Question

A lift of mass $M$ is accelerating upwards with an acceleration $a$. If the tension in the supporting cable is $T_1$ during upward acceleration and $T_2$ when it accelerates downwards with the same acceleration $a$, then the ratio $T_1 / T_2$ is:

• A. $\frac{g+a}{g-a}$
• B. $\frac{g-a}{g+a}$
• C. $\frac{g^2+a^2}{g^2-a^2}$
• D. $1$

Hint:

Moving upwards increases apparent gravity ($g_{\text{eff}} = g+a$), while moving downwards decreases it ($g_{\text{eff}} = g-a$). The ratio of tensions is simply the ratio of their effective gravities.

Answer 1

The Correct Option is A

Step 1: Understand the lift.
A lift of mass $M$ hangs from a cable. The cable pulls up with tension $T$. Gravity pulls down with weight $Mg$. The amount of tension changes when the lift speeds up or slows down.

Step 2: Apply Newton's second law for upward motion.
When the lift moves up with acceleration $a$, the upward pull is bigger than the weight. The extra force gives the upward push. \[ T_1 - Mg = Ma \] So \[ T_1 = M(g+a) \]

Step 3: Apply Newton's second law for downward motion.
When the lift moves down with the same acceleration $a$, now the weight is bigger than the tension. \[ Mg - T_2 = Ma \] So \[ T_2 = M(g-a) \]

Step 4: Form the ratio.
Divide the first result by the second to get the ratio asked. \[ \frac{T_1}{T_2} = \frac{M(g+a)}{M(g-a)} \]

Step 5: Cancel the mass.
The mass $M$ is the same on top and bottom, so it cancels out. \[ \frac{T_1}{T_2} = \frac{g+a}{g-a} \]

Step 6: Check the logic.
Going up needs more tension, so $T_1$ is larger. Going down needs less, so $T_2$ is smaller. That is why the ratio is more than one, which matches $\dfrac{g+a}{g-a}$. \[ \boxed{\dfrac{T_1}{T_2} = \dfrac{g+a}{g-a}} \]