Forces

Recap of National 5

In the National 5 course, Forces were covered in great detail. The following facts were covered :-

1. A Force can be seen as a push, a pull, or a twist.

2. A Force can change the shape, Speed or direction of an object.

3. A Force has the units of Newtons.


Measuring Forces

In National 5, A Force was measured with a Newton-Balance. A Newton Balance consists of a Spring with an attached scale. As Force is applied, the Spring is extended.

Balanced Forces

When two identical Forces act in opposite directions, they cancel each other out.

Newton's First Law

The above understanding of Forces allowed Isaac Newton to develop the 1st of his three laws of Forces :-

"An object will remain at rest or travel in a straight line with a constant speed, unless an unbalanced Force acts upon it"


Terminal Velocity

When a skydiver jumps out of a plane, they accelerate (see unbalanced Forces below). After a short time, however, they reach their maximum Speed. This maximum Speed is called their Terminal Velocity.

This happens because the two forces acting on the skydiver are balanced.

The Air Resistance that they experience is equal in size and in opposite direction to their Weight :-

Note - By changing their body shape (making themselves more streamlined), the skydiver can alter their air resistance, changing their terminal Velocity.


Unbalanced Force

When an object experiences an unbalanced Force, the object will accelerate. This can either be cause by only one Force acting (as in the diagram below), or by two unequal Forces acting in opposite directions.

Newton's Second Law

The above understanding of Forces allowed Isaac Newton to develop the 2nd of his three laws of Forces :-

"An unbalanced Force acting on an object will cause the object to Accelerate. The size of that Acceleration will depend on the size of the Force and the mass of the object."

The above law is more commonly described as the formula below :-

F = m x a

Where :-

F = Unbalanced Force (N)

m = Mass of object (kg)

a = Acceleration of object (ms-2)

Example 1 -

A Truck of mass 2000 kg is acted upon by two Forces as shown in the diagram below:-

Calculate :-

1. The unbalanced Force acting on the object.

2. The Acceleration experienced by the object.


Unbalanced Force = 50000 - 45000

= 5000 N (to the right)

F = m x a

a = F / m

a = 5000 / 2000

a = 2.5 ms-2

Note - Force is a Vector, so a direction must be given


Motion Case Study : Skydiving

The video below shows how the motion of a skydiver is affected by the changing forces they experience:-

The following Velocity-Time graph shows the motion of a skydiver during a jump. By observation of the graph, it is possible to identify key times in the jump process :-


Resultant Vectors

So far in this section, the focus has been on Forces acting in the same plane, however Forces (and all other Vectors) can act in multiple planes. In order to understand the effect of Forces acting in different planes, the components of the Forces must be found.

The diagram above shows how any vector can be split into two components :-

F = Resultant Force


Fhorizontal = Component of Force acting in the horizontal direction

Fvertical = Component of Force acting in the vertical direction


In order to find the components of a vector, Trigonometry can be used.

Example 1 -

What are the components in the horizontal and vertical direction of a Force of 35 N acting at an angle of 25° upwards from horizontal ?

Fhorizontal = 35 cos25

Fhorizontal = 31.7 N


Fvertical = 35 sin25

Fvertical = 14.8 N


Note - The above method can be used to calculate the component of any vector. This will be especially useful in the Projectile motion section.

Forces on a Slope ( Frictionless )

When an object is on a horizontal surface, its Weight acts vertically downwards and the object remains stationary.

If the surface is tilted at an increasing angle, the object will begin to move. This is because when the slope is at an angle, a component of Weight acts down the slope, causing the object to move.

By again applying Trigonometry to find the components of the Weight the following can be found :-

Example 2 -

A 2 kg block is placed on a slope at an angle of 15° to the horizontal. Calculate the Force acting down the slope, and therefore the Acceleration of the block.

Wdown slope = mg sin

Wdown slope = (2x9.8) sin15

Wdown slope = 5.1 N


Acceleration = F / m

Acceleration = 5.1 / 2

Acceleration = 2.6 ms-2


Note - If the dimensions of the ramp are given, then by using the equations of motion, the Speed at various parts of the ramp can be found.

Forces on a Slope ( With Friction )

The above example does not take into account that the block would experience Friction as it slides down the slope.

