In short
When several bodies are connected — by contact, strings, or pulleys — they share a common acceleration. The system method treats them as one combined mass to find that acceleration: a = F_{\text{net}} / (m_1 + m_2 + \cdots). Then you draw an individual free body diagram (FBD) for any single body to find the internal forces (contact forces, tensions). For an Atwood machine with masses m_1 > m_2, the acceleration is a = (m_1 - m_2)\,g\,/\,(m_1 + m_2) and the tension is T = 2m_1 m_2 g\,/\,(m_1 + m_2).
A porter at a railway station pushes a chain of three loaded luggage trolleys. The trolleys are touching each other — the front one pushes the middle one, the middle one pushes the last one. All three accelerate together. But here is the question that trips up most students: if the porter pushes with 600 N, does every trolley feel 600 N? No. The front trolley feels the full push, but it passes on only a fraction to the middle one, and the middle one passes on an even smaller fraction to the last. The force decreases down the chain, even though the acceleration is the same everywhere.
Understanding why — and computing those internal forces — is what this article is about. You will learn two methods that work together: the system method (which gives you the acceleration in one line) and the individual FBD method (which reveals the forces between the bodies). Together, they solve every connected-body problem you will encounter in Class 11 and JEE.
Two blocks in contact — the simplest connected system
Place two blocks on a frictionless floor. Block 1 has mass M_1 and block 2 has mass M_2. You push block 1 with a horizontal force F. The blocks stay in contact and accelerate together.
Step 1 — System method: find the acceleration
Treat both blocks as a single system of mass (M_1 + M_2). The only external horizontal force on the system is F (the normal forces between the blocks are internal to the system, so they cancel in pairs by Newton's third law).
Why: since both blocks move together, they share the same acceleration. The system method lumps them into one mass and applies Newton's second law once. Internal forces (the push between the blocks) do not appear — they are action-reaction pairs and cancel within the system.
Step 2 — Individual FBD: find the contact force
Now isolate block 2 alone. The only horizontal force on it is the contact force N_{12} — the push that block 1 exerts on block 2.
Apply Newton's second law to block 2 alone:
Why: the only horizontal force on M_2 is the contact push N_{12}, and it must produce the same acceleration a you already found.
Substitute a from equation (1):
Why: the contact force is a fraction of the applied force. If M_2 is small compared to M_1, the contact force is small — a heavy block easily accelerates a light one without passing much force. If M_2 is large, the contact force is nearly equal to F — most of the applied force is spent pushing M_2.
Sanity check: If M_2 = 0 (no second block), N_{12} = 0. If M_1 = 0 (no first block, just the force on M_2 directly), N_{12} = F. Both make physical sense.
You can verify the FBD of M_1 as a cross-check. The net force on M_1 is F - N_{12}, which must equal M_1 a:
Why: both FBDs give the same contact force. This consistency check confirms your answer. Always verify from both sides when you can.
The Atwood machine — gravity pulls both sides
An Atwood machine is the simplest pulley system: two masses, m_1 and m_2, connected by a light inextensible string over a massless frictionless pulley. Think of a bucket on a rope in a village well — the heavy bucket goes down, and whatever is tied to the other end goes up. That is an Atwood machine.
Assumptions: The string is massless and inextensible (so the tension is the same throughout the string and both masses have the same magnitude of acceleration). The pulley is massless and frictionless (so it only redirects the string without absorbing energy or needing torque).
System method: find the acceleration
Think of the system as a whole. The net external force driving the motion is the difference in weights: m_1 g - m_2 g. The total mass being moved is m_1 + m_2. (Why add the masses even though they move in opposite directions? Because both masses resist acceleration — inertia does not care about direction.)
Why: the "engine" driving the system is the unbalanced weight (m_1 - m_2)g. The "load" is the total inertia (m_1 + m_2). A bigger mass difference means faster acceleration; a larger total mass means slower acceleration.
Individual FBDs: find the tension
Draw the FBD for each mass separately.
For m_1 (taking downward as positive, since m_1 moves down):
Why: the weight m_1 g pulls m_1 down, the tension T pulls it up. The net downward force is m_1 g - T, and this equals m_1 a by Newton's second law.
For m_2 (taking upward as positive, since m_2 moves up):
Why: the tension T pulls m_2 up, the weight m_2 g pulls it down. The net upward force is T - m_2 g, and this equals m_2 a.
Solve for T from equation (4):
Substitute a from equation (3):
Why: the tension depends on both masses. If either mass is zero, T = 0 (no string tension if there is nothing on one side). If m_1 = m_2, then T = m_1 g = m_2 g — the system is balanced, and the tension simply supports each mass against gravity. The tension is always less than the weight of the heavier mass (otherwise m_1 could not accelerate downward) and always more than the weight of the lighter mass (otherwise m_2 could not accelerate upward).
