3.5 Analyze constrained and connected particle systems | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Particle Kinetics: Force and Acceleration

Use rope-length relationships and Newton’s second law to connect particle motions

A rope or cable can connect particles so their motions are not independent. Analyze such systems by combining two ideas: the rope’s fixed length creates a motion constraint, and Newton’s second law gives a force equation for each particle. First describe the connection and choose coordinates; then write the constraint and the force equations. Unless stated otherwise, assume ideal ropes that do not stretch and ideal pulleys that are massless and rotate without friction. For one continuous rope under these assumptions, the tension magnitude is the same throughout.

What you will learn

Define a system, observer, reference frame, coordinates, and positive directions for connected particles.
Translate an inextensible-rope constraint into relationships between particle positions, velocities, and accelerations.
Apply Newton’s second law to each particle and solve the resulting equations together.
Check calculated directions, units, and limiting behaviour.

1. Describe the system and write the constraint

A system is the particle or group of particles being studied. Take the observer to be fixed to the ground, so the reference frame uses ground-based axes. For each particle, choose a coordinate along its allowed path and state its positive direction. Position gives location; displacement is the change in position; path length is the distance travelled; velocity is the rate of change of position; and acceleration is the rate of change of velocity.

For particles joined by a taut, inextensible rope, the total length of the moving rope segments remains constant. Write this length using coordinates measured in your chosen directions. Fixed portions can be included as constants. Differentiate the length relationship once to obtain the velocity constraint and again to obtain the acceleration constraint. Signs depend on the chosen coordinates.

For a single rope passing over a fixed pulley and joining two particles, one particle’s motion changes one rope segment while the other particle’s motion changes the other segment. With coordinates chosen in opposite physical directions, their signed displacements—and therefore their signed accelerations—are equal. A movable pulley may be supported by multiple changing rope segments, so its coordinate appears multiple times in the rope-length relationship.

L=constant, \dot L=0, \ddot L=0

Write the rope-length relationship from the actual geometry before writing force equations.
A fixed-length rope constrains motion; it does not by itself determine acceleration.
Differentiate the same constraint to relate velocity and acceleration.

2. Apply Newton’s second law to each particle

Draw a separate free-body diagram for each particle. Include its weight, tension, normal force, and any other real force specified by the setup. A free-body diagram contains forces, not acceleration. Choose the positive direction along each coordinate and write the force equation using that direction.

For an ideal continuous rope over a massless, frictionless pulley, the tension magnitude is the same in each segment. Connected particles share that tension, but their acceleration magnitudes need not be equal when a movable pulley is involved. The rope constraint supplies the relationship between their accelerations.

On a smooth surface, friction is absent. If a surface is inclined, resolve weight into components parallel and perpendicular to the surface when helpful. The normal force acts perpendicular to the surface; the parallel component of weight can drive motion along it.

\sum F_x=ma_x

Use one force equation for each particle whose motion is being solved.
Use the rope constraint alongside—not instead of—Newton’s second law.
Keep signs consistent with the selected positive directions.

3. Solve and check the connected motion

List known masses, forces, and initial conditions, then identify the unknown accelerations and tensions. Combine the force equations with the differentiated rope constraint. Solve symbolically before substituting values; this makes the role of each force and mass clear.

A negative acceleration means that the particle accelerates opposite to its chosen positive direction. Check units: mass times acceleration has units of kilograms times metres per second squared, which is a newton. Check the force equations and the rope constraint using the reported results.

When acceleration is constant, initial velocity and elapsed time can be used to find later velocity or displacement. Do not confuse signed displacement with path length: displacement may be positive or negative, while path length is nonnegative.

v=v_0+at,\quad x=x_0+v_0t+\tfrac12at^2

A negative signed answer indicates acceleration opposite to the chosen positive direction.
Check both force equations and the rope constraint after solving.
Use constant-acceleration relations only when acceleration is constant.

Worked example

A block connected to a hanging particle

A 4.00 kg block rests on a smooth horizontal table and is connected over a fixed, ideal pulley to a hanging 2.00 kg particle. Starting from rest, find the acceleration magnitude and rope tension. Use

g=9.81\ \mathrm{m/s^2}

.

Hanging particle free-body diagram

The hanging particle accelerates downward; the table block accelerates toward the pulley.

