5.5 Coupled Systems and Constraint Forces

Two Blocks, One Rope, One Puzzle

Two blocks are connected by a rope that passes over a pulley. The heavier block hangs on the left, the lighter block on the right. You release them from rest. The heavy block accelerates downward; the light block accelerates upward --- at exactly the same rate.

This is the Atwood machine, one of the oldest and most elegant demonstrations in mechanics. It looks simple. But when you try to analyze it, something new happens. Each block has different forces acting on it: different weights, same tension. The blocks accelerate in opposite directions, yet the magnitudes of their accelerations are identical. Why?

The answer is not in the forces alone. It is in the constraint --- the rope. The rope connects the two motions. If block 1 drops by 5 cm, block 2 rises by 5 cm. If block 1 accelerates at 2 m/s$^2$ downward, block 2 accelerates at 2 m/s$^2$ upward. The rope enforces a kinematic relationship between the two objects, and that relationship adds an equation to the problem.

This is your first encounter with multi-object dynamics. Until now, you have analyzed one object at a time: draw the FBD, write $\sum \vec{F} = m\vec{a}$, solve. That approach still works --- but now you need FBDs for each object, and the equations are linked. The new challenge is not harder math. It is choosing the right system boundaries and recognizing which forces are internal and which are external.

Prediction

Before you read on: An Atwood machine has masses $m_1 = 3$ kg and $m_2 = 1$ kg connected by a massless rope over a frictionless pulley. When released from rest, the system accelerates.

Is the acceleration of the system:

(a) Equal to $g$

(b) Less than $g$

(c) Greater than $g$

Commit to your answer before continuing. Think about what would need to be true for the acceleration to equal $g$.

The Guiding Question

How do multiple objects influence one another when their motions are linked?

You already know how to analyze a single object: isolate it, draw its FBD, apply Newton's second law. The question now is what changes when two or more objects are physically connected --- by a rope, by contact, by a rigid rod. The answer: their accelerations are no longer independent. The connection imposes a constraint, and that constraint becomes a new equation in your system.

Exploration: The Atwood Machine

[Interactive: Atwood Machine Explorer. Two masses hang from a massless rope over a frictionless pulley. Sliders allow students to adjust $m_1$ and $m_2$ independently (range: 0.5 kg to 10 kg). The display shows: (1) the acceleration of the system (magnitude and direction for each block), (2) the tension in the rope, (3) animated FBDs for each block and for the whole system. A toggle switches between "individual object view" (showing FBDs for each block separately, with tension appearing as an unknown) and "whole system view" (showing a single FBD with only external forces --- the two weights). Guided prompts below.]

Spend a few minutes with the interactive. Here are some things to try:

Prompt 1: Set $m_1 = m_2 = 2$ kg. What is the acceleration? What is the tension? Why does this make sense?

Prompt 2: Now make $m_1 = 3$ kg and $m_2 = 1$ kg. What is the acceleration? Is it greater or less than $g$?

Prompt 3: Keep $m_1 = 3$ kg and gradually increase $m_2$ from 1 kg toward 3 kg. What happens to the acceleration? What happens to the tension?

Prompt 4: Switch to the whole-system view. Notice that the tension disappears from the FBD. Why? Where did it go?

Prompt 5: Now make $m_2$ very small (say 0.1 kg). What does the acceleration approach? What does the tension approach? Does this limiting case make physical sense?

If you explored carefully, you noticed something important in Prompt 4. When you draw a FBD for the whole system (both blocks plus the rope), the tension appears twice: once pulling up on $m_1$, once pulling up on $m_2$. But these are both internal forces --- forces between objects inside the system boundary. By Newton's third law, the rope pulls on each block, and each block pulls on the rope, but all of these are interactions within the system. They cancel when you sum forces over the entire system. The only forces that survive are the ones from outside the system: the two weights.

Before you read on: Why does the tension disappear in the whole-system view but not in the individual-object view? Can you state this principle in general terms?

Concept Reveal: Constraints and System Boundaries

Here is the key idea of this section, stated in two parts.

Part 1: Constraint forces enforce kinematic relationships. When objects are connected --- by a rope, by a surface, by direct contact --- the connection imposes a relationship between their motions. A taut, inextensible rope forces two objects to have the same speed (and the same magnitude of acceleration). A block sitting on another block forces the two to have the same acceleration (as long as they move together). The force that enforces this relationship is called a constraint force. Tension in a rope is a constraint force. The normal force between two surfaces in contact is a constraint force.