Cross-check from m_2's equation: From equation (5), T = m_2(g + a). Substituting a:
Both FBDs give the same tension. This is a powerful consistency check — always do it when you have time.
Blocks on a table connected by strings
Now extend to multiple blocks. Place three blocks on a frictionless table, connected end-to-end by light strings. A force F pulls the front block. All three accelerate together, but the tension in each string is different.
System method: one equation for acceleration
Why: same logic as before. All three blocks move together, so treat them as one mass. The only external horizontal force is F. All string tensions are internal to the system.
Individual FBDs: find each tension
For the last block M_3 (the one at the back):
The only horizontal force on M_3 is the string tension T_2 pulling it forward.
Why: T_2 alone must accelerate M_3. The further back the block, the smaller the tension — T_2 only needs to pull M_3, not the entire chain.
For the middle block M_2:
M_2 is pulled forward by T_1 and pulled backward by T_2.
Why: the tension T_1 must accelerate everything behind M_1 — that is, both M_2 and M_3. This is a general pattern: each string tension equals the total mass behind that string, multiplied by the common acceleration.
This pattern is clean and general. For n blocks on a frictionless surface pulled by F at the front, the tension in the string just behind block k is:
The force decreases as you go toward the back of the chain — exactly what the porter at the railway station experiences. The first trolley passes less force to the second than the porter applied, and the second passes even less to the third.
The system method vs. individual FBDs — when to use which
You now have two tools. Here is when to reach for each one.
| You want to find... | Use... | Why it works |
|---|---|---|
| The common acceleration of the system | System method | Internal forces cancel; one equation, one unknown |
| A contact force or string tension between two bodies | Individual FBD of one body | The internal force becomes an external force on that single body |
| Whether the system accelerates at all | System method | Quick check: is the net external force zero? |
| All forces in a complex setup | Both, in sequence | System gives a first, then individual FBDs give each internal force |
The decision rule is simple: start with the system method to get a, then use individual FBDs to get the internal forces. You almost never need to solve simultaneous equations if you do it in this order.
What changes if friction is present? The system method still works — just subtract the total friction from the applied force: a = (F - f_1 - f_2 - \cdots)/(M_1 + M_2 + \cdots). Individual FBDs now include friction forces on each block. The pattern is the same, but the algebra is a little longer. See the article on friction for full details.
Worked examples
Example 1: Atwood machine — 3 kg and 5 kg
Two masses, m_1 = 5 kg and m_2 = 3 kg, are connected by a light inextensible string over a massless frictionless pulley. Find the acceleration of the system and the tension in the string. Take g = 10 m/s².
Step 1. Identify knowns.
m_1 = 5 kg (heavier, descends), m_2 = 3 kg (lighter, ascends), g = 10 m/s².
Why: always label which mass is heavier. The heavier mass determines the direction of acceleration.
Step 2. Find acceleration using the system method.
Why: the net driving force is (5 - 3) \times 10 = 20 N, and the total mass being accelerated is 5 + 3 = 8 kg.
Step 3. Find the tension using the formula.
Why: alternatively, use T = m_1(g - a) = 5(10 - 2.5) = 5 \times 7.5 = 37.5 N. Or from the other side: T = m_2(g + a) = 3(10 + 2.5) = 3 \times 12.5 = 37.5 N. All three give the same answer — a reassuring consistency check.
Result: a = 2.5 m/s², T = 37.5 N.
What this shows: The tension (37.5 N) lies between the two weights (30 N and 50 N). It has to — if T were greater than 50 N, the heavier mass could not accelerate downward, and if T were less than 30 N, the lighter mass could not accelerate upward. The tension acts as a mediator, balancing the tug-of-war between gravity on each side.
Example 2: Three blocks on a table — 50 N pull
Three blocks of masses M_1 = 5 kg, M_2 = 3 kg, and M_3 = 2 kg are placed on a frictionless horizontal table and connected by light strings as shown. A force of F = 50 N is applied to the 5 kg block. Find the acceleration of the system and the tension in each string.
Step 1. Find the acceleration (system method).
Total mass: M_1 + M_2 + M_3 = 5 + 3 + 2 = 10 kg.
Why: all three blocks move as one. The only external horizontal force is 50 N, and the total mass is 10 kg.
Step 2. Find T_2 (tension between the 3 kg and 2 kg blocks).
Draw the FBD of the 2 kg block. The only horizontal force on it is T_2 pulling it forward.
Why: T_2 alone is responsible for accelerating the 2 kg block at 5 m/s².
