Resistive Forces
Introduced| English |
|---|
| resistive force/rɪˈzɪstɪv fɔːs/ |
| differential equation/ˌdɪfəˈrenʃl ɪˈkweɪʒn/ |
| terminal velocity/ˈtɜːmɪnl vəˈlɒsɪti/ |
Why a skydiver stops speeding up
- Jump from a plane and you accelerate -- but not forever.
- Air pushes back harder the faster you fall, until it exactly balances gravity.
- From then on you fall at a steady, top speed.
- Understanding this needs a force that grows with velocity -- and a little calculus.
A resistive force
- A resistive force 阻力 (air resistance, drag) opposes motion through a fluid.
- Unlike friction, it grows with speed -- move faster and it fights back harder.
- At rest there is no drag at all; at high speed it can be huge.
A resistive force grows larger as the object moves faster.
Unlike ordinary friction, drag increases with speed -- that is why it can eventually balance gravity.
Modeling the drag
- We often model drag as proportional to speed, $F = -bv$, or to speed squared, $F = -cv^2$.
- The minus sign means it always points against the motion.
- Applying $\sum F = ma$ gives a differential equation 微分方程 for the velocity.
Applying Newton's second law to a velocity-dependent drag gives a ____ equation for v(t).
$m\,dv/dt = mg - bv$ is a differential equation, solved to find how v approaches terminal velocity.
Terminal velocity
- As you speed up, drag grows until it balances the driving force (gravity).
- Then the net force is zero, acceleration stops, and speed levels off.
- That steady top speed is the terminal velocity 终极速度.
Resistive forces and terminal velocity
As a falling body speeds up, drag grows until it balances weight and the body stops accelerating.
At terminal velocity, the object's acceleration is...
Drag has grown to cancel gravity, so the net force -- and the acceleration -- is zero.
Approaching the limit
- Solving $m\dfrac{dv}{dt} = mg - bv$ shows $v$ rising smoothly toward $v_t = mg/b$.
- It gets ever closer but never quite exceeds it.
- That is why a skydiver's speed flattens out instead of climbing without limit.
An object of weight $mg = 60\ \text{N}$ falls with drag $F = -bv$ where $b = 3\ \tfrac{\text{kg}}{\text{s}}$. Find its terminal velocity (in m/s).
At terminal velocity $mg = bv_t$, so $v_t = 60/3 = 20\ \tfrac{\text{m}}{\text{s}}$.
Select all true statements about terminal velocity.
Drag acts the whole way down. Terminal velocity is just where it has grown enough to cancel gravity.
With the same drag coefficient $b$, a heavier object has a terminal velocity that is...
$v_t = mg/b$ grows with mass, so a heavier object falls faster at the top speed.
A falling object of mass $m$ feels drag $F = -bv$.
- Terminal velocity is where $\dfrac{dv}{dt} = 0$, so $mg = b v_t$.
- Solving, $v_t = \dfrac{mg}{b}$ -- heavier objects or weaker drag mean a higher top speed.
Terminal velocity is not the moment drag appears -- drag acts the whole way down. It is the special speed where drag has grown enough to cancel gravity. Below it, you are still accelerating (just less and less).
A resistive force grows with speed ($F = -bv$ or $-cv^2$), so $\sum F = ma$ becomes a differential equation. As speed rises, drag builds until it cancels gravity -- the net force hits zero at the terminal velocity $v_t = mg/b$, which the object approaches smoothly but never exceeds.