Pulley Problems JEE: Two-Block Systems and Friction

How do you solve a two-block pulley problem?

For pulley problems JEE students face, use one fixed process: draw separate free-body diagrams, assume a motion direction, write the string constraint, apply Newton’s second law to each block and solve the equations simultaneously. An ideal fixed pulley with a massless inextensible string gives equal acceleration magnitudes and the same tension on both sides. Test friction before choosing its formula.

Use this five-step sequence every time:

1. Draw separate free-body diagrams. Define the blocks in the system and show only the forces acting on each block.
2. Assume a motion direction. Usually assume the heavier or more strongly driven side moves forward. This is only a sign choice.
3. Write the string constraint. For one taut, inextensible string passing over one fixed pulley, both ends have equal displacement, speed and acceleration magnitudes.
4. Apply Newton’s second law to each block. Choose one positive direction for each block and write the force equation along that direction.
5. Solve the equations simultaneously. Find acceleration first, then substitute it into either block equation to find tension.

The governing law for each block is:

A negative acceleration does not make the solution invalid. It means the actual motion is opposite to the direction you assumed.

Before inserting friction as coefficient times normal reaction, decide whether the friction is static or kinetic. First test whether static friction can maintain equilibrium. Use kinetic friction only after proving that sliding occurs.

Static friction satisfies: $0\le f_s\le {\mu }_{s}N$

Kinetic friction during sliding is: $f_k={\mu }_{k}N$

Why should you build pulley equations from Newton’s laws?

Newton’s laws work for every standard two-block setup, while shortcut formulas work only under specific assumptions. This is the same Class 11 free-body diagram method used throughout mechanics. Define the system, write the string constraint and apply Newton’s second law block by block. The familiar Atwood-machine formulas then follow from the equations instead of being memorised.

What assumptions define an ideal fixed-pulley model?

The standard model uses a light inextensible string, a smooth massless pulley, a taut string and point-like or non-rotating blocks. These conditions create the equal-acceleration and equal-tension results used in basic questions. Check them before using any fixed-pulley shortcut.

Consider two hanging masses connected by one string over a fixed pulley. Let the second mass be greater than the first, so the second mass is assumed to move downward while the first moves upward.

Because the string is inextensible, its total length cannot change. If one end moves downward by a certain distance, the other end must move upward by the same distance. Differentiating this displacement relation with respect to time gives equal speed magnitudes, and differentiating again gives equal acceleration magnitudes.

The directions are opposite in space. Their coordinate signs depend on the positive axis chosen for each block.

Equal tension needs a separate argument. It follows from the string being massless and the pulley being smooth and massless, not merely from the string being continuous. A massive string can have different tensions along its length, while pulley inertia or axle friction can require different tensions on the two sides.

How are the Atwood-machine equations derived?

Choose upward as positive for the lighter first mass and downward as positive for the heavier second mass. Both equations can then use the same positive acceleration magnitude, even though the blocks move in opposite physical directions. Apply Newton’s second law separately before combining the equations.

For the first mass: $T-m_{1}g=m_{1}a$

For the second mass: $m_{2}g-T=m_{2}a$

Add the equations. Tension disappears because it is an internal force for the combined two-block system.

Therefore:

This is the system viewpoint. The net external driving force is the difference between the two weights, while the total accelerated mass is the sum of both masses.

The system equation finds acceleration quickly, but it cannot directly find tension because tension cancels. Use an individual block equation to find tension.

Using the first mass: $T=m_1(g+a)$

This tension formula belongs only to the ideal two-mass Atwood-machine arrangement. Do not use it for a table-plus-hanging-block problem or a setup with pulley inertia.

For a broader treatment, including non-basic constraints, read constraints, tension and friction.

How can you check an Atwood-machine answer quickly?

Use three checks. For two positive hanging masses, acceleration must be less than gravitational acceleration. When the second mass is heavier, tension must lie between the two weights. Both individual block equations must also return the same tension after substitution.

• Check the acceleration: $a<g$
• Check the tension: $m_{1}g<T<m_{2}g$
• Substitute the calculated acceleration into both block equations and confirm that they produce the same tension.

How do you solve an ideal Atwood machine numerically?

For a 2 kg and 3 kg ideal Atwood machine, assume the 3 kg block moves downward and the 2 kg block moves upward. Write one Newton’s-law equation for each block, add them to eliminate tension, and substitute the acceleration back. The acceleration is 2 metres per second squared and the tension is 24 N.

Take:

For the 2 kg block, upward is positive: $T-20=2a$

For the 3 kg block, downward is positive: $30-T=3a$

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

Therefore: $a=2{\text{m s}}^{-2}$

Substitute into the first equation: $T-20=2(2)$ $T=24\text{N}$

Substitution into the second equation gives the same result: $30-T=3(2)$ $T=24\text{N}$

How can you verify the Atwood-machine answer independently?

