Connected bodies and free body diagrams

A block hangs off the edge of a table, tied by a string over a pulley to a second block resting on the table top. Push the whole picture, and both blocks move together — but Newton's second law only ever talks about one object at a time. Apply it to the pair as a lump and you get nonsense; apply it to each block separately, with its own honest list of forces, and the algebra falls into place on its own. That separate list — one object, isolated, every external force drawn as an arrow — is the free body diagram (FBD). It is not a decoration before the "real" physics; it is the real physics, because a mistaken force on a diagram becomes a wrong equation. This article builds the FBD method for the two connected-body setups your ISC syllabus asks for — a string over a single pulley, and two blocks pushed together in contact — and shows how the shared-acceleration constraint turns two separate equations into one solvable system.

The core insight

Isolate each connected body in its own free body diagram, apply $\sum F = ma$ to each one separately, and link the results using the constraint that connected or contacting bodies share the same magnitude of acceleration.

The core idea

What a free body diagram actually is. An FBD shows exactly one chosen object, drawn alone, with every external force acting on it — weight, normal reaction, string tension, applied push or pull, friction if present — represented as a labelled arrow. Nothing from any other body appears except through these forces. Draw a second body's forces on the first body's diagram and the equation built from it is wrong before you write a single symbol.

Why "connected" bodies need more than one diagram. When objects are joined by a string or pushed together in contact, each object still obeys Newton's second law on its own: $\sum F = ma$ applied to that object only, using only the forces that touch that object. Two objects give two separate equations. What ties the two equations together is a geometric fact, not a force: an inextensible string keeps both ends moving at the same speed, and two blocks that stay in contact cannot separate or overlap — so both share one common acceleration magnitude $a$ . A single, light, frictionless pulley changes only the direction a string pulls, never the tension along it, so $T$ is one number, not two.

Case 1 — a hanging mass and a table mass over a single pulley. Mass $m_1$ hangs vertically; mass $m_2$ sits on a smooth horizontal table; an inextensible string joins them over a frictionless pulley at the table's edge. Isolating each block and combining their equations (worked in full below) gives

a = \frac{m_1 g}{m_1+m_2}

T = \frac{m_1 m_2 g}{m_1+m_2}

where: $a$ is the common acceleration of both blocks, $T$ is the string tension (same on both sides of the ideal pulley), $m_1$ is the hanging mass, $m_2$ is the mass on the table, and $g$ is the acceleration due to gravity.

Case 2 — two blocks in contact, pushed by one force. If a force $F$ pushes block $m_1$ , which is in contact with block $m_2$ ahead of it, both blocks share one acceleration $a = F/(m_1+m_2)$ , and the contact (normal) force between them is found by applying $\sum F = ma$ to just one of the two blocks alone.

Figure

Two separate free body diagrams side by side for a hanging block and a table block joined by a string over a pulley

Derivation

Derivation AAcceleration and tension for the hanging-mass over-a-pulley system

The goal is to derive $a = m_1 g/(m_1+m_2)$ and $T = m_1 m_2 g/(m_1+m_2)$ for a hanging mass $m_1$ connected by a string over a single frictionless pulley to a mass $m_2$ resting on a smooth table, using nothing beyond Newton's second law applied once to each isolated block and the shared-acceleration constraint of an inextensible string.

Step 1. Isolate $m_1$ and list only the forces that touch it: its weight $m_1 g$ acting downward, and the string tension $T$ acting upward (the string pulls it up toward the pulley). Taking downward as positive for this block, since it is falling, Newton's second law gives $m_1 g - T = m_1 a$ .

Step 2. Isolate $m_2$ and list only the forces that touch it: weight $m_2 g$ down, the table's normal reaction $N$ up (these cancel vertically since $m_2$ has no vertical acceleration on the table), and the string tension $T$ pulling it horizontally toward the pulley. Taking the direction of pull as positive, Newton's second law along the table gives $T = m_2 a$ .

Step 3. Bring in the constraint the string supplies. Because the string is inextensible and passes over one ideal pulley, the length that shortens on $m_1$ 's side equals the length that lengthens on $m_2$ 's side every instant, so both blocks move with the same magnitude of acceleration $a$ , and the tension is the same single value $T$ throughout its length.

