The unit of force is the newton (N). In S.I. base units, the newton is ${kg⋅m}/{s^2}$

Net force is the resultant force of a vector addition of all forces acting on a body. Traction force is usually opposed by resistance due to surface friction and air resistance. Air resistance is proportional to the speed of the mass, so a vehicle that has constant traction from an engine will accelerate to the point where the traction force and the resistance forces are equal. At this point, the net force is zero and the vehicle moves at constant velocity.

Similarly, a falling object will accelerate till air resistance is equal and opposite in direction, and freefall at constant velocity results.

- A force must be applied to cause work to be done upon a mass, or to change its energy or position in a gravitational field.
- The work done to accelerate a mass is the change in potential, equal to the impulse.
- There is a difference between mass (kg) and weight (N). Gravitational acceleration is the same for all masses at the same point in a gravitational field, but the gravitational force (weight) is proportional to the mass being accelerated.
- There is a contradiction with Special Relativity. According to Newton's Second Law, a mass would have a large but finite momentum as it approaches the speed of light. Einstein's theory demonstrates that the rest mass cannot be assumed for moving objects. The mass increases as velocity increases, requiring exponential amounts of energy to accelerate any massive object towards the speed of light.

Consider the set-up as shown in the illustration. A weight suspended will provide a constant force of:

$$F_g = m.g$$where g is the acceleration due to Earth's gravity, $9.81 m/s^2$.

A string over a pulley will transfer this force to a trolley on a table. What will be the resultant acceleration of the system (the suspended mass and the trolley)?

The force of gravity acts only on the suspended mass, but the mass of the system is the sum of the mass of the trolley and the suspended mass, m + M.

Experimental practice: the friction of the trolley can be calculated by trying different masses m till the trolley just begins to move.

The types of problems free body diagrams are used for include:

Example: In this example, two masses are connected by a string on a surface. A force is applied in the horizontal direction. What is the accceleration?

In this case, there is a mass m on a plane inclined at angle θ to the horizontal. The two forces are gravity F_{g} = mg, and the reaction force, R. If the plane were horizontal, R = F. If the plane were vertical, R would equal 0, since the force of gravity has no component in the x direction.

So, what is the value of R if the plane is between horizontal and vertical? What interests us most is the acceleration down the slope. Therefore, we create a set of axes using the slope surface as the x direction, and the normal to the surface is the y-axis.

Trigonometry tells us that the component of the gravitational force at angle θ is F_{g}.sinθ, and correspondingly the value of R, in the y-direction, is F_{g}.cosθ.

Opposing this motion will be a frictional force, which is the applied force times the coefficient of friction for the surfaces involved. The coefficient for static friction is higher than dynamic friction, so an accurate calculation must take this into account.

Newton's Third Law of Motion means that any force will experience a force which acts equally in the opposite direction. It is why we 'feel' a wall push back at us, or when a skater pushes another, she will be pushed back in the opposite direction.

To conserve momentum, the mass of the projectile times its velocity will equal the mass of the rifle and shooter times their combined velocity backwards.

The fact that every applied force will experience a reaction force leads to the expression 'action-reaction force pairs'. A book on a table, a person walking along the road, a boy jumping out of a boat - these are all examples of action-reaction force pairs.

- Consequences:
- Impacts result in force of equal magnitude being applied to both participants.
- Work is a measure of the energy used to change the condition of an object. When a billiard ball strikes another ball of equal mass in an ideal elastic collision, the second ball is accelerated to the velocity of the first ball, which stops. There is work done on both objects in order to exchange the kinetic energy.
- Momentum is conserved: Projectiles, such as bullets and rockets, cause an equal force of recoil in the opposite direction of their acceleration. This is also the reason that we risk injury stepping off skateboards too quickly, and that small boats are hard to step out of.
- In an accelerating frame, such as a lift, we experience a reaction force from the floor of the lift, which adds or subtracts from our apparent weight, depending on the direction of the lift's acceleration.
- The centripetal force of circular motion results in an equal and opposite centrifugal 'force'. More strictly an 'effect' than a true force, centrifugal force creates artificial gravity in rotating spaceships.

Since Newton's Third Law states that every force will experience an equal and opposite reaction force. This is known as an action-reaction pair of forces. e.g. A man's weight is a reaction force upwards, equal and opposite to the gravitational attraction force downwards.

In the diagram, forces C and D are an action-reaction pair of forces, but A and B are. The friction force is not always equal to the applied force, since it is equal to the mass of the object times the coefficient of friction of the surface. A reaction force is produced by the applied force and is always equal in magnitude and opposite in direction.

F_{applied} = F_{reaction}

Apparent weight = F_{g} + ma

where F_{g} is the force of gravity on a mass (normal weight), m is the mass of the object, and a is the acceleration in the axis of the gravitational field (up or down).

A ball falling along a curved surface has forces and acceleration as follows:

The drag coefficient is a dimensionless quantity and is given by:

$$ C = {2F}/{ρv^2 A}$$- Where F
_{d}= drag force - ρ = mass density of fluid
- v = speed of object through the fluid
- A = reference surface area

The reference surface area is the area of the object's area which contributes to the drag force. It is not the total surface of the object. For example,the reference surface area of a sphere is πr^{2} (as opposed to the surface of a sphere 4πr^{2}).

The drag force on an object moving through a fluid is therefore:

F_{d} = ½ ρv^{2}C_{d} A

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