Treat both blocks as one system. The external driving force is 10 N, and the string accelerates a total mass of 5 kg. Dividing the driving force by the total mass gives the same acceleration of 2 metres per second squared. The tension and acceleration also pass the physical bounds.

Thus:

The physical checks pass:

A tempting wrong equation is: $m_{2}g-m_{1}g=m_{2}a$

It uses the correct driving force but the wrong inertial mass. Both blocks accelerate, so the force difference must accelerate the combined mass, not only the heavier block.

How do you solve a pulley problem with friction on a table?

First test whether static friction can prevent motion. For a 4 kg table block connected to a hanging 3 kg block, the maximum static friction is 20 N, but the hanging weight is 30 N. Equilibrium is impossible. The blocks move, so kinetic friction of 10 N must be used in Newton’s-law equations.

Take:

The table block is pulled horizontally toward the pulley. The hanging block is assumed to move downward.

Can static friction keep the system at rest?

No. The normal reaction on the 4 kg table block is 40 N, so the maximum available static friction is 20 N. Equilibrium would require static friction of 30 N to balance the hanging block’s pull. The required friction exceeds the available maximum.

The table block has no vertical acceleration, so its normal reaction equals its weight. $N=4(10)=40\text{N}$

The maximum available static friction is: $f_{s,max}={\mu }_{s}N$

If the system were at rest, the hanging block would require a tension of 30 N. The table block would therefore need 30 N of static friction to balance that tension. $30\text{N}>20\text{N}$

Equilibrium is impossible, so the system moves.

Which friction should you use after motion starts?

Use kinetic friction after the blocks begin sliding. Maximum static friction no longer applies. Here the kinetic friction is 10 N. Assume the 3 kg block moves downward and the 4 kg block moves toward the pulley, then write one equation for each block. $f_k={\mu }_{k}N$

For the block on the table: $T-10=4a$

For the hanging block: $30-T=3a$

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

Therefore:

Substitute into the table-block equation:

Numerically: $T\approx 21.4\text{N}$

How can you check the rough-table answer?

Treat both blocks as one system. Tension is internal and cancels. The external driving force is the 30 N hanging weight minus 10 N of kinetic friction, giving 20 N. This force accelerates a total mass of 7 kg, so the system method must reproduce the block-by-block acceleration.

Therefore:

If the hanging pull were no greater than 20 N, static friction would adjust to the required value and acceleration would be zero. Static friction would equal 20 N only at limiting equilibrium.

What mistakes do JEE students make in pulley problems?

Most wrong answers come from applying a correct statement outside its conditions. The dangerous errors look reasonable: using maximum static friction automatically, setting tension equal to weight or calculating acceleration with only the heavier mass. Correct them by returning to separate free-body diagrams, the string constraint and Newton’s second law.

• Using maximum static friction automatically: Students remember friction as coefficient times normal reaction and insert it immediately. Static friction instead adjusts through a range: $0\le f_s\le {\mu }_{s}N$

It reaches its maximum only at limiting equilibrium. Test equilibrium first, then decide between static and kinetic friction.

• Giving both blocks the same signed acceleration: Their acceleration magnitudes are equal in the ideal fixed-pulley setup, but their physical directions are opposite. Signs depend on the coordinate axis chosen for each block. Choosing upward positive for one block and downward positive for the other allows both equations to use the same positive magnitude.
• Cancelling tension before drawing separate free-body diagrams: Tension is internal only when both blocks are treated as one combined system. It remains essential for analysing either block and calculating tension. Write individual equations before using the combined-system shortcut.
• Assuming tension equals weight: This is true only when that hanging block has zero acceleration. For an upward-accelerating block: $T>mg$

For a downward-accelerating block: $T<mg$

• Using only the heavier block’s mass: The external driving force accelerates every mass linked by the taut string. Divide it by the total accelerated mass.
• Assuming tension is always identical on both sides: Equal tension requires the ideal assumptions. Pulley rotational inertia, string mass or axle friction can make the two tensions different.
• Writing equal accelerations without checking the constraint: Equal magnitudes require a taut, inextensible string over one fixed pulley. A slack string gives no such relation, while a movable-pulley arrangement can produce a different acceleration ratio.
• Treating a negative acceleration as failure: A negative value means the initial direction assumption was reversed. Keep the magnitude and reverse the stated direction.

Before marking any answer, perform this 15-second audit:

1. Check the units of acceleration and tension.
2. Check whether the final direction agrees with the sign.
3. Test limiting cases, such as equal hanging masses or zero friction.
4. For two freely hanging positive masses, verify: $a<g$
5. Substitute the acceleration into both block equations and confirm that they give the same tension.