Part 2: System boundaries determine which forces appear. When you draw a system boundary around a single object, every interaction with another object shows up as a force on your FBD. When you draw the boundary around multiple objects, forces between those objects become internal forces and cancel in pairs (by Newton's third law). Only forces from objects outside the boundary remain.

This is not a mathematical trick. It is a direct consequence of Newton's third law. If you include both ends of an interaction inside your system, the two forces in the interaction pair add to zero. The force "disappears" --- not because it stops existing, but because its effect is already accounted for within the system.

The strategic question in every coupled-system problem is: where should I draw the boundary?

Whole-system approach: Draw the boundary around everything. Internal forces cancel. You get one equation with one unknown (the acceleration). But you cannot find the tension --- it is hidden inside the system.
Individual-object approach: Draw the boundary around each object separately. Internal forces appear as unknowns (like tension). You get multiple equations with multiple unknowns (acceleration and tension). But you can solve for everything.

Both approaches give the same acceleration. The whole-system approach is faster when you only need the acceleration. The individual-object approach is necessary when you also need the constraint forces (tensions, normal forces between objects).

Worked Example: Atwood Machine, Two Ways

Let's solve the Atwood machine problem from the prediction: $m_1 = 3$ kg, $m_2 = 1$ kg, massless rope, frictionless pulley. Find the acceleration and the tension.

Approach 1: Whole System

Draw one system boundary around both blocks and the rope. The only external forces are:

Weight of $m_1$: $m_1 g = 3(10) = 30$ N downward (on the left side)
Weight of $m_2$: $m_2 g = 1(10) = 10$ N downward (on the right side)

The tension is internal and cancels. Now, what is the "net force" on the system? This requires a moment of care. The two blocks move in opposite directions. If $m_1$ moves down and $m_2$ moves up, then the net effect driving the motion is the difference in their weights:

$$F_{\text{net}} = m_1 g - m_2 g = (m_1 - m_2)g$$

The total mass being accelerated is $m_1 + m_2$. Applying Newton's second law to the whole system:

$$a = \frac{F_{\text{net}}}{m_{\text{total}}} = \frac{(m_1 - m_2)g}{m_1 + m_2}$$

Plug in numbers:

$$a = \frac{(3 - 1)(10)}{3 + 1} = \frac{20}{4} = 5 \text{ m/s}^2$$

The acceleration is $5$ m/s$^2$ --- exactly half of $g$. This confirms the prediction: the acceleration is less than $g$. Answer (b).

Why is it less than $g$? Because gravity is not pulling the system unopposed. Gravity pulls $m_1$ down and $m_2$ down, but the constraint (rope) forces $m_2$ to go up when $m_1$ goes down. The net driving force is only the difference in weights, but the full mass of both blocks must be accelerated. The system is like a tug-of-war where the winning team (heavier block) must drag the losing team (lighter block) along with it.

Approach 2: Individual Objects

Now draw separate FBDs for each block. Take downward as positive for $m_1$ and upward as positive for $m_2$ (so that the positive direction for each block corresponds to the direction of motion --- this is a common and useful convention for Atwood machines).

Block 1 (mass $m_1 = 3$ kg, moving down):

Forces: weight $m_1 g$ downward (positive), tension $T$ upward (negative).

$$m_1 g - T = m_1 a \tag{1}$$

Block 2 (mass $m_2 = 1$ kg, moving up):

Forces: tension $T$ upward (positive), weight $m_2 g$ downward (negative).

$$T - m_2 g = m_2 a \tag{2}$$

Note the constraint: both blocks share the same magnitude of acceleration $a$. This is the kinematic relationship enforced by the inextensible rope.

Solve for $a$: Add equations (1) and (2):

$$(m_1 g - T) + (T - m_2 g) = m_1 a + m_2 a$$

$$m_1 g - m_2 g = (m_1 + m_2)a$$

$$a = \frac{(m_1 - m_2)g}{m_1 + m_2} = \frac{(3 - 1)(10)}{3 + 1} = 5 \text{ m/s}^2$$

Same answer as before. Good.

Solve for $T$: Substitute $a = 5$ m/s$^2$ back into equation (2):

$$T - m_2 g = m_2 a$$

$$T = m_2 g + m_2 a = m_2(g + a) = 1(10 + 5) = 15 \text{ N}$$

You can verify with equation (1): $T = m_1 g - m_1 a = 3(10 - 5) = 15$ N. Consistent.

