Forces, Systems, and Rotational Motion in Physics

Physics Concepts: Forces, Systems, Pulleys, and Torque

Forces in Systems

• Concept: When multiple objects interact, the forces and accelerations need to be considered for the entire system, or for individual components within the system.

• Blocks Pushing on Blocks:

  • If a force ($F$) pushes a system of two blocks (Block 1 with mass $m<em>1$ and Block 2 with mass $m</em>2$), the total force results in an acceleration ($a$) for the combined mass: $F = (m<em>1 + m</em>2)a$.

  • The force that Block 1 exerts on Block 2 is specifically the force required to accelerate Block 2: $F<em>{1 \to 2} = m</em>2 a$.

• Ropes Connecting Objects (e.g., Train Cars):

  • When objects are connected by ropes and pulled, the rope at the front (pulling the entire system) experiences the greatest tension/force.

  • Subsequent ropes experience less force, as they are only pulling the remaining objects behind them.

  • Analogy: If you're at the front of a cargo system, you pull the hardest; the force diminishes down the line.

• Newton's Third Law and Skateboarding:

  • Principle: For every action, there is an equal and opposite reaction.

  • Skateboard Example: When you step on a skateboard and try to walk forward, the force you apply forward causes the skateboard to react oppositely and 'zoom away' due to low friction, often leading to a fall.

  • Friction in Skateboards: The relevant friction for wheel movement is within the axle (where ball bearings are used to minimize it for free spinning), not the friction between the wheels and the ground (which would be skidding).

Pulleys and Atwood Machines

• Setup: An Atwood machine typically consists of two masses ($m<em>1$ and $m</em>2$) connected by a rope over a pulley. For calculations, the rope and pulley are usually assumed to be massless.

• Force of Gravity (Weight): The primary force acting on each mass is gravity ($F<em>g$), calculated as mass times the acceleration due to gravity ($g$): $F</em>g = mg$.

• Direction of Motion: The system will accelerate in the direction of the heavier mass.

• Calculating Acceleration for an Atwood Machine:

  • Step 1: Calculate individual gravitational forces.

    • $F<em>{g1} = m</em>1 g$

    • $F<em>{g2} = m</em>2 g$

  • Step 2: Determine the net force on the system. Since the forces act in opposite directions (one pulling down, the other pulling up through the pulley), the net force is the difference between these two: $F<em>{net} = |F</em>{g1} - F_{g2}|$.

  • Step 3: Calculate the total mass of the system. $M<em>{total} = m</em>1 + m_2$.

  • Step 4: Use Newton's Second Law to find acceleration. $F<em>{net} = M</em>{total} a \implies a = \frac{F<em>{net}}{M</em>{total}}$.

  • Combined Formula: A direct formula for the acceleration of an Atwood machine is $a = \frac{(m<em>1 - m</em>2)g}{m<em>1 + m</em>2}$ (assuming m1 > m2).

• Practice Problem 1:

  • Block A ($m<em>A$) = $25 \text{ kg}$, Block B ($m</em>B$) = $15 \text{ kg}$.

  • Acceleration due to gravity ($g$) = $10 \text{ m/s}^2$.

  • Calculations:

    • Force of gravity on A: $F_{gA} = 25 \text{ kg} \times 10 \text{ m/s}^2 = 250 \text{ N}$.

    • Force of gravity on B: $F_{gB} = 15 \text{ kg} \times 10 \text{ m/s}^2 = 150 \text{ N}$.

    • Net force: $F_{net} = 250 \text{ N} - 150 \text{ N} = 100 \text{ N}$.

    • Total mass: $M_{total} = 25 \text{ kg} + 15 \text{ kg} = 40 \text{ kg}$.

    • Acceleration: $a = \frac{100 \text{ N}}{40 \text{ kg}} = 2.5 \text{ m/s}^2$.

• Practice Problem 2:

  • Mass 1 = $18 \text{ kg}$, Mass 2 = $13 \text{ kg}$.

  • Acceleration due to gravity ($g$) = $9.8 \text{ m/s}^2$.

  • Calculations (using formula):

    • $a = \frac{(18 \text{ kg} - 13 \text{ kg}) \times 9.8 \text{ m/s}^2}{18 \text{ kg} + 13 \text{ kg}}$

    • $a = \frac{5 \text{ kg} \times 9.8 \text{ m/s}^2}{31 \text{ kg}}$

    • $a = \frac{49 \text{ N}}{31 \text{ kg}} \approx 1.58 \text{ m/s}^2$

    • Rounded to $1.6 \text{ m/s}^2$.

• Constraint: The acceleration of a pulley system can never exceed the acceleration due to gravity ($g$) because that would imply an external force is pushing it faster.

Torque

• Concept: Torque is the rotational equivalent of linear force. It's what causes an object to rotate around a pivot point or fulcrum.

• Example: A seesaw. When a force is applied downward on one end, it creates a rotational movement around the central balance point.

• Equation: Torque ($\tau$) is calculated as the product of the force ($F$) and the perpendicular distance ($r<em>{\perp}$) from the fulcrum to the line of action of the force: $\tau = F \times r</em>{\perp}$.

  • The perpendicular distance is crucial; if the force is applied directly at the fulcrum, there is no torque regardless of how long the lever arm is.

• Leverage Implications: Torque demonstrates how a small force applied at a great distance from a fulcrum can lift a very heavy object a short distance, or how slight movements can result in significant rotational force.

  • Examples: Trebuchets (launching objects far), heavy machinery (lifting heavy weights).

Personal Context and Social Observations

Personal Background and Anecdotes

• Hair: The speaker's natural hair color was very blonde, which their mother liked. Initially only allowed partial dyeing (green and blonde), eventually got permission for full hair dye. The speaker dyes their hair approximately once a month. The idea of