Friction always act in the opposite direction to motion, so in the above example Friction would have acted up the slope, reducing the unbalanced Force, and therefore reducing the Acceleration.

Example 3 -

A 2 kg box slides down a ramp with an angle of 20°. If the box accelerates at 2 ms-2, what is the magnitude of the frictional Force acting upon it ?


Unbalanced Force = mg sinθ - Friction

Unbalanced Force = (2 x 9.8 x sin20) - Friction

Unbalanced Force = 6.7 - Friction


Unbalanced Force = m x a

6.7 - Friction = 2 x 2

Friction = 6.7 - 4

Friction = 2.7 N (up the slope)

Pairs of Forces and Newton's Third Law

Newton's first and second laws explain the effect of Forces which act on a single object, whereas Newton's third law applies to two interacting objects.


Newton's Third Law States :-

"If object A exerts a Force on object B, then object B exerts a Force on A which is equal in size but opposite in direction."


In order to find the Force exerted between objects (such as the Tension in a tow bar between a caravan and a car) the following procedure can be used :-

1. Model the two objects as one combined object to calculate the Acceleration of both.

2. Model each object separately to find the unbalanced Force between them.

Example 4 -

A force of 20 N is applied as shown in the diagram above on a frictionless surface. Calculate the Force exerted on the 3 kg mass.


Object combined :-

a = F/m

a = 20 / ( 2 + 3 )

a = 4 ms-2


Object A only :-

Funbalanced = m x a

20 - FB on A = 2 x 4

FB on A = 12 N (to the left)

However, by Newton's Third Law, FB on A = FA on B

Force Exerted on the 3 kg Mass = 12 N (to the right)

Microgravity

In everyday language, astronauts float about in 'Zero Gravity', this is not the case however. The more appropriate way to describe the conditions is Microgravity, meaning that Gravity is small, but still present.

In Low Earth Orbit, for example aboard the International Space Station, the astronauts appear to be weightless, to not experience any Gravity at all. This is incorrect. In Low Earth Orbit, the Acceleration due to Gravity is ~9 ms-2, which is less than at the Earth's surface, but definitely NOT zero Gravity.

The astronauts appear Weightless as both they and the Space Station are falling at identical Speeds in relation to each other. This same effect can be simulated on Earth in the flight profile of Reduced-Gravity Aircraft, part of the training for astronauts.

The video below demonstrates moving in Microgravity and "Burp Thrust"

Moving in Microgravity

Once a spacecraft leaves the Earth's atmosphere, the method of controlling Speed and direction becomes completely different. The major change, however, is that engines are required only for maneuvering, they are not required to maintain speed.

Newton's First Law States :-

"An object will remain at rest or travel in a straight line at a constant Speed, unless an unbalanced Force acts upon it."

In the atmosphere, a vehicle must have its engines active to provide thrust, overcoming the frictional forces which act to slow it down. Without the engine thrust, the vehicle would slow down, due to the unbalanced Force acting against motion, until its forward Velocity became zero.

Once outside the atmosphere, there is no air resistance acting to slow the craft down. So once out of the atmosphere, a spacecraft will travel at a constant Velocity in a straight line,without the need of engine thrust.

Without air resistance to slow down a vehicle, retro rockets must be used. A retro rocket is simply a rocket engine which applies a force upon a rocket in the opposite direction to its motion, causing a negative Acceleration.

The diagram below shows a Command module (on the Left) attached to a satellite (on the right):-

If the combined craft is assumed to be stationary to start, to move the spacecraft from position A to position B requires several engine burns:-

1. Command Module engine burn - Causes the spacecraft to Accelerate towards B.

2. Command Module engine cutoff - Spacecraft travels at a constant Speed towards B.

3. Satellite engine burn - Applies a negative Acceleration to the spacecraft, reducing its Velocity.

4. Satellite Engine cutoff - Spacecraft is now stationary at B.

The above maneuver required only linear engine burns, but in order to turn the spacecraft, small maneuvering thrusters must be used. These are small clusters of rockets which can apply a Force perpendicular to the motion of the spacecraft, to change its direction.

The video below shows how maneuvering thrusters are used on the Apollo spacecraft