7.5 Distinguish internal and external forces in system balances | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Systems of Particles

EN PH 131 Engineering Mechanics: Dynamics — Study topic 7.5

A force is not inherently internal or external: its classification depends on the system you choose. For one object, the contact force from another object is external. If both objects are included in the system, that same interaction is internal. This choice matters because forces exchanged within a system cancel in the whole-system force balance, while forces exerted from outside do not. In this lesson, an observer measures motion from a fixed ground frame. We use planar Cartesian axes, with positive horizontal direction to the right and positive vertical direction upward unless stated otherwise. The system boundary is the imagined line separating the objects included in the balance from everything outside it.

What you will learn

Define a system and decide whether a force is internal or external to it.
Apply Newton’s second law to a system of particles using only the forces external to the chosen system.
Explain why internal force pairs cancel in a whole-system balance but still matter when finding the motion of one part.
Choose a system boundary that makes the requested unknown easier to calculate.

1. Choose the system before listing forces

A system is the object or collection of objects whose motion you are analysing. First name what is inside it. Then classify each force by asking: does the force come from an interaction between parts both inside the system, or from something outside it? The first kind is internal; the second is external. The same physical interaction can change category when you redraw the system boundary.

For a single block, its weight, a rope pull, and a floor’s normal force are external forces on that block. For a system containing two blocks in contact, the force each block exerts on the other is internal. The floor’s normal force and gravity remain external if the floor and Earth are not included.

Newton’s third law says that the two forces in an interaction pair are equal in magnitude and opposite in direction. When both interacting objects are in the system, those forces cancel in the vector sum for the whole system. They do not disappear from the motion of each object considered separately.

\sum \mathbf{F}_{\mathrm{ext}}=M\mathbf{a}_{G}

Internal or external describes a force relative to a stated system boundary.
Forces between two parts of one system cancel in its total force balance.
A force internal to a combined system may be external to either part considered alone.

2. The whole-system force balance

For a collection of particles with constant total mass, the centre of mass, denoted by

G

, moves according to the net external force. Here

M

is the system’s total mass and

\mathbf{a}_{G}

is the acceleration of its centre of mass. Internal forces do not appear in this equation because each internal interaction has an equal-and-opposite partner in the same system.

In a one-dimensional problem, choose a positive direction and give every force component and acceleration a sign in that direction. A negative result means the actual direction is opposite to the direction you chose. The equation applies to the selected system as a whole; it does not generally give the force on each part.

Before calculating, list the known masses and external forces, and identify the requested quantity. If the question asks for the system’s acceleration, the whole-system balance often avoids finding internal contact forces or tension. If it asks for one of those internal forces, analyse an individual part after finding the system’s motion.

\sum F_{x,\mathrm{ext}}=M a_{Gx}

Use the vector sum of external forces for the whole-system balance.
Internal forces cancel as pairs; they can still affect the motion of individual parts.
Keep the chosen axes and signs consistent in every equation.

3. System choice, contact forces, and impulse

A useful force diagram shows the external forces acting on the system, not every interaction inside it. For a combined system of blocks on a smooth horizontal surface, the vertical weight and normal forces can balance, while a horizontal applied pull determines the horizontal acceleration. A contact force between the blocks is omitted from the combined-system diagram because it is internal.

To find that contact force, isolate one block. The contact force from the other block is now external to the smaller system. Apply Newton’s second law to that block, using the acceleration already found for the combined system. This two-stage approach separates the system’s overall motion from the force transfer between its parts.

The same boundary logic works with impulse and momentum. Over a time interval, internal impulses cancel in the total momentum balance. If the net external impulse in a direction is zero, the system’s total momentum in that direction does not change. This statement concerns the total momentum; the individual parts may still change velocity as they exchange momentum.

\mathbf{P}_{2}-\mathbf{P}_{1}=\int_{t_1}^{t_2}\sum\mathbf{F}_{\mathrm{ext}}\,dt

Draw the diagram for the system named in the balance.
Use a smaller system when an internal interaction force is required.
Conservation of total momentum requires zero net external impulse in the direction considered.

4. A reliable balance routine

State the system and the observer’s frame before writing equations. Identify the initial and final states if the problem involves a time interval. Sketch the relevant bodies and external forces, then mark axes and positive directions. Classify each force by the boundary rather than by its name: tension, friction, contact, and gravity can each be internal or external depending on which bodies are included.

Choose a whole-system balance when the unknown is the system’s acceleration or change in total momentum. Write the external-force sum in components, solve symbolically, and substitute SI units. Check that force has units of newtons and that acceleration has units of metres per second squared. For an impulse balance, check that impulse and momentum both have units of kilogram metres per second.

