Solution
Given: $m = 15$ kg, $g = 10$ m/s$^2$.
Initial state (Stationary): Tension in each lower cord $F = 7.5$ N.
Initial Equilibrium Analysis
Upward force: $T_{1,i}$ (Top cord)
Downward forces: $mg$ + $2F$ (Two lower cords)
$T_{1,i} = mg + 2F = 15(10) + 2(7.5) = 150 + 15 = 165$ N.
Elasticity Constraint: The cords are elastic. Let the stiffness be $k$. Since the total height of the elevator is fixed, if the block moves vertically by a distance $\delta$ relative to the elevator, the extension in the top cord changes by $\delta$ and the bottom cords change by $-\delta$ (or vice versa).
Therefore, the change in tension in the top cord ($\Delta T_1$) and the bottom cord ($\Delta T_2$) are related by magnitude:
$|\Delta T_1| = |\Delta T_2|$ (assuming identical cords).
Specifically, if block moves down relative to lift: Top stretches ($\Delta T_1 > 0$), Bottom relaxes ($\Delta T_2 < 0$). Thus $\Delta T_1 = - \Delta T_2$.
General Equation of Motion (Elevator accelerating up with $a$):
$T_1′ – 2T_2′ – mg = ma$
Expressing in terms of changes: $(T_{1,i} + \Delta T_1) – 2(F + \Delta T_2) – mg = ma$.
Since $T_{1,i} – 2F – mg = 0$, this simplifies to:
$\Delta T_1 – 2\Delta T_2 = ma$
Substitute $\Delta T_2 = -\Delta T_1$:
$\Delta T_1 – 2(-\Delta T_1) = ma \Rightarrow 3\Delta T_1 = ma \implies \Delta T_1 = \frac{ma}{3}$
(a) Acceleration for lower cords to become relaxed
Relaxed means $T_2′ = 0$. The change required is $\Delta T_2 = 0 – 7.5 = -7.5$ N.
Using $\Delta T_1 = -\Delta T_2$, we get $\Delta T_1 = +7.5$ N.
Substitute into the motion equation:
$\Delta T_1 – 2\Delta T_2 = ma$
$7.5 – 2(-7.5) = 15a$
$22.5 = 15a \implies a = 1.5$ m/s$^2$ (Upwards).
(b) Elevator moves up with $a = 1.0$ m/s$^2$
Calculate change in tension:
$\Delta T_1 = \frac{ma}{3} = \frac{15(1)}{3} = 5$ N.
New Tension $T_{1,final} = T_{1,i} + \Delta T_1 = 165 + 5 = 170$ N.
(c) Elevator moves up with $a = 2.0$ m/s$^2$
First, check for slackness using the linear model:
$\Delta T_1 = \frac{15(2)}{3} = 10$ N.
Then $\Delta T_2 = -10$ N.
New bottom tension would be $F + \Delta T_2 = 7.5 – 10 = -2.5$ N.
Impossible. Cords cannot push. Thus, the lower cords go slack ($T_2′ = 0$).
We must solve the equation anew with $T_2′ = 0$:
$T_1′ – 0 – mg = ma$
$T_1′ = m(g + a) = 15(10 + 2) = 15(12) = 180$ N.
(d) Acceleration for upper cord to become relaxed
Relaxed means $T_1′ = 0$. Change required $\Delta T_1 = -165$ N.
This implies the block moves significantly upwards relative to the elevator. The bottom cords will stretch.
$\Delta T_2 = -\Delta T_1 = +165$ N.
Equation: $\Delta T_1 – 2\Delta T_2 = ma$
$-165 – 2(165) = 15a$
$-495 = 15a \implies a = -33$ m/s$^2$.
Note: Negative sign indicates acceleration is downwards.