Step 3. Find T_1 (tension between the 5 kg and 3 kg blocks).
T_1 must accelerate everything behind it — the 3 kg block and the 2 kg block.
Why: the string carrying tension T_1 is pulling both the 3 kg and 2 kg blocks. Alternatively, from the FBD of the 3 kg block: T_1 - T_2 = M_2 a, so T_1 = M_2 a + T_2 = 3 \times 5 + 10 = 25 N. Same answer.
Result: a = 5 m/s², T_1 = 25 N, T_2 = 10 N.
What this shows: The tension decreases from front to back. T_1 = 25 N must pull 5 kg of mass behind it (M_2 + M_3). T_2 = 10 N must pull only 2 kg (M_3). The applied force of 50 N is the largest force because it must accelerate the entire 10 kg system. Each string only "sees" the mass behind it.
Common confusions
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"The tension in the string equals the weight of the block." Only when the system is in equilibrium (zero acceleration). In an Atwood machine with unequal masses, T is between the two weights, not equal to either one. The only time T = mg is when a = 0.
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"If I push from the other side, the contact force should be the same." No. If you push M_2 instead of M_1 with the same force F, the contact force between the blocks becomes M_1 F/(M_1 + M_2), not M_2 F/(M_1 + M_2). The contact force depends on which mass is being pushed directly and which is being carried.
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"The system method gives me everything — why bother with individual FBDs?" The system method cannot give you internal forces. It only gives the acceleration. To find the tension in a specific string or the contact force between two blocks, you must draw the FBD of one body and use that body's equation of motion.
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"In an Atwood machine, the heavier mass falls with acceleration g." No — the string and the lighter mass restrain it. The acceleration is (m_1 - m_2)g/(m_1 + m_2), which is always less than g. Only when m_2 = 0 (no mass on the other side, just a free-falling block with a trailing string) does a approach g.
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"I need to solve simultaneous equations for every connected-body problem." Not if you use the system method first. The system method gives a in one step. After that, each internal force comes from a single equation (one FBD, one unknown). Simultaneous equations are a sign you are doing it the hard way.
If you are comfortable finding acceleration and tension for basic systems, you can stop here. What follows covers extensions you will encounter in JEE Advanced and competitive exams.
What changes if the pulley has mass?
The derivation above assumed a massless pulley. If the pulley has mass M and radius R, it has a moment of inertia I = \tfrac{1}{2}MR^2 (for a uniform disc). Now the tensions on the two sides of the string are not equal — the difference in tension provides the torque that angularly accelerates the pulley.
Call the tensions T_1 (on the m_1 side) and T_2 (on the m_2 side). The three equations become:
Why: the angular acceleration \alpha is related to the linear acceleration by \alpha = a/R (the string does not slip on the pulley). The torque on the pulley is (T_1 - T_2)R.
Adding all three equations:
The pulley's moment of inertia appears in the denominator as an extra "effective mass" \tfrac{1}{2}M. A heavier pulley slows the system down — it has to be spun up, and that takes some of the available force.
What changes if the string has mass?
If the string has mass m_s and total length L, then the tension varies along the string. At a point a distance x from the lighter mass, the tension is:
The string's own weight contributes to the tension, and the tension is no longer uniform. In most JEE problems, strings are treated as massless — but when the problem explicitly gives the string's mass, this is the approach.
The Atwood machine as a measuring tool
In a physics lab, the Atwood machine is used to measure g. If you choose masses that are nearly equal (m_1 \approx m_2), the acceleration is small:
A small acceleration means the motion is slow enough to time accurately with a stopwatch. You measure a from the distance and time, then compute g = a(m_1 + m_2)/(m_1 - m_2). This is how early physicists measured g before electronic sensors existed — and it is still a standard experiment in Indian physics labs.
Multiple pulleys and compound Atwood machines
Some JEE problems feature a "double Atwood" — a mass on one side and a second Atwood machine hanging from the other side of the string. The constraint is that the acceleration of the second pulley (as a whole) is determined by the outer system, and the masses on the inner pulley have additional relative accelerations. These problems require careful constraint equations. The key insight: if the outer string accelerates at a_0 and the inner masses have acceleration a' relative to their pulley, then the absolute accelerations are a_0 + a' and a_0 - a'.
Where this leads next
- Free body diagrams — how to draw FBDs correctly, the essential skill underneath everything in this article.
- Newton's second law — the law that drives every equation here: F_{\text{net}} = ma.
- Tension and strings — why massless strings have uniform tension, and what changes when strings have mass.
- Static and kinetic friction — the next step: connected bodies on rough surfaces, where friction complicates (but does not change) the method.
- Motion on inclined planes — connected bodies where one block is on an incline and the other hangs vertically.