Question

If the system of blocks shown in the figure is released from rest, the ratio of the tensions $T_1$ and $T_2$ is (Neglect the mass of the string shown in the figure)

• A. $1:1$
• B. $1:2$
• C. $1:3$
• D. $3:4$

Hint:

For connected bodies, find the common acceleration first using $a = F_{net}/M_{total}$. Then isolate individual bodies to find internal tensions.

Answer 1

The Correct Option is B

Step 1: Identify Masses and Acceleration:
Left Side Mass: $M_L = 4\text{ kg}$. Right Side Total Mass: $M_R = 3\text{ kg} + 3\text{ kg} = 6\text{ kg}$. Since $M_R>M_L$, the right side moves down and the left side moves up with acceleration $a$. \[ a = \frac{(M_R - M_L)g}{M_R + M_L} = \frac{(6-4)g}{6+4} = \frac{2g}{10} = \frac{g}{5} \]
Step 2: Calculate Tension $T_2$ (Main String):
Consider the 4 kg block moving up: \[ T_2 - 4g = 4a \] \[ T_2 = 4(g + a) = 4\left(g + \frac{g}{5}\right) = 4\left(\frac{6g}{5}\right) = \frac{24g}{5} = 4.8g \]
Step 3: Calculate Tension $T_1$ (Lower String):
Consider the bottom-most 3 kg block moving down: \[ 3g - T_1 = 3a \] \[ T_1 = 3(g - a) = 3\left(g - \frac{g}{5}\right) = 3\left(\frac{4g}{5}\right) = \frac{12g}{5} = 2.4g \]
Step 4: Find Ratio:
\[ \frac{T_1}{T_2} = \frac{2.4g}{4.8g} = \frac{1}{2} \] The ratio is $1:2$.