Free-Body Diagrams & Net Force — Print · Grade 12 Science

Chapter: Forces, Rotation & Equilibrium

Free-Body Diagrams & Net Force

Learning goals

Readings for this unit

Before this unit

Vectors and Newton’s second law.

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Start with something to explain

Why can an upward-angled pull change friction?

Illustrative evidence case. A rope pulls a block forward and partly upward while the block stays on a table.

Select the object before collecting force arrowsSelect the object before collecting force arrowsselected bodytable on blockEarth on blockrope on blockBoundary: block only. Include forces ON that body.
Original block-only interaction inventory, not a claim of balanced horizontal forces.

Notice and wonder

Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.

Why can an upward-angled pull change friction?

Your first model

Draw only real interactions and then resolve the tension.

Build the language from the idea

free-body diagram
A simplified sketch of a single object with an arrow drawn for every force acting on it, pointed in the correct direction and sized to show relative strength.
net force (ΣF)
The single overall force found by adding every force on an object as a vector, accounting for direction.
equilibrium
The condition in which the net force on an object is zero, so its velocity does not change.
normal force
The support force a surface pushes back with, always directed perpendicular to the surface.
vector
A quantity that has both magnitude and direction, such as force, velocity, or acceleration.

Real objects rarely feel just one force. A crate being dragged across a warehouse floor feels its own weight pulling down, the floor pushing up, a rope pulling forward, and friction dragging back — four forces at once. To predict how it moves, you cannot juggle all of that in your head; you need a system. That system is the free-body diagram, the single most-used tool in all of mechanics, and it is the bridge between a messy real scene and the clean equation ΣF = ma.

To draw one, you mentally cut the object free from its surroundings and represent it as a simple dot or box. Then you draw an arrow for every force *acting on that object* — and only those. Each arrow starts at the object, points the way the force pushes or pulls, and is drawn longer for a stronger force. The usual cast of characters is small and worth memorizing: weight (gravity, always straight down), the normal force (a surface pushing perpendicular to itself), tension (a rope or cable pulling along its length), an applied force (a direct push or pull), and friction (always opposing sliding). Naming the forces correctly is more than half the battle.

Select the object before collecting force arrowsSelect the object before collecting force arrowsselected bodytable on blockEarth on blockrope on blockBoundary: block only. Include forces ON that body.
Original block-only interaction inventory, not a claim of balanced horizontal forces.

Once every force is on the diagram, you find the net force by adding the arrows as vectors — quantities with both size and direction. Forces along the same line are easy: add them if they point the same way, subtract them if they oppose. The net force is what actually goes into Newton's second law, which in its full form reads ΣF = ma. A common and costly mistake is to plug a single force into that equation; only the *sum* of all forces determines the acceleration, and the object always accelerates in the direction of that net force.

Because forces are vectors, it usually pays to split the problem into directions and handle each independently. Physicists resolve forces into perpendicular components, typically horizontal (x) and vertical (y), then apply Newton's second law separately to each: ΣFₓ = maₓ and ΣF_y = ma_y. For a crate sliding along level ground, the vertical forces (weight down, normal up) cancel, so all the acceleration comes from the horizontal forces. Breaking a two-dimensional problem into two one-dimensional problems is the standard professional move, and it keeps the arithmetic honest.

Write independent equations along the selected axesWrite independent equations along the selected axesAxisForce equationMotion constraintHorizontalTcosθ−f=maxAcceleration can benonzeroVerticalN+Tsinθ−mg=mayay=0 on a level surfaceTwo component equations share one physical diagram.
Original force equations for a level-table contact model with a single angled tension.

When two forces act at a right angle instead of along one line, you combine them with the Pythagorean theorem, exactly as in geometry. A 9 N pull east and a 12 N pull north do not add to 21 N; the true net force is √(9² + 12²) = √225 = 15 N, aimed diagonally between them. This is a beautiful moment where math and physics turn out to be the same tool: the resultant of two perpendicular forces is literally the hypotenuse of a right triangle whose legs are those forces.

