12PHYS - Mechanics

Finn LeSueur

2019

- State what each letter stands for
- Give the units for each letter
- Rearrange the equation for \(m\) and \(a\)
- Derive the SI units for F (not Newtons)

For a car of **mass 1500kg** which is accelerating at \(3.7ms^{-2}\):

- What net force is needed to maintain this acceleration?

- If the engine is producing \(6000N\) of thrust, what is the difference and what happened to it?

Torque can be thought of as the **turning effect** around a **pivot**. Torque is sometimes known as **moment** or **leverage**.

- \(F =\) force in Newtons
- \(d_{\bot} =\) perpendicular distance of force from pivot

- A small force at a small distance produces a small torque,
- the same small force at a larger distance produces a larger torque.

A force of \(9N\) acting up at a distance of \(10cm\) is needed to lift the top off a bottle of soft drink. Start by drawing a rough diagram. **Calculate the torque applied.**

A force of \(9N\) acting up at a distance of \(10cm\) is needed to lift the top off a bottle of soft drink. **Calculate the torque applied.**

Calculate the torque applied if the lever is stretched to \(75cm\).

\[\begin{aligned} & && \text{Knowns} \newline & && \text{Unknowns} \newline & && \text{Formula} \newline & && \text{Sub and Solve} \end{aligned}\]Calculate the torque applied if the lever is stretched to \(75cm\).

\[\begin{aligned} & \tau = Fd_{\bot} \newline & \tau = 9 \times 0.75 \newline & \tau = 6.75 \text{Nm anticlockwise} \newline \end{aligned}\]Calculate the torque applied if the lever is compressed to \(1cm\).

\[\begin{aligned} & && \text{Knowns} \newline & && \text{Unknowns} \newline & && \text{Formula} \newline & && \text{Sub and Solve} \end{aligned}\]Calculate the torque applied if the lever is compressed to \(1cm\).

\[\begin{aligned} & \tau = Fd_{\bot} \newline & \tau = 9 \times 0.01 \newline & \tau = 0.09 \text{Nm anticlockwise} \newline \end{aligned}\]Yes, and you must always state which direction it is acting in.

**Clockwise or Anticlockwise**

*Newton’s First Law* tells us equilibrium is when an object is **at rest** or **moving uniformly**.

For this to occur we need two things:

- Sum of all forces to be 0
- Sum of all torques to be 0

Building bridges, setting up scaffolding, see-saws and more!

\(m_{1}=2kg\), \(d_{1}=15cm\), \(m_{2}=1kg\), \(d_{2}=30cm\)

- Calculate the clockwise and anticlockwise torques
- Are they in balance?

\(m_{1}=7kg\), \(d_{1}=65cm\), \(m_{2}=13kg\), \(d_{2}=35cm\)

- Calculate the clockwise and anticlockwise torques
- Are they in balance?

The rock has mass \(1100kg\) and is at distance \(50cm\) from the pivot. If Ash exerts \(70N\) of downward force at a distance of \(8m\) from the pivot can he move the rock?

Archimedes once said: *“Give me a place to stand and I will move the world”*

**Question**: Assuming the mass of the Earth is \(5.972\times 10^{24} kg\) at a distance of 1km from the pivot and Archimedes’ mass is \(75kg\), how long would his lever have to be?

- Calculate the clockwise torque
- Calculate the anticlockwise torque
- Is it balanced?

The plank may not be massless. You may need to take it into account.

- The mass of the plank acts through its
**center of gravity** - Because the plank is uniform, this is the middle of the plank

- Draw and label all forces on a diagram
- Draw and label the distances between all forces and the
**pivot** - Calculate all clockwise torque
- Calculate all anticlockwise torque
- Balance torques & forces

- \(d_{1}=30cm\), \(d_{2}=70cm\), \(m_{1}=900g\), \(m_{2}=300g\), see-saw mass = \(100g\).
- Calculate the total anticlockwise moment
- Calculate the total clockwise moment
- Is it balanced?

- Assume the system is in equilibrium (\(\tau_{clockwise} = \tau_{anticlockwise}\))
- \(d_{1}=0.5m\), \(d_{2}=1.5m\), \(F_{1}=2.5N\), see-saw mass = \(0.5kg\), \(F_{2}=?\).
- Draw the weight force of the see-saw on your diagram
- Find the unknown force, \(F_{2}\)

Textbook: Force, Equilibrium and Motion - Q7, 8, 10, 11, 12