In this lecture the following are introduced:

•Velocity and speed
• Momentum as the quantity of motion
• Newton's laws of motion
• Internal and external forces

Motion

Speed is obviously a part of motion. Now since the direction of movement will be important, we need to define the related vector quantity called velocity. Velocity is speed in a given direction. To define a velocity, we take a unit vector in the required direction and multiply it by its scalar speed.

The bold face, , indicates that velocity is a vector. The normal face, , indicates that its size (or magnitude) is a scalar. The bold face with a "hat", , is a vector of unit length in the required direction. The notation thus says, that the velocity vector has a scalar size in the direction of the unit vector .

Biographical information on Newton
The poet Alexander Pope, who lived in Newton's time, wrote:
Nature and Nature's laws lay hid in night
God said:"Let Newton be!" and all was light.

How much motion does a body have?
Two bodies moving with the same velocity can produce different effects because of their different masses.
Demonstration: a ping-pong ball and fire-extinguisher, falling from the same height onto a suspended horizontal thin sheet of glass. The ping-pong ball will bounce and the fire-extinguisher will break the glass. The different effect shows that the amount of motion a system has needs to include both the mass and the velocity.

Momentum

Newton defined momentum as the quantity of motion that a system has. (To be precise, Newton called it the quality of motion). From Newton, the momentum of a system is the product of its mass and velocity.

Momentum has the unit of kg.m.s^-1.

Newton's laws of motion

It is not modern scientific custom to speak of a "law of Nature", because the nature of law is that it can be broken. The paradigms or symmetries of nature are not in principle breakable, however most textbooks seem to be happy to write of Newton's "laws" of motion.

Newton's first law of motion.

Bodies remain in a state of rest, or of constant speed in a straight line, unless compelled to change by a push or a pull.
In other words, if things don't change they'll probably remain the same! Less colloquially, with no net force being applied, the momentum remains constant. We will come back to this later, but the converse of this leads to…

Newton's second law of motion

The size of the force on a system is measured by how quickly its momentum changes.

This teaches us how to catch fast moving cricket balls and how to jump off tables. The secret is to increase the time during which the event happens, by letting your hands give way and your knees to bend.

A Russian schoolteacher, Konstantin Tsiolkovsky (1857-1935), investigated what would happen to Newton's law if the mass changed. He became the father of modern rocketry.

However, in this course, we will look at situations where the mass remains constant, so we end up with:

      i.e.  

The force here is the resultant (or "net") force acting on the mass.
The unit of force is the Newton. 1 Newton gives a body of mass 1 kg an acceleration of 1 m.s^-2.

Newton's third law of motion

This is about the origin of contact force. To every action there is an equal and opposite action, i.e. an equal reaction.
It is often stated, "to every action there is an equal and opposite reaction", but this is a double negative - OK, for Shakespeare, but not for modern language. Actually we might say, "If you push me, I'll push you back". It would be interesting if this law was not true, because then we could sail yachts by the crew blowing on the sails.

Puzzle: The farmer and the horse.
A farmer hitches his horse to a cart and orders the horse to pull. The horse says "I have just learnt from Newton's third law, that any force I apply will create an equal but opposite force. If I pull on the cart (red arrow below) then the cart will pull back on me with equal force (mauve arrow). Two equal forces in opposite directions will cancel, so however hard I try it will not move".

The farmer tries a different tack by saying "but you are actually pushing on the ground". The horse replies "same argument, the force that I push against the ground (green arrow below) will be exactly matched by the force of the ground on me (brown arrow). Again there are equal forces in opposite directions which will cancel, so again I cannot move".

The farmer has accepted false logic. In mathematics you can only equate the same kinds of things, otherwise you end up with an argument like this.
½ full glass = ½ empty glass
Multiply both sides by 2
full glass = empty glass, i.e. 1 = 0.
Fullness and emptiness cannot be equated like that. They are different kinds of things.

Internal and external forces

The difficulty with the horse and cart is that the forces act on different things. The red arrow acts on the cart, the mauve arrow acts on the horse, the brown arrow acts on the horse and the green arrow acts on the ground. If you want to know what happens to the cart, then only consider the forces which act on the cart. When the horse and cart are taken to be a single system then the forces are internal and can be added together. In this case, the forces do cancel, which means that horse and cart are locked together.

Unbalanced external forces cause a system to accelerate. When the cart is considered as the system, then the horse supplies an external force (red arrow below) which causes the cart to move.
When the horse is considered as the system, the the ground supplies an external force (brown arrow) and the cart supplies a force (mauve arrow). If the brown arrow is greater than the mauve arrow then there is a resultant external force on the horse which causes the horse to move.

With this in mind, it is necessary to remember that
The force which causes motion of a system is the resultant external force on that system.

Graphical Representation

Using exactly the same ideas as were used in kinematics, but with velocity and acceleration multipled by mass, (momentum = m·v force = m·a) we have the following:

• On the momentum vs time graph, the instantaneous force is given by how quickly the momentum is changing. This is written as . The force is given by the blue slope (or tangent) at that time. It is also the size of the blue line on the force vs time graph. • On the force vs time graph, the effect of force through time is to change the momentum. The effect of force through time is the red area under the graph, which is written as . The change in momentum is also the size of the red arrow on the momentum vs time graph, which is written as . The effect of force through time is to change the momentum, and this is written as .