A special and important case is equilibrium, when all the forces sum to exactly zero. A book resting on a table, a lamp hanging motionless from a cord, and an airplane cruising at a steady altitude and speed are all in equilibrium — not because no forces act, but because the forces perfectly cancel. By the first law, zero net force means the velocity never changes, so an object in equilibrium is either sitting still or moving at constant velocity. Reading a free-body diagram tells you instantly which case you are in: if the arrows balance, motion holds steady; if they don't, the leftover net force sets the acceleration through ΣF = ma.

Model check: Draw only forces exerted on the selected object; motion arrows are not forces. Net 12N right; a=12/4=3 m/s² right.

Build your understanding

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Part 1 · Draw forces on one selected object

A free-body diagram isolates one object and shows the external interactions acting on it. A rope pulls, Earth attracts and a table supports the block; the block’s forces on these agents belong in their diagrams instead. The boundary determines the force list. Motion arrows and acceleration arrows describe responses, so distinguish them from force arrows with labels and units rather than adding them to the net-force sum.

Select the object before collecting force arrowsSelect the object before collecting force arrowsselected bodytable on blockEarth on blockrope on blockBoundary: block only. Include forces ON that body.
Original block-only interaction inventory, not a claim of balanced horizontal forces.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Should “the block pushes the table” appear in this diagram?

Visual reasoning — after your attempt

No. That force acts on the table, outside the selected block boundary.

Choose a body and draw external interactions acting on it. Label each arrow by its agent and direction. Do not add velocity or acceleration as if they were forces. Forces exerted by the body on other objects belong in their diagrams, not this one.

A box on a rough floor may have weight, support, an applied pull, and friction. A motion arrow can be drawn separately with a different label.

Pause and explain

Why should the object boundary be chosen first?

Check your reasoning after answering

It determines which interactions are external and which body receives each force, preventing mixed-body force sums.

Part 2 · Resolve angled forces before balancing

An angled force can be represented by two perpendicular components that together equal the original vector. Use either the original vector or its components in a sum, not all three as separate forces. With θ measured above horizontal, a 10 N tension at 30° has components 8.66 N right and 5.00 N up. The upward part reduces the support needed for vertical equilibrium, so normal force cannot automatically be replaced by mg.

Resolve one angled interaction without counting it twiceResolve one angled interaction without counting it twiceFx = F cosθFy = F sinθA component is a projection, not another interaction.
Original projection triangle; the dashed vertical is not an additional real force.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

For a 2 kg block with no vertical acceleration, what normal force remains?

Visual reasoning — after your attempt

N=mg−Fy=19.6−5.0=14.6 N, assuming those are the only vertical forces.

An angled force can contribute to both horizontal and vertical sums. Resolve components using an explicitly labeled angle. A pulling force angled upward can reduce normal support and thereby change the available friction in a suitable contact model. Normal force is not automatically mg.

For a 20 N pull at 30° above horizontal, upward component is 10 N and horizontal component is about 17.3 N.

Pause and explain

Why does upward pulling change the normal force?

Check your reasoning after answering

If vertical acceleration is zero, the upward pull carries part of the weight, leaving less force required from the floor.

Part 3 · Write independent component equations

A vector equation becomes one scalar equation per perpendicular axis. Independent does not mean unrelated: the same tension angle enters both, and contact constraints can set a component of acceleration. For a block that remains on a horizontal table, ay=0 while ax may be nonzero. Assign signs to every component before substituting values; then solve the vertical support account before using a friction model that depends on N.

Write independent equations along the selected axesWrite independent equations along the selected axesAxisForce equationMotion constraintHorizontalTcosθ−f=maxAcceleration can benonzeroVerticalN+Tsinθ−mg=mayay=0 on a level surfaceTwo component equations share one physical diagram.
Original force equations for a level-table contact model with a single angled tension.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Why must N be found before μN in this pulled-block example?

Visual reasoning — after your attempt

The upward tension component changes N, so using mg directly in μN would use the wrong contact force.

Use ΣFx=max and ΣFy=may with the same selected body and coordinate convention. Zero acceleration in one direction does not require zero acceleration in the other. State the constraint that makes one component zero, such as remaining on a horizontal surface.

A box can accelerate horizontally while vertical forces balance. Solve the vertical balance before using a friction expression that depends on normal force.

Pause and explain

Does vertical equilibrium imply complete equilibrium?

Check your reasoning after answering

No. It only constrains the vertical component; the horizontal net force can remain nonzero.

