Constraint Motion: Pulleys, Strings, and Inclined Planes

Hello! Welcome to the fifth lesson in our study of calculus-based mechanics.

In our last lesson, we explored the impulse-momentum theorem by integrating Newton's second law with respect to time. We saw how this framework is essential for analyzing collisions and variable-mass systems like rockets. Today, we return to the direct application of $F = m a$ , but we'll tackle systems where the motion of one object is linked to, or constrained by, the motion of others.

This lesson addresses the learning outcome: Solve constraint motion problems involving pulleys, strings, and inclined planes using Newtonian dynamics. The core challenge in these problems is that simply writing $F = m a$ for each object often leaves you with more unknowns than equations. The key is to derive additional equations from the physical constraints of the system.

1. The Challenge of Interconnected Systems

When objects are connected by ropes, pulleys, or are in contact with moving surfaces, their motions are not independent. A rope has a fixed length, and a block must remain on the surface of an incline. These are kinematic constraints. Let's see how this plays out in a concrete example.

The following video sets up a moderately complex pulley system. As you watch, pay close attention to the process of drawing free-body diagrams and applying Newton's second law to each component.

12.1 Pulley Problems

This video from MIT OpenCourseWare demonstrates the initial analysis of a pulley system. It perfectly illustrates the problem we need to solve: we end up with fewer equations than unknowns.

Please watch from the beginning (0:03) to 10:26. The instructor identifies the systems, makes simplifying assumptions (massless ropes and pulleys), draws free-body diagrams, and applies Newton's law. Note the conclusion he reaches about the number of equations versus the number of unknowns.

As the video demonstrates, after applying $F = m a$ to masses 1 and 2, we have two equations but three unknowns: the tension $T$ , the acceleration of mass 1 ( $a_{1}$ ), and the acceleration of mass 2 ( $a_{2}$ ). We are missing a crucial piece of the puzzle: a mathematical relationship between $a_{1}$ and $a_{2}$ that arises from the fact that they are connected by a rope of a fixed length. This missing piece is the constraint equation.

2. The Method: Deriving Constraint Equations

The most robust method for finding the constraint equation is to describe the system's geometry with coordinates and then see how those coordinates must relate to each other as the system moves.

The following Olympiad-level handout formalizes this exact strategy. It is a powerful technique that you will use frequently.

Mechanics III: Dynamics

This handout by Kevin Zhou, designed for Physics Olympiad training, introduces the central idea for solving constraint problems.

Read 'Idea 1' at the beginning of the handout (Section 1: Blocks, Pulleys, and Ramps). This is the core method. Then, carefully study the solution to 'Problem 1 (Morin 3.2)', the double Atwood's machine. This is a classic, non-trivial application of the method.

Let's break down the method described in the handout:

Assign Coordinates: Define a coordinate system and assign position variables to each object whose motion you need to track. For vertical motion, a good choice is to measure the downward displacement from a fixed reference point (like the ceiling).
Write the Constraint Equation: Write an equation that captures the physical constraint of the system in terms of these coordinates. For a pulley system, this is usually an equation stating that the length of the rope is constant.
Differentiate: Since velocity is the first time-derivative of position ( $v = d x / d t$ ) and acceleration is the second ( $a = d^{2} x / d t^{2}$ ), we can find the relationship between the accelerations by differentiating the position constraint equation twice with respect to time.

Example: The Double Atwood's Machine (from LINK)

Let's trace the logic for the double Atwood's machine in "Problem 1" of the handout you just read.

Let $x_{1}, x_{2}, x_{3}$ be the downward positions of the three masses from the ceiling.
Let $x_{p}$ be the downward position of the moving pulley.

The system has two ropes, so we have two length constraints:

Upper Rope: Connects $m_{1}$ to the moving pulley. Its length is constant: $x_{1} + x_{p} = L_{1}$ .
Lower Rope: Goes around the moving pulley to connect $m_{2}$ and $m_{3}$ . Its length is constant: $(x_{2} - x_{p}) + (x_{3} - x_{p}) = L_{2}$ .

