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MTH5P2B: Newton’s laws of motion

This unit is about Newton's laws of motion and their application

Forces and Newton’s laws of motion
Nature and Nature’s Laws lay hid in Night.
God said, Let Newton be! and All was Light.

The picture shows crates of supplies being dropped into a remote area by parachute. What forces are acting on a crate of supplies and the parachute?

One force which acts on every object near the earth’s surface is its own weight. This is the force of gravity pulling it towards the centre of the earth. The weight of the crate acts on the crate and the weight of the parachute acts on the parachute.

The parachute is designed to make use of air resistance. A resistance force is present whenever a solid object moves through a liquid or gas. It acts in the opposite direction to the motion and depends on the speed of the object.

The crate also experiences air resistance, but to a lesser extent than the parachute. Other forces are the tensions in the guy lines attaching the crate to the parachute. These pull upwards on the crate and downwards on the parachute. All these forces can be shown most clearly if you draw force diagrams for the crate and the parachute.

Force diagrams are essential for the understanding of most mechanical situations.
A force is a vector: it has a magnitude, or size, and a direction. It also has a line of action. This line often passes through a point of particular interest. Any force diagram should show clearly
● the direction of the force
● the magnitude of the force
● the line of action.

Example: A body of mass 5 kg is acted upon by two forces of 6N and 8N which are perpendicular. Find the resultant force and the direction that it makes with the horizontal.

 

In figures 3.1 and 3.2 each force is shown by an arrow along its line of action. The air resistance has been depicted by a lot of separate arrows but this is not very satisfactory. It is much better if the combined effect can be shown by one arrow.

When you have learned more about vectors, you will see how the tensions in the guy lines can also be combined into one force if you wish. The forces on the crate and parachute can then be simplified.

Centre of mass and the particle model
When you combine forces you are finding their resultant. The weights of the crate and parachute are also found by combining forces; they are the resultant of the weights of all their separate parts. Each weight acts through a point called the centre of mass or centre of gravity.

Think about balancing a pen on your finger. The diagrams show the forces acting on the pen.

So long as you place your finger under the centre of mass of the pen, as in figure 3.5, it will balance. There is a force called a reaction between your finger and the pen which balances the weight of the pen. The forces on the pen are then said to be in equilibrium. If you place your finger under another point, as in figure 3.6, the pen will fall. The pen can only be in equilibrium if the two forces have the same line of action.

If you balance the pen on two fingers, there is a reaction between each finger and the pen at the point where it touches the pen. These reactions can be combined into one resultant vertical reaction acting through the centre of mass.

The behaviour of objects which are liable to rotate under the action of forces is covered in Mechanics 2 Chapter 11. In Mechanics 1 you will only deal with situations where the resultant of the forces does not cause rotation. An object can then be modelled as a particle, that is a point mass, situated at its centre of mass.

Newton’s third law of motion

Sir Isaac Newton (1642–1727) is famous for his work on gravity and the mechanics you learn in this course is often called Newtonian Mechanics because it is based entirely on Newton’s three laws of motion. These laws provide us with an extremely powerful model of how objects, ranging in size from specks of dust to planets and stars, behave when they are influenced by forces.

We start with Newton’s third law which says that
● When one object exerts a force on another there is always a reaction of the same kind which is equal, and opposite in direction, to the acting force.

You might have noticed that the combined tensions acting on the parachute and the crate in figures 3.3 and 3.4 are both marked with the same letter, T. The crate applies a force on the parachute through the supporting guy lines and the parachute applies an equal and opposite force on the crate.

When you apply a force to a chair by sitting on it, it responds with an equal and opposite force on you. Figure 3.8 shows the forces acting when someone sits on a chair.

The reactions of the floor on the chair and on your feet act where there is contact with the floor. You can use R1, R2 and R3 to show that they have different magnitudes. There are equal and opposite forces acting on the floor, but the forces on the floor are not being considered and so do not appear here.

Why is the weight of the person not shown on the force diagram for the chair?

Gravitational forces obey Newton’s third law just as other forces between bodies.
According to Newton’s universal law of gravitation, the earth pulls us towards its centre and we pull the earth in the opposite direction. However, in this book we are only concerned with the gravitational force on us and not the force we exert on the earth.
All the forces you meet in mechanics apart from the gravitational force are the result of physical contact. This might be between two solids or between a solid and a liquid or gas.

Friction and normal reaction
When you push your hand along a table, the table reacts in two ways.
● Firstly there are forces which stop your hand going through the table. Such forces are always present when there is any contact between your hand and the table. They are at right angles to the surface of the table and their resultant is called the normal reaction between your hand and the table.
● There is also another force which tends to prevent your hand from sliding. This is the friction and it acts in a direction which opposes the sliding.

 

Figure 3.9 shows the reaction forces acting on your hand and on the table. By Newton’s third law they are equal and opposite to each other. The frictional force is due to tiny bumps on the two surfaces (see electronmicrograph below). When you hold your hands together you will feel the normal reaction between them. When you slide them against each other you will feel the friction.

When the friction between two surfaces is negligible, at least one of the surfaces is said to be smooth. This is a modelling assumption which you will meet frequently in this book. Oil can make surfaces smooth and ice is often modelled as a smooth surface.
When the contact between two surfaces is smooth, the only forces between them are normal reactions which act at right angles to any possible sliding.
What direction is the reaction between the sweeper’s broom and the smooth ice?
A TV set is standing on a small table. Draw a diagram to show the forces acting on the TV and on the table as seen from the front.
SOLUTION
The diagram shows the forces acting on the TV and on the table. They are all vertical because the weights are vertical and there are no horizontal forces acting.
Draw diagrams to show the forces acting on a tennis ball which is hit downwards across the court
(i) at the instant it is hit by the racket
(ii) as it crosses the net
(iii) at the instant it lands on the other side.
SOLUTION

Force and motion
How are the rails and handles provided in buses and trains used by standing passengers?
Newton’s first law
Newton’s first law can be stated as follows.
● Every particle continues in a state of rest or uniform motion in a straight line unless acted on by a resultant external force.
Newton’s first law provides a reason for the handles on trains and buses. When you are on a train which is stationary or moving at constant speed in a straight line you  can easily stand without support. But when the velocity of the train changes, a force is required to change your velocity to match.
This happens when the train slows down or speeds up. It also happens when the train goes round a bend even if the speed does not change. The velocity changes because the direction changes.
Why is Josh’s car in the pond?
A coin is balanced on your finger and then you move it upwards.
By considering Newton’s first law, what can you say about W and R in each of these situations?
(i) The coin is stationary.
(ii) The coin is moving upwards with a constant velocity.
(iii) The speed of the coin is increasing as it moves upwards.
(iv) The speed of the coin is decreasing as it moves upwards.
SOLUTION
(i) When the coin is stationary the velocity does not change. The forces are in equilibrium and R = W.
(ii) When the coin is moving upwards with a constant velocity the velocity does not change. The forces are in equilibrium and R = W.
(iii) When the speed of the coin is increasing as it moves upwards there must be a net upward force to make the velocity increase in the upward direction so R > W. The net force is R − W.
(iv) When the speed of the coin is decreasing as it moves upwards there must be a net downward force to make the velocity decrease and slow the coin down as it moves upwards. In this case W > R and the net force is W−R.
Driving forces and resistances to the motion of vehicles
In problems about such things as cycles, cars and trains, all the forces acting along the line of motion will usually be reduced to two or three: the driving force forwards, the resistance to motion (air resistance, etc.) and possibly a braking force backwards.
Resistances due to air or water always act in a direction opposite to the velocity of a vehicle or boat and are usually more significant for fast-moving objects.
Tension and thrust
The lines joining the crate of supplies to the parachute described at the beginning of this chapter are in tension. They pull upwards on the crate and downwards on the parachute. You are familiar with tensions in ropes and strings, but rigid objects can also be in tension.
When you hold the ends of a pencil, one with each hand, and pull your hands apart, you are pulling on the pencil. What is the pencil doing to each of your hands? Draw the forces acting on your hands and on the pencil.
Now draw the forces acting on your hands and on the pencil when you push the pencil inwards.
Your first diagram might look like figure 3.18. The pencil is in tension so there is an inward tension force on each hand.
When you push the pencil inwards the forces on your hands are outwards as in figure 3.19. The pencil is said to be in compression and the outward force on each hand is called a thrust.
If each hand applies a force of 2 units on the pencil, the tension or thrust acting on each hand is also 2 units because each hand is in equilibrium.

Which of the above diagrams is still possible if the pencil is replaced by a piece of string?
Resultant forces and equilibrium
You have already met the idea that a single force can have the same effect as several forces acting together. Imagine that several people are pushing a car. A single rope pulled by another car can have the same effect. The force of the rope is equivalent to the resultant of the forces of the people pushing the car. When there is no
resultant force, the forces are in equilibrium and there is no change in motion.
Example 3.4 A car is using a tow bar to pull a trailer along a straight, level road. There are resisting forces R acting on the car and S acting on the trailer. The driving force of the car is D and its braking force is B.
Draw diagrams showing the horizontal forces acting on the car and the trailer
(i) when the car is moving at constant speed
(ii) when the speed of the car is increasing
(iii) when the car brakes and slows down rapidly.
In each case write down the resultant force acting on the car and on the trailer.
SOLUTION
(i) When the car moves at constant speed, the forces are as shown in figure 3.20 (overleaf). The tow bar is in tension and the effect is a forward force on the trailer and an equal and opposite backward force on the car.
There is no resultant force on either the car or the trailer when the speed is constant; the forces on each are in equilibrium.
For the trailer: T − S = 0
For the car: D − R − T = 0
(ii) When the car speeds up, the same diagram will do, but now the magnitudes of the forces are different. There is a resultant forward force on both the car and the trailer.
For the trailer: resultant = T – S
For the car: resultant = D – R – T
(iii) When the car brakes a resultant backward force is required to slow down the trailer. When the resistance S is not sufficiently large to do this, a thrust in the tow bar comes into play as shown in the figure 3.21.
Newton’s second law
Newton’s second law gives us more information about the relationship between the magnitude of the resultant force and the change in motion. Newton said that
● The change in motion is proportional to the force.
For objects with constant mass, this can be interpreted as the force is proportional to the acceleration.
Resultant force = a constant × acceleration
The constant in this equation is proportional to the mass of the object: a more massive object needs a larger force to produce the same acceleration. For example, you and your friends would be able to give a car a greater acceleration than you would be able to give a lorry.
Newton’s second law is so important that a special unit of force, the newton (N), has been defined so that the constant in equation (1) is actually equal to the mass.
A force of 1 newton will give a mass of 1 kilogram an acceleration of 1 m s–2. The equation then becomes:
Resultant force = mass × acceleration
This is written: F = ma
The resultant force and the acceleration are always in the same direction.
Relating mass and weight
The mass of an object is related to the amount of matter in the object. It is a scalar. The weight of an object is a force. It has magnitude and direction and so is a vector.
The mass of an astronaut on the moon is the same as his mass on the earth but his weight is only about one-sixth of his weight on the earth. This is why he can bounce around more easily on the moon. The gravitational force on the moon is less because the mass of the moon is less than that of the earth.
When Buzz Aldrin made the first landing on the moon in 1969 with Neil Armstrong, one of the first things he did was to drop a feather and a hammer to demonstrate that they fell at the same rate. Their accelerations due to the gravitational force of the moon were equal, even though they had very different masses. The same is true on earth. If other forces were negligible all objects would fall with an acceleration g.
When the weight is the only force acting on an object, Newton’s second law means that
Even when there are other forces acting, the weight can still be written as mg. A good way to visualise a force of 1 N is to think of the weight of an apple. 1 kg of apples weighs approximately (1 × 10) N = 10 N. There are about 10 small to medium-sized apples in 1 kg, so each apple weighs about 1 N.


Note
Anyone who says 1 kg of apples weighs 1 kg is not strictly correct. The terms weight and mass are often confused in everyday language but it is very important for your study of mechanics that you should understand the difference.

