## Mahi Tuatahi: Draw a diagram indicating all the forces acting upon you

## Drawing Force Diagrams

**Use these tips to draw a second, improved diagram**- Use arrows (force is a vector)
- Label with names
- Label with numerical values
- Start arrows from center of mass
- Length indicates magnitude
- Use dashes to indicate equally sized vectors

### Muscle Contraction & Light-Headedness

### Pātai: Vector Reminder

On your whiteboards, add these vectors (head to tail) to find the resultant (net) force acting.

### Pātai: Acceleration

For each, calculate the acceleration that a $5kg$ object would experience.

### Equilibrium

A state of zero acceleration (constant velocity) due to $F_{net} = 0$.

- When performing vector addition, equilibrium is seen by the vectors starting and ending at the same point.
- Vector diagrams having different start and end points indicates disequilibrium, and thus $F_{net} \neq 0N$.

### Recall: Newton’s Second Law

Force is equal to the change in momentum per change in time. For a constant mass, force equals mass times acceleration. $F=m \times a$

**TLDR**: The acceleration created by a force depends directly upon the mass of the object.

## Pātai: Terminal Velocity

- What is my net force?
- What is my acceleration?
- Why does it not equal $g$?
- What is my drag and weight forces were balanced?

### Whakatika

- What is my net force?

$620N$ down - What is my acceleration?

$F=ma, a=\frac{F}{m}=\frac{620}{72}=8.61ms^{-2}$ - Why does it not equal $g$?

Because friction is acting against the weight force! - What if my drag and weight forces were balanced?

$F_{net}=0$ and therefore $a=0$ and I would be in equilibrium.

## Akoranga 22 Mahi Tuatahi

## Ngā Whāinga Ako

- Be able to label different forces
- Be able to label forces on an angle

Write the date and ngā whāinga ako in your book

**Friction**: Acts against the direction of motion between two in-contact moving objects.**Weight**: Acts down towards the center of mass of the object attracting you.**Applied**: Often called thrust/push/pull depending on the situation.**Drag**: Acts against the direction of motion when an object moves through a medium (e.g. air)

**Spring**: A compressed or extended spring can exert a force in the opposite direction to its displacement.**Magnetic**: A moving charged object inside an electric field will experience a magnetic force.**Tension**: A force exerted through a non-rigid object like rope**Buoyant**: A force felt due to the displacement of another medium (air, water)**Normal/Support**: A force exerted by a rigid object at $90\degree$ to the surface (equal and opposite to the force being applied to it).

## Support/Normal Force

- If the surface is sufficiently strong, the support/normal force will always oppose the force acting upon it exactly
- It will always act at $90\degree$ to the surface

Draw equal and opposite weight and normal forces on your diagram.

### Whakatika

## Support/Normal Forces on an Angle

- It will always act at $90\degree$ to the surface
- In this case, the normal force does not equal the weight force
- This is because not all of the weight force is acting
**perpendicular**to the surface.

- It instead is
**equal and opposite to the component of the weight force perpendicular to the surface**. - We can find it using the right triangle that is formed between the weight force exerted on the plane, and the total net force.
- $F_{n} = F_{w}cos(\theta)$
- The angle inside this triangle is the same as the angle of the incline.

## Calculating a Missing Force

**If**an object is in equilibrium, we can calculate a missing force by assuming $F_{net} = 0$.- Take the three forces acting upon the object and add them through
**vector addition** - They should form a closed right-angled triangle, allowing you to find the unknown side.

### Task/Ngohe

Worksheet - everything up to but not including the **Work** section.

## Case Study

Step 1. Consider what we know about the motion of the object, and what this implies about the net force acting upon the object.

### Question 1

- What force do we know is
**not**acting due to the cars movement? - Therefore, what three forces
**are**acting? - Draw a force diagram illustrating these forces and their relative magnitude. Ensure you label them!

#### Question 1 Whakatika

- What force do we know is
**not**acting due to the cars movement?

_Thrust, because it is not accelerating._ - Therefore, what three forces
**are**acting?

_Weight, friction and support._ - Draw a force diagram illustrating these forces and their relative magnitude. Ensure you label them!

### Question 2

- What do we know about the motion of the car?
- Therefore, what can we say about its acceleration, forces and state of equilibrium?
- Therefore, what do we know the vector diagram will look like?
- Draw a diagram.

#### Question 2 Whakatika

- What do we know about the motion of the car?

_It is stationary (constant velocity)._ - Therefore, what can we say about its acceleration, forces and state of equilibrium?

_$a = 0, F_{net} = 0$ and is therefore __in equilibrium__._ - Therefore, what do we know the vector diagram will look like?

_The vectors will form a closed loop._ - Draw a diagram.

### Question 3

- We need to calculate the
**weight force**of the car. Hint: $F=ma$. - Recognise & state that not all of the weight force acts directly through the slope.

### Question 3 Continued

- Recgonise and state we need to find component of weight force acting into slope, and is equal in magnitude to the support force. Hint: $F_{n} = F_{w}cos(\theta)$.
- Calculate $F_{f}$ using EITHER Pythagoras OR trigonometry to find the frictional side of the triangle.

#### Question 3 Whakatika

- We need to calculate the
**weight force**of the car. Hint: $F=ma$.

_$W=F=m \times a = 1500 \times 9.8 = 14700N$. Use $a = g$ because weight force is due to gravity._ - Recognise & state that not all of the weight force acts directly through the slope.

_Because the weight force is acting on an angle to the slop, the component perpendicular to the slope is equal & opposite to the normal force._

#### Question 3 Whakatika Continued

- Calculate the normal force. Hint: $F_{n} = F_{w}cos(\theta)$.

$F_{n} = 14700 cos(12) = 14378N$ - Calculate $F_{f}$ using EITHER Pythagoras OR trigonometry to find the frictional side of the triangle.

$F_{f} = F_{w}sin(\theta) = 14700sin(12) = 3056N$

OR

$F_{f} = \sqrt(F_{w}^{2} - F_{n}^{2}) = 3056N$.

## Practise / Whakawai

- Textbook
- New:
*Activity 10A: Newton’s Laws* - Old:
*Activity 9A: Newton’s Laws*

- New:
- Homework Booklet: Q31, Q32, Q34