Now, differentiate each equation twice with respect to time (noting that the derivatives of constants $L_{1}, L_{2}$ are zero):

$x ¨_{1} + x ¨_{p} = 0 ⟹ a_{1} + a_{p} = 0 ⟹ a_{p} = - a_{1}$ .
$(x ¨_{2} - x ¨_{p}) + (x ¨_{3} - x ¨_{p}) = 0 ⟹ a_{2} + a_{3} - 2 a_{p} = 0$ .

Substituting $a_{p} = - a_{1}$ into the second result gives:

a_{2} + a_{3} - 2 (- a_{1}) = 0 ⟹ 2 a_{1} + a_{2} + a_{3} = 0

This is the required constraint equation. Combined with the three force equations from the free-body diagrams, you now have a complete system of four equations for the four unknowns ( $a_{1}, a_{2}, a_{3}$ , and the tension $T$ ). As you saw in the handout, this system is now solvable.

3. Constraints on Inclined Planes

The same logic applies when dealing with inclined planes. The constraints might involve a string going over a pulley, or they might involve a block staying in contact with a moving surface.

System Analysis: Two Inclined Planes with Pulley — A classic problem involving two blocks connected by a string over a pulley, with each block on a separate inclined plane. The constraint is that the length of the string connecting them is fixed.

In the system shown in the image "Two Blocks on Inclined Planes", the constraint is simple. If we define coordinates $s_{1}$ and $s_{2}$ along each incline, the total length of the string is $s_{1} + s_{2} = L$ . Differentiating twice gives $a_{1} + a_{2} = 0$ , or $a_{1} = - a_{2}$ . The rest of the problem involves resolving forces along the tilted axes, a familiar process.

Let's look at a solved problem that combines an incline, a pulley, and friction.

Newtonian Mechanics – Single Particle

This document provides a worked example combining the concepts of inclined planes, pulleys, and friction.

Read through problem 2-32 on pages 10-12. This problem seeks the angle of inclination for which the system moves at a constant speed (i.e., zero acceleration). Follow how the force equations are set up and combined with the condition a=0.

This example reinforces how to combine the force analysis on an incline (decomposing gravity, finding the normal force, calculating friction) with the tension forces from a pulley system.

Test your understanding!

Consider the system in the image "Two Blocks on Inclined Planes" (LINK). Let $m_{1} = 5$ kg, $m_{2} = 10$ kg, $θ_{1} = 3 0^{\circ}$ , and $θ_{2} = 6 0^{\circ}$ . Assume the surfaces are frictionless.

Write down the force equations for each block along its incline.
Using the constraint $a_{1} = - a_{2}$ , solve for the acceleration of the system. Let $a = a_{2}$ .

Show answer

Let's define the positive direction for each block as "down its incline". With this choice, the constraint is $a_{1} = a_{2} = a$ , because if one block moves down its incline, the other must as well. Let the tension in the string be $T$ .

The net force on $m_{1}$ along its incline is:

F_{n e t, 1} = m_{1} g sin θ_{1} - T = m_{1} a (Eq. 1)

The net force on $m_{2}$ along its incline is:

F_{n e t, 2} = m_{2} g sin θ_{2} - T = m_{2} a (Eq. 2)

This is a problem! We defined positive as down the incline for both, but the string connects them. If one goes down, the other must go up. Let's redefine.

Corrected Approach: Let's define the positive direction as $m_{2}$ sliding down its incline ( $θ_{2}$ ) and $m_{1}$ sliding up its incline ( $θ_{1}$ ). The magnitude of their accelerations must be the same, let's call it $a$ .

