Net Force in One and Two Dimensions Revision Notes for HSC SSCE Physics

Net Force in One and Two Dimensions

Introduction to net force

In most real-world situations, multiple forces act on an object simultaneously. When this happens, we need to consider the combined effect of all these forces. This combined effect is called the net force or total force.

The net force is the vector sum of all individual forces acting on an object. It determines the object's acceleration according to Newton's second law. Understanding how to calculate net force is essential for predicting how objects will move when subjected to multiple forces.

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The net force determines an object's motion, not the individual forces acting on it. This is why we must consider all forces together when analyzing motion.

Newton's second law with multiple forces

You may recall Newton's second law in its simple form: $\vec{F} = m\vec{a}$ , which applies when only a single force acts on an object. However, in almost all practical situations, there are at least two forces acting, and often many more.

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The more general form of Newton's second law accounts for multiple forces:

$\vec{a}_A = \frac{\vec{F}_{\text{net on A}}}{m_A} = \frac{\sum \vec{F}_{\text{on A}}}{m_A}$

This equation tells us that the acceleration of an object depends on the net force acting on it, not just individual forces.

The net force equation

The net force on an object is defined mathematically as:

$\vec{F}_{\text{net, A}} = \sum \vec{F}_A$

This means we add all the forces acting on object A as vectors. The method we use to add these forces depends on whether they act in one dimension or two dimensions.

Adding forces in one dimension

When all forces act along a single line (one dimension), we can use simple algebraic addition. The key is to establish a positive direction and treat forces in that direction as positive, whilst forces in the opposite direction are negative.

Forces in the same direction

Consider a block subjected to two forces both acting towards the right.

If we define right as the positive direction, the net force is:

$F_{\text{net}} = (+4.0 \text{ N}) + (+3.0 \text{ N}) = 7.0 \text{ N}$

The block will accelerate to the right. When forces act in the same direction, the net force is simply the sum of the magnitudes, and the object accelerates in that direction.

Forces in opposite directions

Now consider a block subjected to two forces acting in opposite directions.

If right is still the positive direction, the $3.0 \text{ N}$ force acting left must be negative:

$F_{\text{net}} = (+4.0 \text{ N}) + (-3.0 \text{ N}) = +1.0 \text{ N}$

The block will still accelerate to the right, but at a much lower rate than in the previous example. When forces oppose each other, the net force is the difference between their magnitudes, acting in the direction of the larger force.

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When working in one dimension, always establish your positive direction first. This makes calculations straightforward and reduces errors.

Adding forces in two dimensions

When forces act at angles to each other, we cannot simply add them algebraically. We must treat them as vectors and use vector addition techniques.

Forces at right angles

Consider a block subjected to two forces acting at right angles to each other.

We cannot add these forces algebraically because they act in different directions. Instead, we treat them as the two perpendicular components of the net force.

Using Pythagoras's theorem

When two forces are perpendicular, we can find the magnitude of the net force using Pythagoras's theorem:

$F_{\text{net}} = \sqrt{(3.0 \text{ N})^2 + (4.0 \text{ N})^2} = 5.0 \text{ N}$

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Notice that the net force ( $5.0 \text{ N}$ ) is less than when the forces acted in the same direction ( $7.0 \text{ N}$ ), but greater than when they acted in opposite directions ( $1.0 \text{ N}$ ). The relative orientation of forces significantly affects the net force magnitude.

Finding the direction using trigonometry

To find the angle $\theta$ at which the net force acts, we use trigonometry:

$\tan \theta = \frac{4.0 \text{ N}}{3.0 \text{ N}}$

Therefore:

$\theta = \tan^{-1}\left(\frac{4.0 \text{ N}}{3.0 \text{ N}}\right) = 53°$

Worked example: Tugboats pulling a barge

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Worked Example: Tugboats Pulling a Barge

Two tugboats apply forces on a barge through ropes, as shown below. What is the resultant force due to the two tugboats?

Given information:

$\vec{F}_A = 4.0 \times 10^3 \text{ N}$
$\vec{F}_B = 4.0 \times 10^3 \text{ N}$
The two forces are acting at $90°$ to each other

Solution:

Since the forces are perpendicular, we apply Pythagoras's theorem:

$F_{\text{net}} = \sqrt{F_A^2 + F_B^2}$

$= \sqrt{(4.0 \times 10^3 \text{ N})^2 + (4.0 \times 10^3 \text{ N})^2}$

$= 5.66 \times 10^3 \text{ N}$

Therefore, $F_{\text{net}} = 5.7 \times 10^3 \text{ N}$ (to 2 significant figures)

Force diagrams

Force diagrams are visual representations that help us understand the forces acting on an object. They are particularly useful when working with forces in two dimensions.