In the following questions, the following procedure is used:
1. Draw a simple diagram of the situation.
2. For each mass draw all the forces which act on it.
3. For each mass: write resultant external force equals mass times acceleration.
4. Solve the equation(s) for each of the unknowns.

Worked Example

1. A woman of mass 60 kg stands on a weighing machine in a lift.
Find the mass that the machine displays when :-
(a) the lift is stationary.
(b) the lift is accelerating upwards at 2 m.s^-2.
(c) the lift is accelerating downwards at 2 m.s^-2.

The mass recorded on the scale equals the normal reaction of the floor but scaled in kg. i.e.

When the lift is stationary, there is the downward force of gravity on the woman (m·g). This creates an upward force from the floor onto the woman (the normal reaction), as recorded by the scales. With no acceleration, the forces cancel, and the normal reaction force equals the weight force. Thus mass recorded = m·g/g = m = 60 kg

When there is an upwards acceleration, the floor pushes up harder than the weight down.

  

When there is an downwards acceleration, the floor pushes up less hard than the weight down.

  

Questions

2. A block of mass 2 kg is at rest on a smooth horizontal table on the x-axis of a coordinate system and 10 m from the origin. A constant horizontal force of 5 N acts on the block in the positive x direction. Find
(a) the force of the table on the block. (ans 19.6 N up)
(b) the acceleration of the block. (ans 2.5 m.s^-2)
(c) the distance of the block from the origin 4 s after the application of the force. (ans 30 m)

3. A block of mass 7 kg is projected upwards along a smooth inclined plane inclined at 30° with an initial speed of 6 m.s^-1. Find
(a) the deceleration of the mass up the plane. (ans 4.9 m.s^-2)
(b) how far up the plane the mass rises before (momentarily) coming to rest. (ans 3.65 m)

4. A 2 kg mass and a 1 kg mass lie in contact on a smooth horizontal plane. A constant horizontal force of 6 N presses on the 2 kg mass in the direction of the 1 kg mass. Find
(a) the values of all the forces. (ans F₁₂ = F₂₁ = 2 N)
(b) the common acceleration of the blocks. (ans 2 m.s^-2)

5. A 6 kg mass and a 4 kg mass are tied together with a light inexstensible string and suspended around a fixed pulley. The system is at rest at the moment it is released. Find
(a) the acceleration of the masses after release. (ans 1.96 m.s^-2)
(b) the tension in the string after release. (ans 47 N)

6. A 7 kg block lies on a smooth horizontal table. A light inextensible string attached to the block passes over a light pulley at the edge of the table and a 3 kg mass is hung on the other end of the string. Assume that the system is at rest at the moment it is released. Find
(a) the tension in the string after release. (ans 20.6 N)
(b) how fast the system is moving 5 s after release. (ans 14.7 m.s^-1)

7. An 18 kg block lies on a smooth plane which is inclined upwards at an angle of 30° to the horizontal. A light inextensible string attached to the block passes up over a light pulley at the top of the plane and then down to a 7 kg mass at the other end. Assume that the system is at rest at the moment it is released. Find
(a) the acceleration of the system after release. (ans 0.78 m.s^-2)
(b) the tension in the string after release. (ans 74 N)

8. A 6 kg block lies on a smooth plane which is inclined upwards at an angle of 30° to the horizontal. A light inextensible string attached to the block passes up over a light pulley at the top of the plane and then down to a 4 kg mass at the other end. Assume that the smaller mass is initially moving upwards at 2 m.s(-1.
Find the time for the 4 kg mass to come (momentarily) to rest. (ans 2.04 s)

9. A 7 kg block lies at rest on a rough horizontal table. The table is raised slowly on one side till it makes an angle of 36°52' with the horizontal. At this angle the block starts to move with constant speed down the table.
Find the coefficient of friction for the block on the plane. (ans 0.75)

10. A 9 kg block lies on a rough plane (coefficient of friction 0.09) which is inclined upwards at an angle of 30° to the horizontal. A light inextensible string attached to the block passes up over a light pulley at the top of the plane and then down to a 0.8 kg mass at the other end. Assume that the system is at rest at the moment it is released. Find
(a) the acceleration of the system after release. (ans 3 m.s^-2)
(b) the tension in the string after release. (ans 10.2 N)

11. A mass of 6 kg is on a rough horizontal table (coefficient of friction 0.5). A light inextensible string is attached to the mass and passes over a light pulley to a second mass of 3.8 kg. The system moves from rest under the influence of gravity and travels 2.5 m.
Find the speed of the system after travelling this distance. (ans 2 m.s^-1)

Summarising:

The velocity of a system is its speed in a given direction.  
The momentum of a system is the product of its mass and velocity.  

Newton's laws of motion:
1. Bodies remain in a state of rest, or of constant speed in a straight line, unless compelled to change by a push or a pull.
2. The size of the force on a system is measured by how quickly its momentum changes.  
3. To every action there is an equal and opposite action, i.e. an equal reaction.
Internal forces are one part of a system acting on another part of the system, they do not cause motion, they keep a system together.
The force which causes the motion of a system is the resultant external force.
The effect of resultant external force through time is to change the momentum: