In the lectures up till now we have been discussing kinematics, the description of motion. Now we start to explore the dynamics of motion, introducing the key concept of force and Newton's Laws of Motion.

*Godfrey Kneller's 1689 portrait of Isaac Newton
* (From Wikipedia)

What is a force?

Not this.

In a somewhat circular definition, a force is something that when applied to an object can cause it to change it's motion. As we'll see however it is the **sum** of forces on an object that actually determines it's motion.

Force is a vector, and so those sums will be vector sums!

Is motion of an object fundamentally different to it being at rest? Our experience with reference frames should tell us otherwise. An object with a *constant* velocity in one reference frame would be at rest in a reference frame which had the same *constant* velocity as the object. It follows then that motion with a *constant* velocity is not intrinsically different to an object being at rest. This idea, which begins with Galileo, is the basis for Newton's first law of motion.

*Every object continues in its state of rest, or of uniform velocity in a straight line, as long as no net force acts on it.*

This law essentially describes the property of an object that we call **inertia**.

Newton's first law only applies in **inertial reference frames**, which are frames which are not accelerating. Bear in mind that acceleration can mean change of direction as well as magnitude of velocity, so a rotating reference frame is not an inertial frame.

An example from Physclips.

In fact, although we usually treat it as one, the Earth is not an inertial reference frame, as it is rotating. The effect of the Earth's rotation on a moving object over time can be measured using a Foucalt Pendulum

How much inertia does an object have? This is determined by it's **mass**, for which the SI unit is kg. The more mass an object has the harder it is to get it to change it's motion.

For an object to change it's velocity and thus have an acceleration, it needs to be subjected to a net force. The degree of acceleration produced by the sum of the forces is determined by the size of the force and the mass of the object according to the formula.

$\large\Sigma\vec{F}=m\vec{a}$

Dimensional analysis shows us that force must have units $\mathrm{kgms^{-2}}$. These units are normally expressed as Newtons and denoted by the symbol $\mathrm{N}$.

*Whenever one object exerts a force on a second object, the second exerts an equal force in the opposite direction on the first.*

Another way of saying this: *Forces come in action-reaction pairs and the sum of the forces in a pair is zero.*

Be careful with the third law, the equal and opposite forces are acting on **different** objects. The motion of an object is determined only by the sum of the forces acting on **that** object.

All objects on Earth (or near to it) are constantly subjected to gravitational force. The weight of an object is equal to the force which acts on it due to gravity.

As we established earlier that all objects in free fall have acceleration $\vec{g}=9.81\mathrm{ms^{-2}}$ it follows from the second law that the force due to gravity on an object is

$\vec{F}_{G}=m\vec{g}$

Each object subject to gravitational force exerts an equal and opposite force on the Earth to the one it exerts on us. However the Earth's mass is quite large!

When an object is supported by a surface it is subject to a force that we call the normal force.
In the case where the gravitational force is the only other force acting on the object and the surface is flat with respect to the surface of the Earth the normal force is equal and opposite to the gravitational force. This is **not** a result of the 3rd law, the diagram below shows the right way to think about the various action and reaction pairs.

When you stand on a scale it claims to measure your mass in kg or lbs. In fact it is measuring the force exerted on it (equal and opposite to the normal force it exerts), assuming that this is your weight and then showing the value of weight divided by the acceleration due to gravity.

As you can see the normal force on an object is only equal to the gravitational force when there is no other force on the object. If you stand on a set of scales in an elevator your apparent mass as displayed by the scale will thus be affected by the acceleration of the elevator.

Frequently in physics problems you will see ropes, strings, cords, wires, etc. Typically they are assumed to be mass-less. This means that the net force on any section of a rope when extended is considered to be zero independent of the acceleration of the rope ($\Sigma\vec{F}=m\vec{a}$ with $m=0$), allowing a perfect transmission of force from one end of the rope to another. The force transmitted from one end of the rope to the other is the **tension** in the rope, and it should be noted that all real ropes have a maximum tension above which they break.

