Three masses $m_1 = 4$ kg, $m_2 = 4$ kg and $m_3 = 6$ kg are suspended from a fixed smooth frictionless pully as shown in the figure below. The value of $T_1/T_2$ is ______ (take $g = 10\ \text{m/s}^2$)
Answer: (A) $5/3$
Concept. All three masses move with the same acceleration magnitude: $m_1$ on one side, $m_2$ and $m_3$ (hanging one below the other) on the other. Find $a$ from the whole system, then isolate single masses to get each tension.
Step 1. Read the figure. $T_1$ is the tension in the string over the pulley (it holds $m_1$ on the left and $m_2$ on the right). $T_2$ is the tension in the short string between $m_2$ and $m_3$.
Step 2. Acceleration. Left side $4$ kg, right side $m_2+m_3=10$ kg, so the right side goes down:
$$a=\frac{(10-4)g}{4+10}=\frac{60}{14}=\frac{30}{7}\ \text{m s}^{-2}$$
Step 3. Isolate $m_1$ (accelerating up). Forces: $T_1$ up, $m_1g$ down:
$$T_1-m_1g=m_1a\ \Rightarrow\ T_1=4\left(10+\frac{30}{7}\right)=4\times\frac{100}{7}=\frac{400}{7}\ \text{N}$$
Step 4. Isolate $m_3$ (accelerating down). Forces: $T_2$ up, $m_3g$ down:
$$m_3g-T_2=m_3a\ \Rightarrow\ T_2=6\left(10-\frac{30}{7}\right)=6\times\frac{40}{7}=\frac{240}{7}\ \text{N}$$
Step 5. Ratio.
$$\frac{T_1}{T_2}=\frac{400}{240}=\frac53$$
Answer. Option A. Check with $m_2$ (moving down, pulled up by $T_1$ and down by $T_2$ and its weight): $m_2g+T_2-T_1=40+\dfrac{240-400}{7}=\dfrac{120}{7}$, and $m_2a=4\times\dfrac{30}{7}=\dfrac{120}{7}$. Correct.
Solution by Sreeraj P, M.Sc Physics
