Pulley Constraint Equations for JEE: String Length, Tension and Friction

How do you solve a two-block pulley problem?

Pulley problems in JEE are solved most reliably by starting with the string constraint. Draw a separate free-body diagram for each block. Choose one expected direction of motion and assign a common acceleration variable only after using the string-length constraint. Write

along each block’s direction of motion. Use the same tension on both sides only for a massless string passing over an ideal, frictionless pulley. Solve the simultaneous equations. A negative acceleration means the actual motion is opposite to the assumed direction.

Use this sequence:

1. Draw each block separately, away from the pulley diagram.
2. Mark weight, normal reaction, tension and friction where applicable.
3. Write the fixed-length string constraint.
4. Decide which end is expected to move down, right or up.
5. Assign positive acceleration along each block’s assumed motion.
6. Write the force equation for each block.
7. Solve the equations together.
8. Check the sign, units and physical limits.

Equal acceleration magnitude comes from the string constraint, not from visual similarity or Newton’s third law.

How do Newton’s laws and the fixed string length give the pulley equations?

Newton’s laws connect forces to acceleration, while the string constraint connects the accelerations of the blocks. Keep these two jobs separate. First use geometry to relate the motion. Then draw a separate free-body diagram for each mass and apply Newton’s second law. This order prevents sign errors and shows when a common acceleration is valid.

For an inextensible string over one fixed pulley, measure the downward coordinates of the two ends from the pulley. The exposed string length is $L=y_1+y_2+\text{constant}$

Because the total length does not change, $y_1+y_2=\text{constant}$

Differentiating once gives $v_1+v_2=0$

Differentiating again gives $a_1+a_2=0$

If one end moves downward by a distance of magnitude $x$ the other moves upward by the same distance. Their speed magnitudes and acceleration magnitudes are equal: $a_1,=,a_2$

The signs are opposite because both coordinates were defined downward. If each block’s positive direction is chosen along its expected motion, both equations can use the same positive acceleration variable.

Equal acceleration is caused by the fixed length of the string. It is not caused by Newton’s third law.

Which forces should you draw on each block?

Draw only the forces acting directly on the chosen block. In standard two-block systems, these are weight, normal reaction, tension and friction where applicable. Do not include forces exerted by the block on another object. Check the direction of each force before writing Newton’s second-law equation.

• Weight: Vertically downward, with magnitude $mg$
• Normal reaction: Perpendicular to the contact surface.
• Tension: Along the string and away from the block.
• Friction: Along the contact surface, opposing relative slipping or impending slipping.

Do not draw the block’s pull on the string as a force on the block. That force acts on the string and is the third-law partner of the string’s pull on the block.

When can both sides of the string have the same tension?

Both sides have equal tension only when the string is massless and the pulley is ideal. An ideal pulley has no rotational resistance, and its axle is frictionless. Under these assumptions, use one tension variable for both blocks. A massive pulley requires a net torque, so its two tensions are generally different. $T_1\ne T_2$

The torque equation is $(T_1-T_2)R=I\alpha$

The tension difference provides the torque needed to accelerate the pulley. Using one tension would remove that torque from the model.

What sign convention should you use in pulley problems?

Write each block’s equation as positive in its assumed direction of motion. If the heavier hanging block is expected to move downward, take downward as positive for that block. If the lighter block is expected to move upward, take upward as positive for it. This gives both blocks the same positive acceleration variable after the string constraint has been applied.

Do not force one global vertical direction onto two blocks moving oppositely unless you are comfortable with signed coordinates.

How do you derive the ideal Atwood-machine formulas?

For two hanging masses connected by a massless string over an ideal fixed pulley, the heavier mass is expected to move downward because its weight is greater. The lighter mass moves upward. The fixed-string constraint gives equal acceleration magnitudes, while the ideal pulley gives one common tension. Let the masses satisfy

The force equations are $m_{1}g-T=m_{1}a$ $T-m_{2}g=m_{2}a$

Add them. Tension cancels because it is internal to the two-block system:

Therefore,

Substitute this into either block equation: $T=m_2(g+a)$

After simplification,

Treating both blocks as one system gives acceleration quickly because internal tension cancels. I use this method first for acceleration. Separate block equations are still required to find tension.

Do movable pulleys also give equal accelerations?

No. Equal acceleration magnitudes apply to the two ends of one inextensible string passing over a fixed pulley. A movable pulley changes how several string segments share displacement, so the acceleration ratio need not be one to one. The ratio must come from that system’s string-length equation, not from the fixed-pulley result.

The constraint-first method still works, but the constraint changes. Master the fixed-pulley case before using it for movable pulleys.

