two objects connected together. You could call them connected particles. This term is also used to describe objects connected by strings passing over pulleys or other supports.
For problems involving pulleys, the usual modelling assumptions are that:
• The objects are particles, so you can ignore air resistance and their dimensions
• The string is light, so its weight can bee ignored
• The string is inextensible, so the two ends of the string have the same speed and acceleration
• The pulley is smooth, so the tension is the same throughout the string
• The pulley is light, so no force is required to turn it.

Example 10
Particles of mass 3kg and 5kg are attached to the ends of a light, inextensible string passing over a smooth pulley. The system is released from rest. Find the acceleration of the system and the tension in the string.
a
T T a
Considering 3kg particle
F = ma
T – 3g = 3a (1)
Considering 5kg particle
F = ma
5g – T = 5a (2)
3g 5g (1) + (2) 2g = 8a a = g /
4 ms -2
Sub into (1)
T – 3g = 3( g /
4
)
T = 15g /
4
N

Example 11
A block of mass 4kg rests on a rough horizontal table, with coefficient of friction 0.5. It is attached by a light, inextensible string to a particle of mass 9kg. The string passes over a smooth pulley at the edge of the table and the 9kg mass hangs freely. Find the acceleration of the system, the tension in the string and the resultant force acting on the pulley.
0.5R
R
4g
T a
9g
T
Considering 9kg mass
F = ma
9g – T = 9a (1)
Considering 4kg (vertically)
R = 4g
Considering 4kg (horizontally)
F = ma
T – 0.5R = 4a
T – 2g = 4a (2)
(1) + (2) 7g = 13a a = 7g /
13 ms -2
Sub into (2) T – 2g = 4( 7g /
13
)
T = 54g /
13
N

Example 11
A block of mass 4kg rests on a rough horizontal table, with coefficient of friction 0.5. It is attached by a light, inextensible string to a particle of mass 9kg. The string passes over a smooth pulley at the edge of the table and the 9kg mass hangs freely. Find the acceleration of the system, the tension in the string and the resultant force acting on the pulley.
T 54g /
13
N
T
𝑇
𝑅
=
T
R
54g /
13
N
54𝑔
13
2
+
54𝑔
13
2
𝑻
𝑹
= 𝟓𝟕. 𝟔𝑵

Example 12
R
A
A
2kg
30°
T a
3m
2g
(a) Considering A
(parallel to plane)
F = ma
T – 2gsin30 = 2a
T = g + 2a
T
3kg B
3g
1m
The diagram shows a block, A of mass 2kg, 3m from the top of a smooth slope inclined at 30° to the horizontal. It is connected to a block B, of mass 3kg, by a light inextensible string passing over a smooth pulley at the top of the slope. Block B hangs freely a distance of 1m above a horizontal plane. The system is released from rest
Find:
(a) The speed with which B hits the plane
(b) How close A gets to the top of the slope.
Considering B
F = ma
3g – T = 3a
3g – g – 2a = 3a a = 0.4g
a = 0.4g
u = 0 s = 1 v = ?
v 2 = u 2 + 2as v 2 = 0 + 2(0.4g)(1) v = 2.8ms
-1

Example 12
A
2kg
30°
3m
(b) R
A a = ?
1m
30° u = 2.8ms
-1 v = 0 s = ?
2g
3kg B
1m
The diagram shows a block, A of mass 2kg, 3m from the top of a smooth slope inclined at 30° to the horizontal. It is connected to a block B, of mass 3kg, by a light inextensible string passing over a smooth pulley at the top of the slope. Block B hangs freely a distance of 1m above a horizontal plane. The system is released from rest
Find:
(a) The speed with which B hits the plane
(b) How close A gets to the top of the slope.
F = ma
-2gsin30 = 2a a = -½g v 2 = u 2 + 2as
0 = 2.8
2 + 2(- ½g)s s = 0.8m
Distance from top = 1.2m