Define coordinates and constraint
Use a ground-fixed observer. Let the block’s positive coordinate point toward the pulley and the hanging particle’s positive coordinate point downward. The rope length is fixed, so coordinate changes, velocities, and accelerations have equal magnitudes in these chosen positive directions.
$a_1=a_2=a$
Write force equations
For the table block, tension is the only horizontal force. For the hanging particle, weight acts downward and tension upward. Apply Newton’s second law along each positive coordinate.
$T=m_1a,\quad m_2g-T=m_2a$
Solve and substitute
Add the equations to eliminate the internal rope tension. Substitute the masses and gravitational acceleration to find the common acceleration, then use the first equation to find tension.
$a=\frac{m_2g}{m_1+m_2}=3.27\ \mathrm{m/s^2},\quad T=m_1a=13.1\ \mathrm{N}$

Answer: The block accelerates toward the pulley and the hanging particle accelerates downward, both at

3.27\ \mathrm{m/s^2}

. The rope tension is

13.1\ \mathrm{N}

.

Check: The hanging particle’s weight is

19.6\ \mathrm{N}

, greater than the

13.1\ \mathrm{N}

tension, so its downward acceleration has the correct sign. The table block’s tension gives

13.1\ \mathrm{N}/4.00\ \mathrm{kg}=3.27\ \mathrm{m/s^2}

.

Worked example

Two hanging particles over a fixed pulley

Two particles of masses 3.00 kg and 5.00 kg are joined by one ideal rope passing over a fixed, frictionless pulley. They start from rest. Find their acceleration and the rope tension using

g=9.81\ \mathrm{m/s^2}

.

Heavier particle free-body diagram

The heavier particle’s positive direction is downward.

Set directions and relate accelerations
Use a ground-fixed observer. Choose downward as positive for the 5.00 kg particle and upward as positive for the 3.00 kg particle. The fixed-rope constraint makes their signed accelerations equal in these selected coordinates.
$a_1=a_2=a$
Apply Newton’s second law
For the heavier particle, weight minus tension is positive. For the lighter particle, tension minus weight is positive. Adding the equations removes tension and determines acceleration.
$m_2g-T=m_2a,\quad T-m_1g=m_1a$
Calculate the unknowns
The positive result confirms acceleration in the chosen directions: the heavier particle moves down and the lighter one moves up. Substitute the acceleration into either force equation to find tension.
$a=\frac{(m_2-m_1)g}{m_1+m_2}=2.45\ \mathrm{m/s^2},\quad T=m_1(g+a)=36.8\ \mathrm{N}$

Answer: The 5.00 kg particle accelerates downward and the 3.00 kg particle upward, each with acceleration magnitude

2.45\ \mathrm{m/s^2}

. The rope tension is

36.8\ \mathrm{N}

.

Check: The tension lies between the particles’ weights: it is less than

49.1\ \mathrm{N}

and greater than

29.4\ \mathrm{N}

. This is consistent with the heavier particle accelerating down and the lighter particle accelerating up.

Worked example

A movable pulley and a two-segment constraint

A 4.00 kg particle A hangs from the free end of an ideal rope. The rope passes over a fixed pulley, under a movable pulley carrying a 3.00 kg particle B, and then attaches to a fixed support. Starting from rest, find the accelerations and rope tension. Use

g=9.81\ \mathrm{m/s^2}

. Assume the movable pulley is massless.

Particle A free-body diagram

Schematic free-body diagram for A; the rope constraint also accounts for the two segments supporting B.

Define coordinates and constraint
Use a ground-fixed observer and choose downward as positive for both A and B. When B moves down, each of its two supporting rope segments lengthens. A must move upward by twice that distance to keep the total rope length fixed. Thus A’s signed acceleration is minus twice B’s.
$a_A+2a_B=0$
Write force equations
For A, weight acts downward and tension upward. Particle B and its massless movable pulley are supported by two upward tensions, so their total upward force is twice the tension. Apply Newton’s second law with downward positive for both particles.
$m_Ag-T=m_Aa_A,\quad m_Bg-2T=m_Ba_B$
Solve the coupled equations
Substitute the constraint, $a_A=-2a_B$ , into A’s force equation. This gives $T=m_Ag+2m_Aa_B$ . Substitution into B’s equation gives the signed accelerations; use A’s equation to calculate the tension.
$a_B=-\frac{5g}{19}=-2.58\ \mathrm{m/s^2},\quad a_A=\frac{10g}{19}=5.16\ \mathrm{m/s^2},\quad T=\frac{36g}{19}=18.6\ \mathrm{N}$

Answer: A accelerates downward at

5.16\ \mathrm{m/s^2}

. B accelerates upward at

2.58\ \mathrm{m/s^2}

. The rope tension is

18.6\ \mathrm{N}

.

