Pulley System PYQ | JEE Main Physics 2025 | Constraint Relation Trick

 

❓ Question

In the pulley system shown, the block A moves with acceleration $a_2$ and the block B moves with acceleration $a_1$.

Find the relation between $a_1$ and $a_2$.

---

✍️ Solution

For an ideal pulley system,

$$\sum \vec a_i\cdot \vec T_i=0,$$

or equivalently,

$$\sum a_iT\cos\theta=0.$$

---

Contribution of Block $A$

Block $A$ is connected to one tension segment.

The acceleration is along the direction of tension, so

$$\theta=0^\circ.$$

Hence,

$$a_A\,T\cos0^\circ=a_2T.$$

Therefore,

$$\boxed{a_A\cdot T_A=a_2T.}$$

---

Contribution of Block $B$

Block $B$ is supported by six tension segments.

Its acceleration is opposite to the direction of tension, so

$$\theta=180^\circ.$$

Hence,

$$a_B\,(6T)\cos180^\circ = -6a_1T.$$

Therefore,

$$\boxed{a_B\cdot T_B=-6a_1T.}$$

---

Applying the Constraint Equation

Adding the contributions,

$$a_2T-6a_1T=0.$$

Cancelling $T$,

$$a_2=6a_1.$$

---

✅ Final Answer

$$\boxed{a_2=6a_1}$$

or equivalently,

$$\boxed{a_1=\frac{a_2}{6}}.$$