What the Tension Tells You

The tension is 15 N. Notice that this is not equal to the weight of either block. It is not $m_1 g = 30$ N. It is not $m_2 g = 10$ N. This is because the system is accelerating --- if the system were in equilibrium (equal masses), the tension would equal the common weight. With unequal masses, the tension falls between $m_2 g$ and $m_1 g$.

You can write the tension in general form:

$$T = \frac{2m_1 m_2 g}{m_1 + m_2}$$

This formula has two useful limiting cases. If $m_1 = m_2 = m$, then $T = mg$ --- the tension equals the weight, and the system is in equilibrium. If $m_2 \to 0$, then $T \to 0$ --- block 1 is essentially in free fall, and there is nothing on the other end of the rope to create tension.

Pause and check: Go back to the prediction. The acceleration is $g/2 = 5$ m/s$^2$, which is less than $g$. Can you now explain, in one or two sentences, why the acceleration must be less than $g$ for any Atwood machine with two unequal masses?

The Constraint Equation

In the Atwood machine, the constraint was straightforward: the rope does not stretch, so both blocks have the same magnitude of acceleration. But constraints can be more subtle. The general approach is:

Identify the physical connection. What object links the two (or more) bodies? A rope? A surface? Direct contact?
Write the kinematic relationship. How does the motion of one body relate to the motion of the other? If a rope is inextensible, the total length of rope is constant. If two objects are in contact, they share a velocity component along the contact direction.
Differentiate to get the acceleration constraint. If positions are linked, velocities are linked, and accelerations are linked. Each differentiation preserves the relationship.

For the Atwood machine, the total length of rope is constant:

$$\ell_1 + \ell_2 = L = \text{constant}$$

where $\ell_1$ and $\ell_2$ are the lengths of rope on each side. Differentiating twice:

$$\ddot{\ell}_1 + \ddot{\ell}_2 = 0$$

If "down" is positive for both, $\ddot{\ell}_1 = a_1$ (block 1 descending increases $\ell_1$) and $\ddot{\ell}_2 = -a_2$ if we define $a_2$ as the upward acceleration of block 2. So $a_1 = a_2$: same magnitude, opposite directions. This is the constraint equation.

Faded Example: Two Blocks on a Surface

A block of mass $m_1 = 4$ kg sits on a frictionless horizontal surface. A rope connects it to a block of mass $m_2 = 2$ kg that hangs over the edge of the table via a frictionless pulley. The rope is taut and massless. Find the acceleration and the tension.

Step 1: Identify the constraint. The rope is inextensible. Both blocks have the same magnitude of acceleration $a$: $m_1$ accelerates horizontally (toward the pulley), $m_2$ accelerates downward.

Step 2: Draw FBDs.

Block 1 (on table): Tension $T$ pulls it horizontally toward the pulley. Normal force $N_1$ and weight $m_1 g$ act vertically and cancel (no vertical acceleration).

Block 2 (hanging): Weight $m_2 g$ pulls it down. Tension $T$ pulls it up.

Step 3: Write Newton's second law for each block.

Block 1 (horizontal): $T = m_1 a$ ... (fill in)

Block 2 (vertical, taking downward as positive): __ $= m_2 a$

Complete the equation for block 2, then add the two equations to eliminate $T$ and solve for $a$. Finally, find $T$.

Check your answer

**Block 2:** $m_2 g - T = m_2 a$ **Add the two equations:** $$T + (m_2 g - T) = m_1 a + m_2 a$$ $$m_2 g = (m_1 + m_2) a$$ $$a = \frac{m_2 g}{m_1 + m_2} = \frac{2(10)}{4 + 2} = \frac{20}{6} \approx 3.33 \text{ m/s}^2$$ **Find tension:** From the block 1 equation: $$T = m_1 a = 4(3.33) = 13.3 \text{ N}$$ **Check with block 2:** $T = m_2 g - m_2 a = 2(10) - 2(3.33) = 20 - 6.67 = 13.3$ N. Consistent. **Limiting cases:** If $m_1 \to 0$ (no block on the table), $a \to g$ and $T \to 0$: the hanging block is in free fall. If $m_1 \to \infty$ (immovable block on the table), $a \to 0$ and $T \to m_2 g$: the hanging block is in equilibrium, and the tension equals its weight. Both limits make physical sense.

