Connected Bodies and the Atwood Machine

Two bodies on one string share one acceleration and one tension, and that is the whole method: write Newton’s second law once for each body, add the equations, and the tension cancels. This does both standard arrangements — both masses hanging, or one on a table with friction — and prints the tension, which is never the weight of either body.

Last updated 11 Nov 2026

Two masses, one string, one acceleration

Acceleration and tension for a connected pair

ArrangementThe two arrangements every syllabus uses. In the first, both weights are in play and the system accelerates on the DIFFERENCE between them, which is what makes an Atwood machine a way of measuring g slowly enough to time by hand. In the second, only the hanging mass drives the motion and the mass on the table contributes inertia and friction but no driving weight. The method is identical: one equation per body, then add.

Mass 1 — the rising mass, or the one on the tableIn kilograms. In the Atwood arrangement it does not matter which mass you call which: the page reports the acceleration as a magnitude and names the descending side for you. In the table arrangement it matters a great deal — mass 1 is the one ON the table, whose weight is carried by the table and never appears in the driving force.

Mass 2 — the descending or hanging massIn kilograms. In the table arrangement this is the hanging mass, and its full weight m₂g is the only thing driving the system.

Coefficient of friction under mass 1Dimensionless, and used only in the table arrangement — it is locked out when both masses hang, because nothing is rubbing on anything. Zero means a smooth table, which is what most first questions assume. The value has to come from the question: μ belongs to a pair of surfaces in a particular state and published tables for the same pair disagree too widely for this page to print one.

Distance the masses move, for the speed and time rowsIn metres, measured along the string — both bodies cover the same distance, because the string does not stretch. Used only for the last two rows, which apply the constant-acceleration equations from rest: v = √(2as) and t = √(2s/a). Leave it at any value you like if you only want the acceleration and the tension.

Acceleration due to gravity, gTake whichever value the question printed, because the three real ones are not interchangeable in the third figure. Standard gravity is exactly 9.806 65 m/s² — NIST’s 2022 CODATA listing marks it “(exact)”, because it is a conventional value fixed by the 3rd CGPM in 1901 rather than a measurement. The Cambridge 9702 data sheet gives 9.81; NCERT and most CBSE papers use 9.8. Between 9.8 and 9.81 there is 0.10 per cent, which is invisible at two significant figures and visible at four. The 10 is there for mental arithmetic and costs you 2 per cent, so do not hand in an answer computed with it. None of these is the local g, which varies with latitude and altitude by a few tenths of a per cent and which no school question means.

2.4500m/s²Example

An Atwood machine with 3 kg and 5 kg, moving 1.2 m from rest, g = 9.8 m/s²

One equation per body, added — and the tension that cancels when you add them

Atwood: a = (m₂ − m₁)g / (m₁ + m₂) · T = 2m₁m₂g / (m₁ + m₂) · Table: a = (m₂ − μm₁)g / (m₁ + m₂)

the method: write ΣF = ma for each body separately, taking the direction of motion as positive for each of them, then ADD the two equations. The tension appears with opposite signs and cancels, leaving one equation in a. Substitute a back into either body’s equation to recover T. This is the whole technique and it scales to three or four connected bodies unchanged
a: one acceleration for both bodies, because the string does not stretch. In the Atwood arrangement it is the weight DIFFERENCE divided by the TOTAL mass, which is why an Atwood machine with nearly equal masses falls so slowly that g can be measured with a stopwatch
T: one tension, because the pulley is taken as light and frictionless. It lies strictly between the two weights in the Atwood case and equals m₁(a + μg) in the table case. It is never equal to either weight unless the acceleration is zero
μm₁g: the friction under the body on the table, which uses the normal force m₁g — the table body’s own weight, since the surface is horizontal and the string pulls along it. Subtract it from the driving weight, not from the acceleration
2T: what the pulley and its support have to hold in the Atwood arrangement, and it is LESS than the combined weight whenever the system is moving. A running Atwood machine weighs less than a stationary one, which sounds wrong and is a direct consequence of one mass accelerating downwards

