Mahi Tuatahi ^🔗

$$ \begin{aligned} F=ma \end{aligned} $$

1. State what each letter stands for
2. Give the units for each letter
3. Rearrange the equation for $m$ and $a$
4. Using the formula, what are the SI units for F (not Newtons)?

For a car of mass 1500kg which is accelerating at $3.7ms^{-2}$:

1. What net force is needed to maintain this acceleration?

$$ \begin{aligned} & && \text{Knowns} \cr & && \text{Unknowns} \cr & && \text{Formula} \cr & && \text{Sub and Solve} \end{aligned} $$

2. If the engine is producing $6000N$ of thrust, why is this different from your calculation?

Torque ($\tau$) – Tōpana Whakahuri ^🔗

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

$$ \begin{aligned} \tau &= Fd_{\bot} \cr torque &= Newtons \times metres \cr torque &= \text{Newton meters (Nm)} \end{aligned} $$

• $F =$ force in Newtons
• $d_{\bot} =$ perpendicular distance of force from pivot

Torque ($\tau$) – Tōpana Whakahuri ^🔗

• A small force at a small distance produces a small torque,
• the same small force at a larger distance produces a larger torque.
• The force is in a straight line
• The torque is circular (CW or CCW)

Pātai: Individual Torques ^🔗

$$ \begin{aligned} & && \text{Knowns} \cr & && \text{Unknowns} \cr & && \text{Formula} \cr & && \text{Sub and Solve} \end{aligned} $$

Pātai Whā: Does torque have a direction? ^🔗

Yes, and you must always state which direction it is acting in.

Clockwise or Anticlockwise

Torque & Equilibrium ^🔗

Equilibrium, Expanded ^🔗

Newton’s First Law tells us equilibrium is when an object is at rest or moving uniformly ($F_{net}=0$).

For this to occur we need two things to be true. You must state these assumptions when doing any equilibrium question.

1. Sum of all forces to be 0 ($\sum F = 0N$)
2. Sum of all torques to be 0 ($\sum\tau = 0Nm$)

Pātai Tahi ^🔗

• $m_{1}=2kg$, $d_{1}=15cm$, $m_{2}=1kg$, $d_{2}=30cm$
• Calculate the clockwise and anticlockwise torques
• Is the system in equilibrium?

Pātai Rua ^🔗

• $m_{1}=7kg$, $d_{1}=36cm$, $m_{2}=13kg$, $d_{2}=65cm$
• Calculate the clockwise and anticlockwise torques
• Is the system in equilibrium?

Mahi Tuatahi ^🔗

1. Calculate the clockwise torque
2. Calculate the anticlockwise torque
3. Is it balanced?

Torque & Equilibrium ^🔗

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

• The mass of the plank acts through its centre of gravity
• Because the plank is uniform, this is the middle of the plank

How To Solve A Torque Problem ^🔗

1. Draw and label all forces on a diagram
2. Draw and label the distances between all forces and the pivot
3. Calculate all clockwise torque
4. Calculate all anticlockwise torque
5. Balance torques & forces
6. Find the unknown value

Pātai: Balanced or Unbalanced? ^🔗

• $d_{1}=30cm$, $d_{2}=70cm$, $m_{1}=900g$, $m_{2}=300g$, \cr see-saw mass = $100g$.
• Is the system in equilibrium?
• Step 1: Calculate the total anticlockwise moment
• Step 2: Calculate the total clockwise moment
• Check if they are balanced.

Pātai: Unknown Force ^🔗

• 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, $F_{2}$

Whakatika ^🔗

$$ \begin{aligned} \tau_{CW} &= \tau_{ACW} \cr \tau_{p} + \tau_{2} &= \tau_{1} \cr F_{p}d_{p} + F_{2}d_{2} &= F_{1}d_{1} \cr ((0.5 \times 9.8)\times 0.5) + F_{2}\times 1.5 &= 2.5 \times 0.5 \cr 2.45 + F_{2} \times 1.5 &= 1.25 \cr F_{2} &= \frac{1.25 - 2.45}{1.5} = -0.8N \end{aligned} $$

This is an interesting answer - it implies that $F_{2}$ is actually acting up.

Pātai: Finding the Supports ^🔗

• Given that the bridge weighs $10,000kg$, find the support force provided by A and B.
  • To find A, we make B the pivot. We also ignore $F_{B}$ because it acting at the pivot and cannot provide any rotational forces.
  • To find B, we make A the pivot. We also ignore $F_{A}$ because it acting at the pivot and cannot provide any rotational forces.
• Solve for each support independently.

Whakatika: Support A ^🔗

We are assuming the bridge is in equilibrium. This means that both net torque and net force are zero.

$$ \begin{aligned} \tau_{CW} &= \tau_{ACW} \cr \tau_{P} + \tau_{C} &= \tau_{A} \cr F_{P}d_{P} + F_{C}d_{C} &= F_{A}d_{A} \cr (m_{p}g)d_{p} + (m_{c}g)d_{C} &= F_{A}d_{A} \cr ((10,000 \times 9.8) \times 25) + ((1,000 \times 9.8) \times 40) &= F_{A} \times 50 \cr 2,450,000 + 392,000 &= F_{A} \times 50 \cr \frac{2,842,000}{50} &= F_{A} = 56,840N \end{aligned} $$

Whakatika: Support B ^🔗

We are assuming the bridge is in equilibrium. This means that both net torque and net force are zero.

$$ \begin{aligned} \tau_{CW} &= \tau_{ACW} \cr \tau_{B} &= \tau_{P} + \tau_{C} \cr F_{B}d_{B} &= F_{P}d_{P} + F_{C}d_{C} \cr F_{B}d_{B} &= (m_{P}g)d_{P} + (m_{C}g)d_{C} \cr F_{B} \times 50 &= (10,000 \times 9.8 \times 25) + (1,000 \times 9.8 \times 10) \cr F_{B} \times 50 &= 2450000 + 98000 \cr F_{B} &= \frac{2,548,000}{50} = 50,960N \end{aligned} $$

Ngā Whakaaro / Thoughts ^🔗

Because the car is closer to support B, it makes logical sense that support B should feel more of the weight force. Therefore, it needs to provide more support force in order to stay in equilibrium.

Whakawai / Practise ^🔗

Textbook: Force, Equilibrium and Motion - Q7, 8, 10, 11, 12 Homework: Q41, Q43