Irodov 1.71: An Elevator Makes Gravity Stronger – Amit & Deepali Sinha Foundation

Problem Irodov 1.71

A single pulley is mounted on the ceiling of an elevator car. A thread passes over it, carrying a load of mass $m_1$ on one end and a load of mass $m_2$ on the other. The pulley and thread are massless and frictionless. The car begins to accelerate upward with a constant acceleration $\vect w_0$ . Find (a) the acceleration $\vect w_1$ of load $m_1$ as seen from the fixed elevator shaft (the ground), and its acceleration $\vect w_1'$ as seen from inside the car; (b) the force $\vect F$ the pulley exerts on the car’s ceiling.

The pulley fixed to the ceiling of the elevator car, carrying m_1 and m_2 on a common thread, while the car accelerates upward at w_0. — The pulley fixed to the ceiling of the elevator car, carrying $m_1$ and $m_2$ on a common thread, while the car accelerates upward at $\vect w_0$ .

Given

$m_1$
$m_2$
$\vect w_0$ (car’s upward acceleration)
massless pulley and thread, no friction

Find

(a)

\vect w_1

and

\vect w_1'

, (b)

\vect F

on the ceiling

Solution

Step 1The car’s frame sees a stronger effective gravity

A generic load m_i on the thread: tension T toward the pulley, weight m_i g. — A generic load $m_i$ on the thread: tension $\vect T$ toward the pulley, weight $m_i\vect g$ .

In the elevator’s own (non-inertial) frame, every mass feels an extra pseudo-acceleration equal to minus the car’s own acceleration, $-\vect w_0$ . Combined with true gravity, each load behaves as if it hung in a stationary field of the effective strength

\vect g' = \vect g - \vect w_0

(1)

Step 2An ordinary Atwood machine in that field

Relative to the car, this is a plain two-mass Atwood machine in the effective field $\vect g'$ from Step 1. Write $g'$ for that field’s magnitude, $g'=g+w_0$ , and use it (a scalar) for the rest of this step; the thread constraint makes $m_1$ and $m_2$ ’s accelerations relative to the car equal and opposite. Newton’s second law for each load, taking the direction of $m_1$ ’s own descent as positive for both,

T = m_1(g'-w_1'), \qquad T = m_2(g'+w_1').

Equating the two and solving for $w_1'$ , then substituting back for $T$ ,

\begin{align*}m_1(g'-w_1') &= m_2(g'+w_1') \\ g'(m_1-m_2) &= w_1'(m_1+m_2) \\ w_1' &= \frac{m_1-m_2}{m_1+m_2}\,g'\end{align*}

T = m_1(g'-w_1') = m_1\!\left(g' - \frac{m_1-m_2}{m_1+m_2}g'\right) = \frac{2m_1m_2}{m_1+m_2}\,g'

T = \dfrac{2m_1m_2}{m_1+m_2}\,g'

(2)

Writing $w_1'$ as a vector along $\vect g'$ and substituting (1) for $\vect g'$ ,

\vect w_1' = \dfrac{m_1-m_2}{m_1+m_2}(\vect g-\vect w_0)

(3)

Step 3Back to the shaft-fixed frame

The ground-frame acceleration of $m_1$ is the car’s own acceleration plus $m_1$ ’s acceleration relative to the car (plain vector addition of frame and relative motion),

\vect w_1 = \vect w_0 + \vect w_1' = \vect w_0 + \frac{m_1-m_2}{m_1+m_2}(\vect g-\vect w_0).

Expanding and collecting the $\vect g$ and $\vect w_0$ terms over the common denominator $m_1+m_2$ ,

\begin{align*}\vect w_1 &= \frac{(m_1+m_2)\vect w_0+(m_1-m_2)\vect g-(m_1-m_2)\vect w_0}{m_1+m_2} \\ &= \frac{(m_1-m_2)\vect g + 2m_2\vect w_0}{m_1+m_2}\end{align*}

\vect w_1 = \dfrac{(m_1-m_2)\vect g+2m_2 \vect w_0}{m_1+m_2}

(4)

Step 4Tension, and the force on the ceiling

Both strands pull down on the massless pulley with the same tension $T$ from (2), so the ceiling must push up on the pulley with $2T$ for the (massless) pulley’s net force to stay zero; by Newton’s third law the pulley pulls down on the ceiling with that same magnitude.

F = 2T = \frac{4m_1m_2}{m_1+m_2}\,g'

$F$ acts straight down, the direction of $\vect g'$ , so the vector $\vect g-\vect w_0$ from (1) supplies both its magnitude $g'$ and its direction at once,

\vect F = \dfrac{4m_1m_2}{m_1+m_2}(\vect g-\vect w_0)

(5)

Check

As $w_0\to0$ (car at rest or moving at constant velocity), Step 2 and Step 4 both fall back to the ordinary suspended-Atwood results, $w_1'=w_1=(m_1-m_2)g/(m_1+m_2)$ and $F=4m_1m_2g/(m_1+m_2)$ , as they must once the car is no longer accelerating. And at $m_1=m_2$ , (4) gives $w_1=w_0$ : with balanced loads nothing slides relative to the car, so $m_1$ simply rides along with the car’s own acceleration, exactly as the boxed result says.

Remark

$F$ in Step 4 exceeds the ordinary, non-accelerating ceiling force $4m_1m_2g/(m_1+m_2)$ by exactly the factor $(g+w_0)/g$ , the same ratio by which $g'$ in (1) exceeds true $g$ : the pulley, not just the two loads, transmits the elevator’s own acceleration straight into the force it puts on the ceiling.