How Does a Screw Fall in an Accelerating Elevator?

A screw released inside an elevator does not always fall toward the floor at the usual rate of $g$. What matters for its motion inside the cabin is the screw’s acceleration relative to the elevator, not just its acceleration relative to the ground.

Assume the screw starts at rest relative to the elevator, the elevator’s acceleration is constant during the fall, and air resistance is negligible. With those assumptions, you can calculate the screw’s time to reach the floor directly.

Calculate acceleration relative to the elevator

Choose upward as the positive direction. Let $A$ be the elevator’s acceleration, positive upward. After release, the screw is in free fall, so its acceleration relative to the ground is $-g$. The elevator’s acceleration relative to the ground is $A$. Subtract the elevator’s acceleration from the screw’s:

$$a_{\text{relative}}=a_{\text{screw}}-a_{\text{elevator}}=-g-A.$$

A negative result means the screw accelerates downward relative to the cabin. Its downward acceleration relative to the elevator is therefore $g+A$ when $A$ is positive. If the screw begins at rest relative to the elevator and the floor is a distance $h$ below it, the time to reach the floor is

$$t=\sqrt{\frac{2h}{g+A}}.$$

This time formula applies when $g+A>0$ and the elevator’s acceleration is constant over the fall. The screw’s initial velocity relative to the cabin is taken to be zero.

What happens in different elevator motions?

Elevator motion	Screw’s downward acceleration relative to the cabin	Result
Constant velocity, or at rest	$g$	Same fall time as in a stationary room
Accelerating upward at $A$	$g+A$	Reaches the floor sooner
Accelerating downward at magnitude $b<g$	$g-b$	Reaches the floor later
Accelerating downward at $g$	$0$	Stays at the same height relative to the cabin if released from rest

For a downward-accelerating elevator, set $A=-b$. Then the screw’s downward relative acceleration is $g-b$. When $b=g$, the cabin and screw both accelerate downward at $g$, so the screw appears weightless inside the cabin. If the elevator accelerates downward faster than $g$, the screw accelerates upward relative to it; it will not fall to the floor from rest relative to the cabin.

Worked example: elevator accelerating upward

Suppose the floor is $2.0\,\text{m}$ below the screw, and the elevator accelerates upward at $2.0\,\text{m/s}^2$. Use $g=9.8\,\text{m/s}^2$. The screw’s downward acceleration relative to the elevator is

$$g+A=9.8+2.0=11.8\,\text{m/s}^2.$$

Since the screw starts at rest relative to the cabin, its relative displacement magnitude is $h=\tfrac12(g+A)t^2$. Solving for time gives

$$t=\sqrt{\frac{2(2.0)}{11.8}}\approx 0.58\,\text{s}.$$

In a stationary elevator, the same $2.0\,\text{m}$ drop would take

$$t=\sqrt{\frac{2(2.0)}{9.8}}\approx 0.64\,\text{s}.$$

The upward acceleration makes the cabin move upward while the screw falls, increasing their relative closing acceleration and shortening the time.

The same result in the elevator’s frame

An accelerating elevator is a non-inertial frame. To use Newton’s laws in that frame, introduce a fictitious acceleration opposite to the elevator’s acceleration. If the elevator accelerates upward, this fictitious acceleration points downward, adding to gravity. The screw’s effective downward acceleration is then $g+A$, matching the result from subtracting accelerations in the ground frame.

This frame-based explanation is useful for understanding why objects seem heavier or lighter in an accelerating elevator. The relative-acceleration calculation is often the simplest way to get the screw’s position and fall time.

Common mistakes

• Using $g$ in every case. Gravity gives the screw’s ground-frame acceleration, but the question concerns its motion relative to the moving cabin.
• Adding accelerations without a sign convention. With upward positive, the screw has acceleration $-g$ and the elevator has acceleration $A$. Subtracting gives $-g-A$; interpret the sign before using a distance formula.
• Confusing velocity with acceleration. A fast-moving elevator with constant velocity does not change the relative fall acceleration. Its acceleration, not its speed, changes the answer.
• Assuming the screw starts at rest in every frame. The calculation above assumes it is released from rest relative to the elevator. If it has initial relative velocity $v_0$, use $y(t)=y_0+v_0t+\tfrac12 a_{\text{relative}}t^2$ with a consistent upward-positive coordinate, then solve for when the screw and floor have the same position.

If the elevator’s acceleration varies during the fall, use the relative acceleration $a_{\text{relative}}(t)=-g-A(t)$ and integrate it over time rather than applying the constant-acceleration formula.