Worked example

An Atwood machine with 3 kg and 5 kg, moving 1.2 m from rest, g = 9.8 m/s²
ONE EQUATION FOR EACH BODY, each with its own direction of motion as positive. For the rising 3 kg: T − 3g = 3a. For the descending 5 kg: 5g − T = 5a. Both use the SAME a and the SAME T, and that is the only physical input — it follows from the string being inextensible and the pulley light.
ADD THEM, AND T CANCELS. 5g − 3g = 3a + 5a, so 2g = 8a and a = 2 × 9.8 / 8 = 2.4500 m/s². That is 25 per cent of g, which is the useful way to read it: the system accelerates on the weight difference while carrying the total mass.
RECOVER THE TENSION from either equation. T = 3(g + a) = 3(9.8 + 2.45) = 36.7500 N. Check it against the other: 5g − 5a = 49 − 12.25 = 36.75 N. Agreed. Now sanity-check the SIZE: the two weights are 29.4 N and 49 N, and 36.75 sits between them, as it must.
WHAT THE PULLEY FEELS. The string pulls down on both sides, so the support carries 2T = 73.5000 N. The two bodies weigh 78.4 N between them. The running machine is 4.9 N lighter than its own weight, and the missing 4.9 N is exactly (m₁ + m₂)a. Hang the whole thing on a balance and you can watch the reading drop when you let go.
AND THE MOTION. From rest over 1.2 m: v = √(2 × 2.45 × 1.2) = 2.4249 m/s and t = √(2 × 1.2 / 2.45) = 0.9897 s. Nearly a second to fall 1.2 m, against 0.49 s in free fall — which is the point of the apparatus. Slowing gravity down by a known factor is what lets a stopwatch measure it.

Add the two equations, cancel the tension, and check that T sits between the weights

Connected bodies look like two problems and are one. The string gives you two facts for free: both bodies have the same acceleration in magnitude, because an inextensible string means one moves exactly as far as the other; and the tension is the same throughout, because the string is light and the pulley is frictionless. With those, the method is mechanical. Write Newton’s second law for each body, taking that body’s own direction of motion as positive. Add the two equations. The tension appears once with each sign and cancels, which leaves one equation in the acceleration. Put the acceleration back into either body’s equation to get the tension. There is no third step and the method does not change when you add a third body.

The tension is the part people get wrong, and it has a built-in check. In the Atwood arrangement the tension lies strictly between the two weights. It must: the rising body is accelerating upwards, so the string has to out-pull its weight; the descending body is accelerating downwards, so its weight has to out-pull the string. With 3 kg and 5 kg the weights are 29.4 N and 49 N and the tension is 36.75 N. If your answer puts the tension at 29.4, at 49, or outside that interval, it is wrong and you can know that before checking the algebra. The one exception is equal masses, where the acceleration is zero and the tension equals the single weight.

The table arrangement is not symmetric, and that is the thing to notice. With one mass on a horizontal table and one hanging, only the hanging weight drives the system. The table body’s weight is carried by the table and never enters the driving force — but it does enter twice in other ways: as inertia, since it is part of the total mass being accelerated, and as the normal force that sets the friction, μm₁g. So making the table body heavier slows the system down on both counts, while making the hanging body heavier speeds it up. In the Atwood arrangement, by contrast, both weights count and the acceleration depends only on their difference over their sum.

A running Atwood machine weighs less than a stationary one. The force the pulley and its support must carry is 2T, and with 3 kg and 5 kg that is 73.5 N against a combined weight of 78.4 N. The missing 4.9 N is (m₁ + m₂)a, and it is missing because part of the system is accelerating downwards. This is the same physics as a lift: stand on bathroom scales in a lift that starts descending and the reading drops. It is worth working out once, because it is the clearest demonstration that the weight of a system and the force it exerts on its support are different quantities.

What the model leaves out. A real pulley has mass, so it needs a torque to spin up; that makes the two tensions unequal and reduces the acceleration, and turning it into a rotational-dynamics problem is what τ = Iα is for. A real string has mass, so the tension varies along it. A real bearing has friction. None of these appear in a school question unless it says so, and the phrase that excludes them is “light, inextensible string over a smooth pulley”. The CBSE Class XI treatment sits inside Laws of Motion, where the whole topic is Newton’s laws applied to systems of this kind; the rotating-pulley version belongs to the rigid-body unit.

Frequently asked questions

Why do both masses have the same acceleration?

Because the string does not stretch. If one body moves 10 cm, the string on its side lengthens or shortens by 10 cm, so the other side must change by the same 10 cm and the other body moves exactly as far. Same displacement in the same time means the same speed and the same acceleration — equal in magnitude, though opposite in direction when one rises as the other falls. Inextensibility is a modelling assumption, and it is the one that makes two unknowns into one.