How do you solve two hanging blocks on an ideal pulley?

For masses of 3 kg and 2 kg, the 3 kg block is expected to move downward because it is heavier, while the 2 kg block moves upward. The fixed-string constraint gives equal acceleration magnitudes. Separate force equations give $a=2{\text{m/s}}^2$ and $T=24\text{N}$

Take

Assume the 3 kg block moves downward. Its weight is 30 N downward, while tension acts upward: $30-T=3a$

Assume the 2 kg block moves upward. Tension acts upward, while its 20 N weight acts downward: $T-20=2a$

Add the equations: $30-T+T-20=3a+2a$ $10=5a$

Hence, $a=2{\text{m/s}}^2$

Substitute into the lighter block’s equation: $T-20=2(2)$ $T=24\text{N}$

The lighter block accelerates upward only if tension exceeds its weight. The heavier block accelerates downward only if tension is below its weight:

Tension cannot be ignored. Without tension, the heavier block would be in free fall with acceleration $g=10{\text{m/s}}^2$

Its actual acceleration is only $2{\text{m/s}}^2$

The string’s upward pull reduces the heavier block’s acceleration.

How do you solve one block on a rough table and one block hanging?

Compare the driving force with friction before choosing the expected direction. Here, the hanging weight is 30 N, while the resisting friction is 10 N. The hanging block is therefore expected to move downward and pull the table block rightward. Newton’s second law then gives

and

A 4 kg block lies on a horizontal table. It is connected over an ideal fixed pulley to a hanging 3 kg block. Use

The table block has no vertical acceleration, so $N=4g=40\text{N}$

The friction magnitude used in this model is $f=\mu N$

Compare the hanging weight with the resisting friction:

The expected motion is rightward for the table block and downward for the hanging block.

For the 4 kg table block, take rightward as positive: $T-10=4a$

For the 3 kg hanging block, take downward as positive: $30-T=3a$

Add the equations: $T-10+30-T=4a+3a$ $20=7a$

Therefore,

Substitute into the first equation:

For a whole-system check, tension is internal. The net external driving force is

The total moving mass is

Thus,

A real start-from-rest problem must distinguish static friction from kinetic friction when separate coefficients are supplied. Static friction adjusts as needed up to its limiting value. Only after slipping starts should kinetic friction be used.

What mistakes do students make in pulley problems?

Most errors in pulley problems for JEE come from the model, not the algebra. Students assign signs before deciding the expected motion, assume equal tension or acceleration without checking the pulley, or treat friction as a fixed force. Fix the free-body diagrams and string constraint before calculating. Correct algebra cannot repair a wrong model.

• Writing the same force equation for both hanging blocks: Students write $mg-T=ma$

for both blocks, even though one accelerates upward. This happens when an equation is copied instead of built from a sign convention. For the upward-moving block, write $T-mg=ma$

• Assuming tension always equals weight: The equality $T=mg$

holds only when that particular block has zero vertical acceleration. An accelerating block generally has tension different from its weight.

• Using the same acceleration in every pulley diagram: Equal magnitudes follow for the two ends of one inextensible string over a fixed pulley. Movable pulleys can produce other acceleration ratios because several string segments change length.
• Calling the two tension forces an action-reaction pair: Tension acts on each block, but these forces are not the Newton’s-third-law pair of one interaction. The partner of the string’s pull on a block is the block’s pull on the string.
• Setting friction equal to the limiting value automatically: Static friction satisfies $0\le f_s\le {\mu }_{s}N$

It adjusts to prevent slipping and need not equal its maximum value. The equation $f_s={\mu }_{s}N$ applies only at impending slip.

• Choosing friction opposite to velocity without examining the contact: Friction opposes relative slipping or the tendency to slip between surfaces. Identify that tendency first, especially when a surface itself moves or accelerates.
• Adding equations with inconsistent positive directions: Tension may appear to double rather than cancel. Rewrite each equation as positive along that block’s assumed motion before adding.
• Using equal tension with a massive pulley: Rotational inertia generally requires unequal tensions: $(T_1-T_2)R=I\alpha$

Combine the blocks’ translational equations with this torque equation and the no-slip relation.

• Ignoring a negative acceleration: A negative result means the actual motion is opposite to the assumed direction. Reverse the arrows and report the positive magnitude with the correct direction.
• Skipping limiting-case checks: For equal hanging masses, the acceleration formula must give $a=0$

If one hanging mass becomes negligible compared with the other, the acceleration should approach $a\to g$

When practising pulley problems, write the string constraint before any force equation, even when the acceleration relation looks obvious. That ten-second habit prevents sign errors and false equal-acceleration assumptions.

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