Forces

12PHYS - AS91171

Finn LeSueur

2020

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

Source

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.

Source

Pātai: Terminal Velocity

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  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

Source
  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.

Akoranga 22 Mahi Tuatahi

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

Source
  • 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