What is the weight of
(i) a baby of mass 3 kg
(ii) a golf ball of mass 46 g?
SOLUTION
Most weighing machines have springs or some other means to measure force even though they are calibrated to show mass. Would something appear to weigh the same on the moon if you used one of these machines? What could you use to find the mass of an object irrespective of where you measure it?
Pulleys
In the remainder of this chapter weight will be represented by mg. You will learn to apply Newton’s second law more generally in the next chapter.
A pulley can be used to change the direction of a force; for example it is much easier to pull down on a rope than to lift a heavy weight. When a pulley is well designed it takes a relatively small force to make it turn and such a pulley is modelled as being smooth and light. Whatever the direction of the string passing over this pulley, its tension is the same on both sides.
Figure 3.22 shows the forces acting when a pulley is used to lift a heavy parcel.
Note
The rope is in tension. It is not possible for a rope to exert a thrust force.
In this diagram the pulley is smooth and light and the 2 kg block, A, is on a rough surface.
(i) Draw diagrams to show the forces acting on each of A and B.
(ii) If the block A does not slip, find the tension in the string and calculate the magnitude of the friction force on the block.
(iii) Write down the resultant force acting on each of A and B if the block slips and accelerates.


SOLUTION

Note
The masses of 2 kg and 5 kg are not shown in the force diagram. The weights 2g N and 5g N are more appropriate.
(ii) When the block does not slip, the forces on B are in equilibrium so
The tension throughout the string is 5g N.
For A, the resultant horizontal force is zero so
The friction force is 5g N towards the left.
(iii) When the block slips, the forces are not in equilibrium and T and F have different magnitudes.
The resultant horizontal force on A is (T – F) N towards the right.
The resultant force on B is (5g – T) N vertically downwards.
Reviewing a mathematical model: air resistance
In mechanics you express the real world as mathematical models. The process of modelling involves the cycle shown in Figure 3.25 and this is used in the example that follows.
Why does a leaf or a feather or a piece of paper fall more slowly than other objects?
Model 1: The model you have used so far for falling objects has assumed no air resistance and this is clearly unrealistic in many circumstances. There are several possible models for air resistance but it is usually better when modelling to try simple models first. Having rejected the first model you could try a second one as follows.
Model 2: Air resistance is constant and the same for all objects.
Assume an object of mass m falls vertically through the air.
The model predicts that a heavy object will have a greater acceleration than a
This seems to agree with our experience of dropping a piece of paper and a book, for example. The heavier book has a greater acceleration.
However, think again about air resistance. Is there a property of the object
other than its mass which might affect its motion as it falls? How do people and
animals maximise or minimise the force of the air?
Try dropping two identical sheets of paper from a horizontal position, but fold one of them. The folded one lands first even though they have the same mass.
This contradicts the prediction of model 2. A large surface at right angles to the motion seems to increase the resistance.
Model 3: Air resistance is proportional to the area perpendicular to the motion.
Assume the air resistance is kA where k is constant and A is the area of the surface perpendicular to the motion.
According to this model, the acceleration depends on the ratio of the area to the mass.
Experiment
Testing the new model
For this experiment you will need some rigid corrugated card such as that used for packing or in grocery boxes (cereal box card is too thin), scissors and tape.
Cut out ten equal squares of side 8 cm. Stick two together by binding the edges with tape to make them smooth. Then stick three and four together in the same way so that you have four blocks A to D of different thickness as shown in the diagram.
Cut out ten larger squares with 12 cm sides. Stick them together in the same way to make four blocks E to H.
Observe what happens when you hold one or two blocks horizontally at a height of about 2 m and let them fall. You do not need to measure anything in this experiment, unless you want to record the area and mass of each block, but write down your observations in an orderly fashion.
1 Drop each one separately. Could its acceleration be constant?
2 Compare A with B and C with D. �Make sure you drop each pair from the same height and at the same instant of time. Do they take the same time to fall? Predict what will happen with other combinations and test your predictions.
3 Experiment in a similar way with E to H.
4 Now compare A with E, B with F, C with G and D with H. Compare also the two blocks whose dimensions are all in the same ratio, i.e. B and G.
Do your results suggest that model 3 might be better than model 2?
If you want to be more certain, the next step would be to make accurate measurements. Nevertheless, this model explains why small animals can be relatively unscathed after falling through heights which would cause serious
injury to human beings.
All the above models ignore one important aspect of air resistance. What is that?

 
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