For this system (Atwood's Machine) we need to consider free-body diagrams for two objects. $\vec{a}$ is the same in magnitude for each weight, the sign relative to gravity must be opposite on one side from another. The weights are both connected by the same rope and thus the force due to tension is the same for each object. The concept behind this device is used in elevators and funiculars |

$\large m_{1}\vec{a}=\vec{T}-m_{1}\vec{g}$ $\large m_{2}\vec{a}=m_{2}\vec{g}-\vec{T}$

$\large \vec{T}=m_{1}\vec{a}+m_{1}\vec{g}$ $\large \vec{T}=m_{2}\vec{g}-m_{2}\vec{a}$

$\large m_{1}\vec{g}+m_{1}\vec{a}=m_{2}\vec{g}-m_{2}\vec{a}$

$\large (m_{1}+m_{2})\vec{a}=(m_{2}-m_{1})\vec{g}$

$\large \vec{a}=\vec{g}\frac{m_{2}-m_{1}}{m_{2}+m_{1}}$

$\large \vec{T}=\vec{g}\frac{2m_{2}m_{1}}{m_{2}+m_{1}}$

Two blocks are connected by a rope which runs over a frictionless pulley of negligible mass. One block hangs from the rope, while the other rests on a frictionless plane inclined at an angle of $\theta$ to the horizontal.

Add arrows indicating the direction of all of the forces acting on both $m_{1}$ and $m_{2}$ to the diagram.

Find an expression for the acceleration, $a$, of $m_{1}$ as a function of $m_{1}$, $m_{2}$, $g$ and $\theta$. Specify which direction is positive (i.e up or down).

Skilled sailors can sail at a broad range of angles to the wind and with modern boats it is possible to go substantially faster than the windspeed.

Sailing a boat close to the wind is in fact a very neat physics trick. A very nice explanation is here.

A simplified version:

Note that to balance the sideways force a boat requires some lateral resistance from the water and the correct positioning of the weight of the crew.

Finding the net force on an object allows us to determine it's acceleration, which from which we can, given supplementary information, deduce the motion of an object.

$\large v= v_{0}+at$

$\large x= x_{0}+v_{0}t+\frac{1}{2}at^2$

$\large v^{2}=v_{0}^2+2a(x-x_{0})$

or

$\large \vec{v}(t)=\vec{v}_{0}+\int^{t}_{0}\vec{a}(t)\,dt$

$\large \vec{x}(t)=\vec{x}_{0}+\int^{t}_{0}\vec{v}(t)\,dt$

A weight on a scale in a elevator type problem. If the scale reads a smaller number than the actual weight it means that the net force on it is less because it is exerting an upward force on the object in response to additional force exerted on it by the accelerating person. Express the sum of the forces on the scale as $m\vec{g}-m\vec{a}$ to form the equation you can solve for $\vec{a}$.

An inclined plane problem. Find the angle $\theta$ using the equation for the force perpendicular to the plane. You know $m$, you know the maximum acceptable force, and of course you know $g$, so all you need to do is solve for $\theta$.

Draw a diagram in which you sum all the forces on the speaker together. This should make a closed triangle because the net force is zero as the speaker is not accelerating (Newton's Second Law). This will not be a right angled triangle, but will be made up of two adjacent right angled triangles with the angle $\theta$ opposite to $\frac{mg}{2}$ and adjacent to T. the angle $\theta$ can be deduced from $h$ and $l$, and the weight of the object $mg$ is known.

An Atwood's Machine..with a twist. You can replace the second weight with the force the window washer exerts. However, if the window washer exerts a force on the rope, what does Newton's Third Law say about the force exerted back on her?

A different modification of Atwood's Machine. (Physicist's are abnormally interested in Atwood's machine!). Replace $m_{2}\vec{g}$ by $m_{A}\vec{g}\sin\theta$. You may note that in the limit $\theta=90^{o}$ the problem reverts to a plain vanilla Atwood's machine.