Finally, check whether the sign matches the chosen direction and whether the result makes physical sense. If an internal force was omitted from the system equation, verify that both objects producing the interaction really are included. If only one is included, the force from the other object belongs in the external-force sum.

Name the system, frame, coordinates, signs, knowns, and unknowns before solving.
Check every force against the system boundary.
Use units and direction checks to catch errors in the balance.

Worked example

Two carts pushed together

Two carts of masses

2.0\ \mathrm{kg}

and

3.0\ \mathrm{kg}

touch on a smooth, horizontal track. A horizontal force of

20\ \mathrm{N}

pushes the

2.0\ \mathrm{kg}

cart to the right. Find the acceleration of the two-cart system and the contact force on the

3.0\ \mathrm{kg}

cart. The carts start from rest; analyse their acceleration at the instant described.

Combined carts and external push

For the combined system, the cart-to-cart contact forces are internal and are not shown.

Set the system and signs
The system is both carts, observed from the fixed track frame. Take right as positive. The known total mass is $5.0\ \mathrm{kg}$ , and the unknowns are the system acceleration and the contact-force magnitude.
Balance the combined system
The smooth track supplies no horizontal friction. The horizontal contact forces between the carts are internal and cancel in the combined-system balance. The applied push is the only horizontal external force.
$20\ \mathrm{N}=(5.0\ \mathrm{kg})a$
Find the acceleration
Solving gives a positive acceleration, so the carts accelerate to the right, as expected from the applied force.
$a=4.0\ \mathrm{m/s^2}$
Isolate the second cart
For the $3.0\ \mathrm{kg}$ cart alone, the contact force from the first cart is external. It is the only horizontal force on this cart, so its magnitude follows from its acceleration.
$F_{\mathrm{contact}}=(3.0\ \mathrm{kg})(4.0\ \mathrm{m/s^2})=12\ \mathrm{N}$

Answer: The two carts accelerate at

4.0\ \mathrm{m/s^2}

to the right. The contact force on the

3.0\ \mathrm{kg}

cart is

12\ \mathrm{N}

to the right.

Check: The acceleration has units of force divided by mass, and the contact force has units of newtons. The remaining

8\ \mathrm{N}

of the applied push accelerates the

2.0\ \mathrm{kg}

cart:

20\ \mathrm{N}-12\ \mathrm{N}=(2.0\ \mathrm{kg})(4.0\ \mathrm{m/s^2})

. The two-cart balance correctly contains the full

20\ \mathrm{N}

external push.

Worked example

A rope pulls two blocks

Two blocks of masses

4.0\ \mathrm{kg}

and

2.0\ \mathrm{kg}

are connected by a light rope on a smooth, horizontal surface. A horizontal force of

18\ \mathrm{N}

pulls the

4.0\ \mathrm{kg}

block to the right. Find the acceleration and the rope tension. Initially both blocks are at rest.

Two-block system and applied pull

The rope tension is internal to the combined system, so it is omitted from this diagram.

Define the system
Take both blocks as the system and use the fixed surface as the observer’s frame. Right is positive. The total mass is $6.0\ \mathrm{kg}$ . The rope tension is internal; the applied force is the only horizontal external force.
Find the common acceleration
Because the rope remains taut, both blocks have the same horizontal acceleration. Apply the horizontal force balance to the combined system.
$18\ \mathrm{N}=(6.0\ \mathrm{kg})a$
Find the tension using one block
Now select the $2.0\ \mathrm{kg}$ block alone. The rope pulls it to the right, and this tension is external to that smaller system. Its acceleration is the value found for the pair.
$T=(2.0\ \mathrm{kg})(3.0\ \mathrm{m/s^2})=6.0\ \mathrm{N}$

Answer: The blocks accelerate at

3.0\ \mathrm{m/s^2}

to the right, and the rope tension is

6.0\ \mathrm{N}

.

Check: For the

4.0\ \mathrm{kg}

block alone, the net force to the right is

18\ \mathrm{N}-6.0\ \mathrm{N}=12\ \mathrm{N}

, giving

12\ \mathrm{N}/(4.0\ \mathrm{kg})=3.0\ \mathrm{m/s^2}

. The equal acceleration is consistent with a taut, light rope.

Worked example

Two skaters push apart

Two skaters are initially at rest on nearly frictionless ice. Their masses are

30\ \mathrm{kg}

and

45\ \mathrm{kg}

. During a brief push, the net external horizontal impulse on the two-skater system is negligible. The

30\ \mathrm{kg}

skater receives a horizontal impulse of

90\ \mathrm{N\,s}

to the left. Find both final horizontal velocities.