Is the tension equal to the weight of the hanging mass?

Only if nothing is accelerating. In the Atwood arrangement the tension sits strictly between the two weights; in the table arrangement it is less than the hanging weight while the system runs, and exactly equal to it when friction holds everything still. The physical reason is the same in both cases: a body accelerating downwards must have a net downward force, so the string cannot be pulling up on it as hard as gravity is pulling down.

What happens if the two masses are equal?

The acceleration is zero and the system stays however you leave it — it is in equilibrium at every position, not just one. The tension is then exactly mg, the only case in which it equals a body’s weight. This is what makes a balanced Atwood machine a measuring instrument: add a small rider to one side and the acceleration that results, which is slow enough to time by hand, gives you g or the rider’s mass. The speed and time rows on this page go blank, because with zero acceleration a body released from rest never covers the distance.

Which mass should I call mass 1?

In the Atwood arrangement it does not matter: the page reports the acceleration as a magnitude and the tension formula is symmetric, so swapping them changes nothing. In the table arrangement it matters entirely — mass 1 must be the one ON the table, because its weight is carried by the table and its normal force sets the friction. Swap them there and you will get the wrong answer with no warning.

Why is the force on the pulley less than the total weight?

Because one of the bodies is accelerating downwards, and a body that is accelerating downwards is not pulling on its support with its full weight. The support carries 2T, and 2T = (m₁ + m₂)g − (m₁ + m₂)a. With 3 kg and 5 kg that is 73.5 N instead of 78.4. It is the same effect as the dip in a bathroom-scale reading when a lift starts down, and it disappears the moment the system is held still.

Related calculators

References

PHYSICS (042) Curriculum, Class XI–XII, Central Board of Secondary Education, session 2025–26 (read 11 November 2026). Unit III, Laws of Motion, reads in part: “Intuitive concept of force, Inertia, Newton’s first law of motion; momentum and Newton’s second law of motion; impulse”, “Equilibrium of concurrent forces, Static and kinetic friction, laws of friction, rolling friction, lubrication” and “Dynamics of uniform circular motion: Centripetal force, examples of circular motion (vehicle on a level circular road, vehicle on a banked road)”. Unit V, Motion of System of Particles and Rigid Body, reads in part: “Moment of a force, torque, angular momentum, law of conservation of angular momentum and its applications” and “Moment of inertia, radius of gyration, values of moments of inertia for simple geometrical objects (no derivation)”. Quoted to establish what is in scope, not to claim any endorsement by the board.
Fundamental Physical Constants — Complete Listing, NIST Reference on Constants, Units and Uncertainty, 2022 CODATA adjustment (read 11 November 2026). The line this batch depends on is “standard acceleration of gravity 9.806 65 (exact) m s⁻²”. The word exact is the point: g_n is a conventional value, not a measured one, so it carries no uncertainty and it is not the g where you are standing. The same listing gives the Newtonian constant of gravitation as 6.674 30 × 10⁻¹¹ m³ kg⁻¹ s⁻² with an uncertainty of 0.000 15 × 10⁻¹¹, which is 22 parts per million — the worst-known of the constants a school paper quotes, and a useful reminder that “constant” and “exact” are different claims.
The International System of Units (SI), 9th edition, 2019, Bureau International des Poids et Mesures (read 11 November 2026). Table 4 gives the newton as kg m s⁻², which is the only definition of force this site uses: a newton is not a unit of mass and a kilogram is not a unit of force. Appendix 1 lists the 3rd CGPM of 1901, whose declaration both separated mass from weight and fixed the conventional value of g_n — the two things a force question most often conflates. Cited, not reproduced.
Cambridge International AS & A Level Physics 9702, syllabus for examination in 2025, 2026 and 2027, Cambridge Assessment International Education (read 11 November 2026). Two exclusions on this syllabus are worth more than most inclusions, because they explain why a question from one board looks unanswerable to a student taught on another. Topic 3.2 asks only for “a qualitative understanding of frictional forces and viscous/drag forces including air resistance” and states that “no treatment of the coefficients of friction and viscosity is required”. Topic 3.3 states that “knowledge of the concept of coefficient of restitution is not required”, while still requiring that for an elastic collision “the relative speed of approach is equal to the relative speed of separation”. Its data sheet gives the acceleration of free fall as 9.81 m s⁻².