Step 4. Solve the two equations together. Adding the Step 1 and Step 2 equations eliminates $T$ : $(m_1 g - T) + T = m_1 a + m_2 a$ , which simplifies to $m_1 g = (m_1+m_2)a$ , so

a = \frac{m_1 g}{m_1+m_2}.

Substituting this back into $T = m_2 a$ gives

T = \frac{m_1 m_2 g}{m_1+m_2}.

Both results came from two separate, honest FBDs and one geometric fact — nothing more.

Worked example

A block of mass $m_1 = 3\ \text{kg}$ hangs from a light, inextensible string over a smooth, frictionless pulley fixed at the edge of a smooth horizontal table, connected to a block of mass $m_2 = 5\ \text{kg}$ resting on the table.

Step 1. FBD of $m_1$ (downward positive): $m_1 g - T = m_1 a$ , i.e. $3(9.8) - T = 3a$ .

Step 2. FBD of $m_2$ (pull direction positive, table smooth so no friction term): $T = m_2 a = 5a$ .

Step 3. Add the two equations to eliminate $T$ : $3(9.8) = (3+5)a$ , so $a = 29.4/8 = 3.675\ \text{m/s}^2$ .

Step 4. Substitute back: $T = 5 \times 3.675 = 18.375\ \text{N}$ .

The system accelerates at $3.675\ \text{m/s}^2$ , with a string tension of $18.375\ \text{N}$ on both sides of the pulley.

Competitive stretch

Two blocks of mass $m_1 = 4\ \text{kg}$ and $m_2 = 2\ \text{kg}$ sit in contact on a smooth floor, $m_1$ behind $m_2$ , pushed by a horizontal force $F = 18\ \text{N}$ on $m_1$ .

Step 1. Treat both blocks as one system to get the common acceleration: $a = F/(m_1+m_2) = 18/6 = 3\ \text{m/s}^2$ .

Step 2. Isolate $m_2$ alone; the only horizontal force on it is the contact push $N$ from $m_1$ : $N = m_2 a = 2 \times 3 = 6\ \text{N}$ .

Step 3. Check with $m_1$ 's own FBD: forces are $F$ forward and the reaction $N$ pushing back (Newton's third law), so $F - N = m_1 a \Rightarrow 18 - 6 = 12 = 4 \times 3$ , consistent.

The contact force between the blocks is $6\ \text{N}$ , confirming $a = 3\ \text{m/s}^2$ .

Common traps

Drawing one combined diagram with forces from both bodies mixed together, instead of a clean, separate FBD for each — the single biggest source of sign and force-counting errors.
Assuming the string tension differs on either side of an ideal (light, frictionless) pulley — at this level tension is the same single value throughout.
Forgetting that connected or contacting bodies share the same magnitude of acceleration, and trying to solve one block's equation alone.
Putting a force that belongs to one body — like the applied force $F$ or a tension $T$ — directly onto a different body's FBD.

Ask the tutor: Want to work through a trickier connected-body setup — like an incline, or three blocks in a row — body by body?

Why cannot I just write one equation for the whole two-block system instead of two separate FBDs

For the two-blocks-in-contact case you actually can find the common acceleration from one system-level equation, $a = F/(m_1+m_2)$ , because both blocks move together and the internal contact force cancels within the system. But that single equation cannot tell you the contact force between the blocks — for that you must isolate one block and apply Newton's second law to it alone. For the pulley case, tension is an internal force too, but it acts along different directions on the two blocks (down-then-up on one side, horizontal on the other), so a single combined equation is not meaningful; you need the two separate FBDs to set up and solve for both $a$ and $T$ .

Does the tension change if the pulley has some mass or friction

Not in this syllabus. Here every pulley is explicitly light (massless) and frictionless, so it only redirects the string without absorbing any force, and the tension is one single value throughout the string's length. A pulley with mass or friction would need extra physics — rotational dynamics — that lies outside the ISC Class 11 syllabus for this topic, and two-pulley systems are explicitly excluded too.

How do I decide which direction is positive for each block

Pick whichever direction each block actually accelerates in and call that positive for that block's own equation — for the hanging mass, downward (since it falls); for the table mass, the direction it gets pulled (since the string drags it that way). The two choices do not need to match a single global axis; what matters is that each equation is internally consistent, and that the shared-acceleration constraint links the same magnitude $a$ across both equations once you know which way each block moves.