Connection to Earlier Work

In Section 5.1, you learned to draw FBDs for a single object: isolate it, identify every interaction, draw the forces. That tool is not being replaced --- it is being extended. In a coupled system, you draw a separate FBD for each object. The new ingredient is the constraint equation that links the accelerations.

In Section 5.4, you studied force models: weight, tension, normal force, friction. Those models still apply. What changes is that tension, which seemed like a simple "given" in single-object problems (hang a mass from a rope, $T = mg$ in equilibrium), now becomes an unknown to be solved for. Tension is a constraint force --- its value is determined by the requirement that the connected objects satisfy the constraint.

The whole-system approach also connects back to a deeper principle. When you sum Newton's second law over all objects in a system, the internal forces cancel by Newton's third law, and you get:

$$\sum \vec{F}{\text{external}} = M$$}} \vec{a}_{\text{cm}

where $\vec{a}_{\text{cm}}$ is the acceleration of the center of mass. For the Atwood machine, this reduces to the whole-system equation we wrote earlier. This is a preview of ideas you will encounter more fully in later chapters.

Spaced Retrieval

Before moving to practice, test your recall of earlier material.

Recall prompt 1: What is the purpose of a free-body diagram? What does it show and what does it not show? (Section 5.1)

Recall prompt 2: Newton's third law says forces come in pairs. If block A pushes on block B with force $F$, what is the force of B on A? Do these forces act on the same object or different objects? (Section 5.2)

Recall prompt 3: A student says the tension in a rope holding a stationary 5 kg mass is "always 50 N." Under what condition is this true, and when does it fail? (Section 5.4)

Practice Layers

Layer 1: Concrete

Problem 1. An Atwood machine has $m_1 = 5$ kg and $m_2 = 3$ kg. The rope is massless and the pulley is frictionless.

(a) Find the acceleration of the system using the whole-system approach.

(b) Find the tension in the rope using the individual-object approach.

Check your answer

**(a) Whole-system approach:** $$a = \frac{(m_1 - m_2)g}{m_1 + m_2} = \frac{(5-3)(10)}{5+3} = \frac{20}{8} = 2.5 \text{ m/s}^2$$ **(b) Individual-object approach:** Block 2 (lighter, accelerating up): $T - m_2 g = m_2 a$ $$T = m_2(g + a) = 3(10 + 2.5) = 3(12.5) = 37.5 \text{ N}$$ **(c) Check with block 1 (heavier, accelerating down):** $$m_1 g - T = m_1 a$$ $$T = m_1(g - a) = 5(10 - 2.5) = 5(7.5) = 37.5 \text{ N}$$ Both equations give $T = 37.5$ N. Consistent. Note that $T$ is between $m_2 g = 30$ N and $m_1 g = 50$ N, as expected.

Layer 2: Pattern

Problem 2. Two blocks are on a frictionless horizontal surface, connected by a massless rope. Block A ($m_A = 6$ kg) is on the left. Block B ($m_B = 4$ kg) is on the right. A horizontal force $F = 30$ N is applied to the right side of block B, pulling both blocks to the right.

(a) Find the acceleration of the system.

(b) Find the tension in the rope between the blocks.

Check your answer

**(a) Without friction, whole-system approach:** The rope tension is internal. The only external horizontal force is $F = 30$ N acting on the total mass $m_A + m_B = 10$ kg. $$a = \frac{F}{m_A + m_B} = \frac{30}{10} = 3 \text{ m/s}^2$$ **(b) Find tension using block A's FBD:** The only horizontal force on block A is the tension $T$ pulling it to the right. $$T = m_A \cdot a = 6(3) = 18 \text{ N}$$ Check with block B: Forces on B are $F = 30$ N to the right and $T = 18$ N to the left (the rope pulls B back toward A). $$F - T = m_B a \implies 30 - 18 = 12 = 4(3). \quad \checkmark$$ **(c) With friction:** Kinetic friction acts on each block: $f_A = \mu_k m_A g = 0.1(6)(10) = 6$ N and $f_B = \mu_k m_B g = 0.1(4)(10) = 4$ N, both opposing the motion (to the left). Whole-system: $F - f_A - f_B = (m_A + m_B)a$ $$30 - 6 - 4 = 10a \implies a = 2 \text{ m/s}^2$$ Block A: $T - f_A = m_A a \implies T = m_A a + f_A = 6(2) + 6 = 18$ N. Interestingly, the tension is the same as before in this case. But the acceleration is smaller. Check block B: $F - T - f_B = m_B a \implies 30 - 18 - 4 = 8 = 4(2)$. Consistent.

