An ideal inextensible rope has constant total length. When different parts of the rope move, that fixed-length condition creates a kinematic constraint relating their positions, velocities, and accelerations.

The key method is to write the variable part of the rope length in terms of coordinates and set its time derivative to zero.

A movable-pulley constraint

Suppose one end of an ideal rope is fixed to the ceiling, the rope passes down around a movable pulley, then back up over a fixed pulley, and its free end supports another mass. Let

• $x_1$ be the downward position of the free end;
• $x_2$ be the downward position of the movable pulley.

Ignoring constant lengths around the pulley rims, the variable rope length is

$$L=x_1+2x_2+\text{constant}.$$

The movable pulley contributes twice because two rope segments change length when it moves.

Since the rope is inextensible,

$$\frac{dL}{dt}=0,$$

so

$$\boxed{v_1+2v_2=0}.$$

Differentiating again,

$$\boxed{a_1+2a_2=0}.$$

Thus if the movable pulley rises with speed $v$, the free end moves downward with speed $2v$.

The coefficients come from geometry, not from force balance. A different pulley arrangement can produce a different constraint equation.

Constraint equations and Newton's laws are different equations

The rope constraint tells how the motions are related. Newton's second law tells how forces produce those motions. Both are needed to solve a pulley dynamics problem.

For an ideal massless rope passing over frictionless pulleys, the tension magnitude is the same along that rope. If mass $m_1$ hangs from the free end and mass $m_2$ is attached to the movable pulley, taking downward as positive for both coordinates gives

$$m_1g-T=m_1a_1,$$

and the movable pulley assembly is supported by two rope segments, so

$$m_2g-2T=m_2a_2.$$

The kinematic constraint is

$$a_1=-2a_2.$$

Substituting this into the force equations gives

$$a_2=\frac{(m_2-2m_1)g}{m_2+4m_1},$$

and

$$a_1=-2a_2.$$

The sign of the result determines which way the actual motion occurs.

Worked example

Let

$$m_1=4.0,\mathrm{kg},\qquad m_2=6.0,\mathrm{kg}.$$

Then

$$a_2=\frac{(6.0-8.0)g}{6.0+16.0} =-\frac{g}{11} \approx-0.89,\mathrm{m/s^2}.$$

Because downward was chosen positive, the movable pulley and $m_2$ accelerate upward at about

$$0.89,\mathrm{m/s^2}.$$

The free end satisfies

$$a_1=-2a_2\approx1.78,\mathrm{m/s^2},$$

so $m_1$ accelerates downward twice as fast.

Using the first force equation,

$$T=m_1(g-a_1) \approx(4.0)(9.81-1.78) \approx32.1,\mathrm N.$$

General procedure

For an ideal rope-and-pulley system:

1. choose coordinates for the independently moving bodies or pulleys;
2. express the total rope length using those coordinates;
3. differentiate the constant-length equation to obtain velocity or acceleration constraints;
4. draw the free-body diagram of each moving body;
5. solve the force equations together with the constraint equations.

Pulley ratios are therefore consequences of geometry plus inextensibility, not extra force laws.