How to draw a force diagram

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When drawing a force diagram, follow these rules:

Draw the force arrow so its tail is at the point where the force acts
Make the length of the arrow proportional to the magnitude of the force
Point the arrow in the direction the force acts
Only show forces acting on the object of interest
Do not show forces exerted by the object on other things

What not to include

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Newton's third law pairs: You cannot show both forces in a Newton's third law pair on a single force diagram, because one acts on the object and the other acts on something else.

Net force: The net force is not an actual force—it's a sum of forces. If you do include it on your diagram, use a different style of arrow to distinguish it from actual forces.

Resolving forces into components

When forces are not perpendicular to each other, we need to break them down into perpendicular components before adding them. This process is called resolving forces into components.

Horizontal and vertical components

Consider a person pulling a suitcase at an angle $\theta$ to the horizontal with force $\vec{F}$ of magnitude $F$ .

Using trigonometry, we can decompose this force into horizontal and vertical components:

Horizontal component: $F_x = F \cos \theta$

Vertical component: $F_y = F \sin \theta$

The two components add vectorially to give the total force $\vec{F}$ .

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The components $F_x$ and $F_y$ are perpendicular to each other, which makes them easier to add. This is the key advantage of resolving forces into components.

Finding magnitude from components

If we know the components, we can find the magnitude of the force using Pythagoras's theorem:

$F = \sqrt{F_x^2 + F_y^2}$

Worked example: Components of a force on a suitcase

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Worked Example: Components of a Force on a Suitcase

Phil is racing to get to the airport luggage check-in before his flight closes. He exerts a force of $30 \text{ N}$ at an angle of $60°$ to the horizontal on his suitcase. Calculate the horizontal and vertical components of this force.

Given information:

$\theta = 60°$
$F = 30 \text{ N}$

Solution:

Horizontal component: $F_x = F \cos \theta = 30 \text{ N} \cos 60° = 15 \text{ N}$

Vertical component: $F_y = F \sin \theta = 30 \text{ N} \sin 60° = 26 \text{ N}$

Therefore, $F_x = 15 \text{ N}$ and $F_y = 26 \text{ N}$

Adding forces using perpendicular components

When multiple forces act at various angles, we use a systematic method to find the net force. This involves breaking each force into components, adding all components in each direction separately, then recombining them.

The component addition method

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Step 1: Break each force into its perpendicular components

For each force, calculate: $F_{1x} = F_1 \cos \theta_1, \quad F_{2x} = F_2 \cos \theta_2, \ldots$ $F_{1y} = F_1 \sin \theta_1, \quad F_{2y} = F_2 \sin \theta_2, \ldots$

Step 2: Add all the $x$ components to get the net $x$ component

$F_{\text{net, x}} = F_{1x} + F_{2x} + F_{3x} + \ldots$

Step 3: Add all the $y$ components to get the net $y$ component

$F_{\text{net, y}} = F_{1y} + F_{2y} + F_{3y} + \ldots$

Step 4: Find the magnitude of the net force using Pythagoras's theorem

$F_{\text{net}} = \sqrt{(F_{\text{net, x}})^2 + (F_{\text{net, y}})^2}$

Step 5: Find the direction using trigonometry

$\tan \theta_{\text{net}} = \frac{F_{\text{net, y}}}{F_{\text{net, x}}}$

Worked example: Boat on a river

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Worked Example: Boat on a River

Eleanor is taking her boat out on the river to do some fishing. Due to the river current, she has to steer the boat at an angle to the direction she actually wants to go. The propeller exerts a force $\vec{F}_P$ on the boat, with magnitude $250 \text{ N}$ pointing N $40°$ E. The water exerts a force $\vec{F}_W$ with magnitude $400 \text{ N}$ pointing directly west. What is the magnitude of the net force exerted on the boat?

Solution approach:

Choose north as the positive $y$ direction and east as the positive $x$ direction.