Layer 3: Structure

Problem 3. When is the whole-system approach faster, and when must you use individual FBDs?

Consider three scenarios:

(a) "Find the acceleration of an Atwood machine."

(b) "Find the tension in the rope of an Atwood machine."

(c) "Two blocks are stacked. Find the friction force between them."

For each, state which approach you would use and why.

Check your answer

**(a)** The whole-system approach is faster. You only need the acceleration, and the whole-system approach eliminates the tension in one step: $$a = \frac{(m_1 - m_2)g}{m_1 + m_2}$$ No need to solve simultaneous equations. **(b)** You must use individual FBDs. The tension is an *internal* force --- it disappears in the whole-system view. To find it, you need to isolate one block and write Newton's second law for that block alone. (Once you know $a$ from part (a), a single equation gives $T$.) **(c)** You must use individual FBDs. The friction between the blocks is an internal force. It vanishes in the whole-system analysis. To find it, isolate one of the two blocks and apply Newton's second law to it. **General rule:** Use the whole-system approach when you need only the overall acceleration and no internal forces. Use individual FBDs when you need any internal force (tension, contact force, friction between objects in the system). A common strategy is to use the whole-system approach *first* to find $a$ quickly, then switch to an individual FBD to find the internal force.

Layer 4: Debug

Problem 4. A student analyzes an Atwood machine with $m_1 = 4$ kg and $m_2 = 2$ kg. They write:

"The net force on the system is $m_1 g - m_2 g + T = (m_1 + m_2)a$."

They get $a = \frac{(m_1 - m_2)g + T}{m_1 + m_2}$ and then say: "I can't solve this because I have two unknowns."

What went wrong? What is the correct equation?

Check your answer

The student included the tension $T$ as a force on the whole system. But tension is an **internal force** --- it is an interaction between two objects that are both inside the system boundary. In the whole-system view, the tension on block 1 (pulling up) and the tension on block 2 (pulling up) are not external forces. They are part of the internal interaction between the blocks via the rope. More precisely: the rope pulls up on $m_1$ with force $T$, and the rope pulls up on $m_2$ with force $T$. But $m_1$ pulls down on the rope, and $m_2$ pulls down on the rope. When you sum over all objects in the system (including the rope), these internal forces cancel by Newton's third law. The correct whole-system equation is: $$m_1 g - m_2 g = (m_1 + m_2)a$$ $$a = \frac{(m_1 - m_2)g}{m_1 + m_2} = \frac{(4-2)(10)}{4+2} = \frac{20}{6} \approx 3.33 \text{ m/s}^2$$ The student's mistake is a common one: **when using the whole-system approach, do not include internal forces.** If you find yourself including a tension that acts between objects inside your system, you have crossed system boundaries --- you are mixing the whole-system and individual-object approaches. Pick one.

Reflection

Why does the system boundary you choose determine which forces appear in your equations?

Think about what a "system boundary" really means. It is a conceptual line you draw around one or more objects. Every force is an interaction between two objects. If both objects in the interaction are inside your boundary, the force is internal and cancels. If one object is inside and the other is outside, the force is external and appears in your equation.

This is not just bookkeeping. It reflects something physical: internal forces cannot change the total momentum of a system (a concept you will formalize in later chapters). Only external forces can change how the system as a whole moves. The system boundary is a choice you make --- but the physics does not depend on that choice. Both approaches (whole-system and individual-object) give the same physical predictions. The choice affects only the difficulty of the algebra and which quantities you can solve for directly.

Choosing system boundaries strategically is one of the most important skills in physics. It is the difference between a three-line solution and a page of algebra.

Looking Ahead

You have now extended Newton's second law from single objects to coupled systems. The key ideas --- constraint forces, system boundaries, internal vs. external forces --- will reappear throughout mechanics. In later chapters, when you study momentum and energy, the concept of "system" becomes even more central: what you include inside your system boundary determines which forces do work, which impulses matter, and which conservation laws apply.

But there is one special case of coupled systems that deserves its own attention: the case where the acceleration is zero. In Section 5.6, you will study equilibrium --- the situation where all forces balance and nothing accelerates. This is not a separate topic from dynamics. It is $\sum \vec{F} = m\vec{a}$ with $\vec{a} = \vec{0}$. The same tools apply, but the equations simplify, and a new class of problems opens up: hanging signs, bridges, cables at angles. The constraint ideas from this section will help you analyze systems in equilibrium where multiple objects interact.