Check: The negative value of

a_B

means B accelerates upward, opposite to its positive downward coordinate. The constraint is satisfied because

5.16+2(-2.58)=0

. Also,

T=m_A(g-a_A)=4.00(9.81-5.16)\ \mathrm{N}\approx18.6\ \mathrm{N}

, consistent with A accelerating downward.

Common mistakes and how to avoid them

Assuming connected particles always have equal acceleration magnitudes.

Correction: Derive the acceleration relationship by differentiating the rope-length equation. A movable pulley can make one particle’s acceleration a multiple of another’s.

Using the same positive direction for every particle without accounting for rope geometry.

Correction: Choose directions explicitly, write the signed rope constraint, and interpret negative results relative to those choices.

Treating tension as the same as weight or as an extra force on the whole system.

Correction: Draw separate free-body diagrams. Tension is a force exerted by the rope; solve for it using the particle force equations.

Putting acceleration on a free-body diagram.

Correction: A free-body diagram contains forces. State acceleration separately and use it in Newton’s second law.

Lesson summary

Define the observer, reference frame, coordinates, and positive directions before solving.
Use constant rope length to relate connected particles’ positions, velocities, and accelerations.
Apply Newton’s second law separately to each particle and combine those equations with the constraint.
Check signs, units, force equations, and the rope constraint before reporting directions.

Check your understanding

Question 1

A single inextensible rope passes over a fixed pulley and connects two particles. If one particle’s acceleration coordinate is positive downward and the other’s is positive upward, what relationship holds between their signed accelerations?

They are equal.
They are negatives of each other.
One is twice the other.
Their sum is the gravitational acceleration.

Show answer and explanation

They are equal.

With these opposite physical positive directions, the fixed-rope constraint gives equal signed accelerations. If both coordinates instead used downward as positive, their signed accelerations would be negatives of each other.

Question 2

A movable pulley is supported by two segments of the same inextensible rope. If its downward displacement is positive, what is the magnitude and direction of the free end’s displacement needed to keep the rope length fixed?

Half the pulley displacement, in the same direction.
Equal to the pulley displacement, in the opposite direction.
Twice the pulley displacement, in the opposite direction.
It is unrestricted by rope length.

Show answer and explanation

Twice the pulley displacement, in the opposite direction.

The two supporting segments each change length by the pulley displacement, so the free end must move twice that amount in the opposite direction.

Question 3

For 3.00 kg and 5.00 kg particles hanging on opposite sides of a fixed pulley, which direction does the heavier particle accelerate?

Upward, because the lighter particle has less weight.
Downward, because its weight exceeds the opposing particle’s weight.
There is no acceleration because the rope has fixed length.
The direction cannot be determined without knowing the rope tension first.

Show answer and explanation

Downward, because its weight exceeds the opposing particle’s weight.

The fixed rope constrains the particles to move together but does not prevent acceleration. The greater hanging weight drives the 5.00 kg particle downward.

Key terms

Inextensible rope: An ideal rope whose length does not change, creating a constraint between the motions of connected particles.
Tension: The pulling force transmitted along a taut rope.
Constraint: A relationship that limits how a system’s positions or motions can change.
Signed acceleration: Acceleration measured along a chosen coordinate; a negative value means it points opposite to the positive direction.

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3.5 · Analyze constrained and connected particle systems

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Use rope-length relationships and Newton’s second law to connect particle motions

What you will learn

1. Describe the system and write the constraint

2. Apply Newton’s second law to each particle

3. Solve and check the connected motion

A block connected to a hanging particle

Two hanging particles over a fixed pulley

A movable pulley and a two-segment constraint

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through EN PH 131

About this lesson and its review