Step 1: Decompose the forces

For the propeller force ( $50°$ from the positive $x$ -axis): $F_{Px} = F_P \cos 50° = 250 \text{ N} \cos 50° = 161 \text{ N}$ $F_{Py} = F_P \sin 50° = 250 \text{ N} \sin 50° = 192 \text{ N}$

For the water force (pointing west, which is $180°$ from positive $x$ -axis): $F_{Wx} = F_W \cos 180° = 400 \text{ N} \cos 180° = -400 \text{ N}$ $F_{Wy} = F_W \sin 180° = 400 \text{ N} \sin 180° = 0 \text{ N}$

Step 2: Add the components

$F_{\text{net, x}} = F_{Px} + F_{Wx} = 161 \text{ N} + (-400 \text{ N}) = -239 \text{ N}$ $F_{\text{net, y}} = F_{Py} + F_{Wy} = 192 \text{ N} + 0 \text{ N} = 192 \text{ N}$

Step 3: Find the magnitude

$F_{\text{net}} = \sqrt{(-239 \text{ N})^2 + (192 \text{ N})^2} = 307 \text{ N}$

Therefore, $F_{\text{net}} = 310 \text{ N}$ (to 2 significant figures)

Applying Newton's third law

Remember that forces are interactions, so forces always come in pairs. Whenever an object exerts a force on something, it experiences an equal and opposite force. This is Newton's third law:

$\vec{F}_{\text{A on B}} = -\vec{F}_{\text{B on A}}$

Real-world example: Boat propeller

Newton's third law explains how a propeller makes a boat move. The turning blades of the propeller push on the water, pushing it backwards away from the boat. According to Newton's third law, the water exerts an equal and opposite force on the propeller, pushing it forwards. Since the propeller is attached to the boat, this pushes the boat forwards. The propeller wouldn't work unless it had something to push against.

Real-world example: Walking

Newton's third law helps us understand why we need friction between our feet and the ground to walk. When you push backwards against the ground with your foot, you exert a friction force against the ground. From Newton's third law, the ground exerts an equal and opposite friction force forwards on you. It is this friction force that pushes you forwards when you walk. Without friction between your feet and the ground, you wouldn't be able to walk.

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Newton's third law force pairs always act on different objects. This is why both forces in the pair cannot appear on the same force diagram.

Contact forces and their components

When your foot touches the ground, the total contact force can be considered as the sum of two components:

Normal force: perpendicular to the surface
Friction force: parallel to the surface

Worked example: Ground reaction force

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Worked Example: Ground Reaction Force

A foot exerts a normal force of $600 \text{ N}$ downwards on the ground and a friction force of $300 \text{ N}$ to the left on the ground. Calculate the total contact force exerted by the ground on the foot, assuming the foot does not slip.

Solution:

Take the positive $x$ direction to be right and the positive $y$ direction to be up.

Given forces exerted by foot on ground: $\vec{F}_{\text{foot on ground, friction}} = F_x = -300 \text{ N}$ $\vec{F}_{\text{foot on ground, normal}} = F_y = -600 \text{ N}$

Apply Newton's third law: $\vec{F}_{\text{ground on foot}} = -\vec{F}_{\text{foot on ground}}$

Therefore: $\vec{F}_{\text{ground on foot, friction}} = -F_x = +300 \text{ N}$ $\vec{F}_{\text{ground on foot, normal}} = -F_y = +600 \text{ N}$

Calculate total contact force: $F_{\text{contact}} = \sqrt{F_x^2 + F_y^2}$

$= \sqrt{(300 \text{ N})^2 + (600 \text{ N})^2}$

$= 670.8 \text{ N}$

Therefore, $F_{\text{contact}} = 671 \text{ N}$ (to 3 significant figures)

Remember!

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Key Points to Remember:

The net force is the vector sum of all forces acting on an object: $\vec{F}_{\text{net}} = \sum \vec{F}$ . It determines the object's acceleration according to Newton's second law: $\vec{a} = \vec{F}_{\text{net}}/m$ .
In one dimension, add forces algebraically by choosing a positive direction. Forces in that direction are positive; forces in the opposite direction are negative.
In two dimensions, forces must be added vectorially. For perpendicular forces, use Pythagoras's theorem: $F_{\text{net}} = \sqrt{F_x^2 + F_y^2}$ .
To add non-perpendicular forces, resolve each force into perpendicular components using $F_x = F \cos \theta$ and $F_y = F \sin \theta$ , add components separately, then recombine using Pythagoras's theorem.
Newton's third law states that if object A exerts a force on object B, then object B exerts an equal and opposite force on object A: $\vec{F}_{\text{A on B}} = -\vec{F}_{\text{B on A}}$ . These force pairs act on different objects and cannot both appear on a single force diagram.

Net Force in One and Two Dimensions (HSC SSCE Physics): Revision Notes