Drag, Free-Form vs Non-Free-Form Fall, and Terminal Velocity
Free-form vs Non-free form (drag-influenced) motion
Free-form fall (no air resistance):
- Net force F_net equals weight W
- W = m g
- Acceleration a = F_net / m = g (direction: downward)
- In this case, weight is the only force acting, so the object accelerates at g
Non-free form (with air resistance/drag):
- Forces: weight W downward, drag D upward (opposes motion)
- Net force F_net = W − D
- Acceleration a = F_net / m = (W − D) / m
- Drag increases with speed, so as speed increases, D grows and a decreases
- The net force is the difference between two forces, weight and drag; it must be smaller than the weight itself (since D > 0)
Weight and drag in the context of the given numbers
- Given: Weight W = 20 N, Drag D = 5 N
- Net force F_net = W − D = 20 − 5 = 15 N
- Mass m can be found from W = m g, so m = W / g
- With g ≈ 9.8 m/s^2, m ≈ 20 / 9.8 ≈ 2.04 kg
- Therefore, acceleration a = F_net / m ≈ 15 / 2.04 ≈ 7.35 m/s^2 (downward)
Formula recap (key relationships)
- Net force in non-free form:
- Weight:
- Free-form (no drag):
- Non-free form (with drag):
- Terminal velocity condition: drag balances weight, so
- If drag depends on velocity as (quadratic drag), then terminal velocity is
- If drag is linear in velocity, e.g., , then at terminal velocity
Drag and velocity dynamics
- Drag grows with speed, so as v increases, D increases and F_net = W − D decreases
- Acceleration a = Fnet / m decreases as Fnet decreases
- When D approaches W, F_net → 0 and a → 0; velocity approaches a constant: terminal velocity
- The direction of acceleration during fall remains downward until terminal velocity is reached
Terminal velocity in practice: feather vs coin (mass effect)
- With the same shape/drag characteristics, heavier object has larger terminal velocity because weight is larger, so it takes a larger drag force to balance it
- Consequence: heavier object reaches a higher terminal speed, and (depending on drag law) may reach it later or sooner depending on how acceleration changes over time
- In many simple drag models, a lighter object reaches its (lower) terminal velocity sooner; a heavier object takes longer to accelerate to a higher terminal velocity
Worked scenario: two objects of different weights in non-free fall
- First conclusion: lighter object reaches terminal velocity first (has a lower v_t)
- Heavier object: reaches terminal velocity later but at a higher v_t
- Explanation: terminal velocity is set by the condition D(vt) = W = m g; heavier object has larger W, so vt must be larger to make D(v) equal to W
Practical questions and quick checks
Question: Who reaches terminal velocity first?
Answer: The lighter object reaches terminal velocity first, but its terminal velocity is lower; the heavier object reaches its terminal velocity later but with a higher terminal velocity
Question: An object is said to fall at a constant velocity of 1000 km/h when the forces balance (i.e., zero acceleration). How is the air resistance determined?
At constant velocity, a = 0, so F_net = 0
Therefore, Drag must balance Weight: D = W = m g
To find D at any given v, you would use the drag model D(v) and solve D(v) = m g for v
Notes on speed regimes and relationships
- In free-form (no drag): acceleration is constant and equal to g, regardless of speed
- In non-free form (with drag): acceleration decreases as speed increases, eventually reaching zero at terminal velocity
- Terminal velocity depends on mass, cross-sectional area, drag coefficient, fluid density, and drag-law exponent (e.g., quadratic vs linear drag)
Connections to foundational principles
- Newton's laws: F = m a governs both free and drag-influenced falls
- Weight as the marshalling force: W = m g
- Drag as a resistive force arising from fluid dynamics, increasing with velocity
Final takeaways
- Free form: a = g; net force equals weight
- Non-free form: a = (W − D)/m; drag reduces acceleration as speed grows
- Terminal velocity occurs when D(v_t) = W
- Heavier objects tend to have higher terminal velocities because a larger weight requires a larger drag to balance it; lighter objects reach smaller terminal velocities
- If you know the drag law, you can compute vt; if D = c v^2, then