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## 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 1. What is my net force?

2. What is my acceleration?

3. Why does it not equal $g$?

4. What is my drag and weight forces were balanced?

### Whakatika 1. What is my net force?
$620N$ down
2. What is my acceleration?
$F=ma, a=\frac{F}{m}=\frac{620}{72}=8.61ms^{-2}$
3. Why does it not equal $g$?
Because friction is acting against the weight force!
4. What if my drag and weight forces were balanced?
$F_{net}=0$ and therefore $a=0$ and I would be in equilibrium.

Quizizz!

## Ngā Whāinga Ako

1. Be able to label different forces
2. 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
• Homework Booklet: Q31, Q32, Q34