Set the system and states
The system contains both skaters. The observer is fixed to the ice, and right is positive. Initially the total horizontal momentum is zero. The push forces are internal to this system; the stated negligible external horizontal impulse permits a total horizontal momentum balance.
Use the system impulse balance
With zero net external horizontal impulse, total horizontal momentum is unchanged. The $30\ \mathrm{kg}$ skater’s final momentum is leftward, so the other skater must have equal rightward momentum.
$0=(30\ \mathrm{kg})v_1+(45\ \mathrm{kg})v_2$
Find the first velocity
The impulse on the first skater equals its momentum change. Its initial velocity is zero, and left is negative under the chosen sign convention.
$v_1=\frac{-90\ \mathrm{N\,s}}{30\ \mathrm{kg}}=-3.0\ \mathrm{m/s}$
Find the second velocity
Substitute the first velocity into the system momentum equation. The positive result means the $45\ \mathrm{kg}$ skater moves to the right.
$v_2=2.0\ \mathrm{m/s}$

Answer: The

30\ \mathrm{kg}

skater moves at

3.0\ \mathrm{m/s}

to the left, and the

45\ \mathrm{kg}

skater moves at

2.0\ \mathrm{m/s}

to the right.

Check: The final total horizontal momentum is

(-90+90)\ \mathrm{kg\,m/s}=0

, matching the initial value. The impulse unit

\mathrm{N\,s}

is equivalent to

\mathrm{kg\,m/s}

. The skaters’ individual momenta change, while their sum remains constant because the external horizontal impulse is negligible.

Common mistakes and how to avoid them

Calling tension or contact force internal in every problem.

Correction: Classify the force relative to the chosen system. It is internal only when both objects in the interaction are included.

Putting an internal force into the combined-system force balance.

Correction: For a whole-system balance, include external forces only. Internal interaction pairs cancel in the total.

Assuming that internal forces have no effect on the parts.

Correction: Internal forces can accelerate individual parts. They cancel only when balancing the selected system as a whole.

Using zero external force to claim every object keeps its velocity.

Correction: A zero net external force means the system’s centre of mass has no acceleration; individual parts can still change motion through internal interactions.

Lesson summary

State the system boundary before classifying any force.
For a whole system, internal forces cancel in pairs; external forces determine the change in total motion.
Use an individual part as the system when you need an interaction force such as tension or contact force.
A momentum balance over a time interval depends on external impulse; internal impulses cancel in the total-system balance.

Check your understanding

Question 1

Two blocks touch and are treated as one system. Is the force they exert on each other internal or external?

Internal, because both blocks are inside the system.
External, because it is a contact force.
External, because it acts horizontally.
Neither; contact forces do not enter mechanics balances.

Show answer and explanation

Internal, because both blocks are inside the system.

The blocks are both inside the chosen system, so their mutual interaction is internal. Its equal-and-opposite forces cancel in the whole-system balance.

Question 2

A rope connects two blocks, but the system contains only the block at the rope’s far end. How should the rope tension be classified for this system?

Internal, because the rope connects the blocks.
External, because the rope is outside the selected system.
Internal, because tension always cancels.
It cannot be classified without knowing the acceleration.

Show answer and explanation

External, because the rope is outside the selected system.

The selected system contains only one block, so the rope is outside it. The rope’s pull is an external force on that block.

Question 3

A two-part system starts with zero total horizontal momentum and receives negligible net external horizontal impulse. What happens to its total horizontal momentum?

It remains zero, although the parts may move in opposite directions.
It must increase because the parts exert forces on each other.
It becomes equal to the momentum of the more massive part.
It remains zero only if each part stays at rest.

Show answer and explanation

It remains zero, although the parts may move in opposite directions.

With zero net external horizontal impulse, the system’s total horizontal momentum is conserved. Internal interactions can change the parts’ separate momenta without changing their sum.

Key terms

System: The object or collection of objects selected for a mechanics balance.
Internal force: A force from an interaction between parts that are both included in the selected system.
External force: A force exerted on the selected system by something outside it.
External impulse: The time integral of the net external force on a system over an interval; it equals the change in the system’s total momentum.

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7.5 · Distinguish internal and external forces in system balances

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

EN PH 131 Engineering Mechanics: Dynamics — Study topic 7.5

What you will learn

1. Choose the system before listing forces

2. The whole-system force balance

3. System choice, contact forces, and impulse

4. A reliable balance routine

Two carts pushed together

A rope pulls two blocks

Two skaters push apart

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