Force Equation for $m_{2}$ : (positive is down the incline)
$m_{2} g sin θ_{2} - T = m_{2} a (Eq. A)$
Force Equation for $m_{1}$ : (positive is up the incline)
$T - m_{1} g sin θ_{1} = m_{1} a (Eq. B)$

We have two equations and two unknowns ( $T$ and $a$ ). Let's solve for $a$ by adding the two equations to eliminate $T$ :

(m_{2} g sin θ_{2} - T) + (T - m_{1} g sin θ_{1}) = m_{2} a + m_{1} a

g (m_{2} sin θ_{2} - m_{1} sin θ_{1}) = (m_{1} + m_{2}) a

a = g \frac{m _{2} sin θ _{2} - m _{1} sin θ _{1}}{m _{1} + m _{2}}

Now plug in the numbers (using $g \approx 9.8 m / s^{2}$ ):

a = 9.8 \frac{10 sin ( 6 0 ^{\circ} ) - 5 sin ( 3 0 ^{\circ} )}{5 + 10} = 9.8 \frac{10 ( \frac{3}{2} ) - 5 ( \frac{1}{2} )}{15} = 9.8 \frac{5 3 - 2.5}{15}

a \approx 9.8 \frac{8.66 - 2.5}{15} = 9.8 \frac{6.16}{15} \approx 4.03 m / s^{2}

Since $a$ is positive, our initial assumption was correct: $m_{2}$ accelerates down its incline and $m_{1}$ accelerates up its incline.

4. Advanced Constraints & Engineering Applications

Constraint motion problems are fundamental to mechanical engineering. Any device with interconnected moving parts—from an engine to a crane to robotic arms—relies on these principles.

Constraint Relations Problem with Pulleys and Inclined Plane — Constraint problems can become quite complex, involving multiple pulleys and inclined planes. The methods we've learned provide a systematic way to solve them.

The Movable Wedge

A classic advanced problem is a block sliding down a wedge that is itself free to slide on a frictionless surface. Here, the constraint is that the block must remain in contact with the wedge's surface. The LINK handout you reviewed earlier contains this exact problem.

Mechanics III: Dynamics

Let's revisit the Olympiad handout to analyze a problem with a different kind of constraint: a block on a movable wedge.

Read the setup and solution for part (a) of 'Problem 4' on page 4. Notice how the analysis is simplified by working in the non-inertial frame of the wedge. This is a powerful technique that we will explore fully in the next lesson.

Engineering Application: Mechanical Advantage

The primary engineering application of pulley systems is to achieve mechanical advantage, which is the amplification of force. The constraint relations directly tell you the mechanical advantage.

In the "Spanish burton" pulley system from LINK (Example 1), we found the constraint $a = - a^{'} /8$ . This means the mass on the right moves 8 times farther than the mass on the left. In an idealized, slow-moving system, the work done on both sides must be equal ( $F \cdot d = const$ ). Since the distance moved is reduced by a factor of 8, the force required to lift the mass is also reduced by a factor of 8. This is an 8:1 mechanical advantage. Such systems are crucial in cranes, elevators, and complex rigging on sailing ships.

Conclusion

Today we've developed a systematic and powerful method for analyzing systems with interconnected parts. You've learned that the key is to supplement the force equations with constraint equations derived from the system's geometry.

Key Takeaways:

Constraint motion problems involve objects whose motions are interdependent.
Solving these requires deriving constraint equations that relate the accelerations of the different objects.
The general method is to:
1. Define coordinates for the objects.
2. Write an equation for the geometric constraint (e.g., constant rope length).
3. Differentiate this equation twice with respect to time to relate the accelerations.
This procedure applies to a wide variety of systems, including complex pulley arrangements and blocks on moving surfaces.

Preview of the Next Lesson:
In the movable wedge problem, we saw a hint of a powerful strategy: analyzing the motion from an accelerating (non-inertial) reference frame. In our next lesson, we will formalize this concept. We will learn how to derive equations of motion in non-inertial reference frames, which involves the introduction of fascinating concepts known as "pseudo-forces" or "fictitious forces" like the centrifugal and Coriolis forces.

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