Part 4 · Check a diagram against the physical contact

A simple support can push normal to a surface but cannot pull an object toward it without adhesion or another mechanism. Similarly an ideal flexible rope transmits tension but cannot push. If an assumed contact solution requires negative N, the starting constraint has failed and the object may leave the surface. Return to the geometry and interactions rather than accepting an algebraic negative contact force as a valid physical result.

Check calculated forces against actual contactCheck calculated forces against actual contactObservationAllowed model conclusionLimitNo surface contactN=0Contact cannot pull abody downRope slackT=0No compressive pushfrom ropeContact maintainedNormal may be positiveCheck geometricconstraintA negative required support signals a failed contact assumption.
Qualitative contact checks; adhesion and other interactions would require explicit additional models.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

What changes if the upward pull exceeds the weight of an unrestrained block?

Visual reasoning — after your attempt

Contact is lost in the simple model; set N=0 and allow upward acceleration rather than retaining an impossible negative normal force.

The result must respect physical constraints. A floor cannot provide a negative upward normal force in the simple nonadhesive contact model; such a result can indicate lost contact or an invalid assumption. Check directions and contact state before accepting algebra.

If a calculated support becomes negative, reconsider whether the object remains on the floor rather than reporting a downward floor support.

Pause and explain

What should be done with an impossible support result?

Check your reasoning after answering

Revisit the assumed contact and constraints, then choose the appropriate motion model; an algebraic answer alone cannot preserve an invalid physical assumption.

Apply the whole unit · Analyze an angled pull

Illustrative 4.00 kg box is pulled by 20.0 N at 30° above horizontal. Kinetic friction coefficient is 0.200; g=10.0 m/s². Assume sliding contact and zero vertical acceleration.

Your task

Draw and resolve the forces, calculate support and friction, find horizontal acceleration, and explain why using μmg gives the wrong friction here.

What your response should show

  • Define the system, quantities, units, and relevant conditions before selecting a model.
  • Show intermediate calculations or a labeled evidence account, including signs or denominators.
  • Explain how the supplied evidence supports the conclusion rather than reporting only an answer.
  • Identify a meaningful uncertainty, model limit, or measurement needed to test the conclusion.
Response guidance — after your attempt
  1. Weight=40.0 N; pull upward component=10.0 N; N=30.0 N.

  2. Friction=0.200×30.0=6.00 N; horizontal pull≈17.32 N.

  3. Net horizontal≈11.32 N; a≈11.32/4.00=2.83 m/s².

The inclined pull reduces floor support, so friction in this model is less than μmg. These equations assume the box is already sliding and retains contact. For a stationary box, static friction must be evaluated instead; substituting a kinetic coefficient would describe the wrong regime.

Scientific references and source-model foundations

Parent / teacher teaching notes

Require a diagram before numbers and ask which force changes when the pull angle changes. Check whether friction is static or kinetic before evaluating it. Ask the learner to explain each symbol, axis, unit and arrow before calculating. Calculated examples, supplied illustrative records and actual historical observations are labeled separately. Selected public OpenSciEd middle-school source figures are foundation examples, not complete high-school editions or an endorsement; all advanced explanations are supplied locally.

Free-Body DiagramN (normal)W = mgF (applied)f (friction)Net force = vector sum of all arrows
A free-body diagram isolates one object and draws every force as a labeled vector — weight, normal force, applied force, and friction. Their vector sum is the net force.

The key idea

A free-body diagram includes external forces on one selected body. Their vector sum determines acceleration; static rigid-body equilibrium also requires torque balance.

Net Force (same line)

ΣF = F₁ + F₂ + …

add forces pointing the same way; subtract forces pointing opposite ways, to get the net force along one direction.

Resultant of Perpendicular Forces

F net = √(Fₓ² + F_y²)

Fₓ and F_y are two forces (or force components) at right angles; their vector sum has this magnitude.

Worked examples

Example 1

An 8 kg crate is pulled right with 60 N while friction drags left with 20 N. Its 78.4 N weight is balanced by an equal normal force. Find the net force and the crate's acceleration.
  1. Vertical forces cancel: normal force up equals weight down, so no vertical acceleration.
  2. Horizontal net force = 60 − 20 = 40 N to the right.
  3. a = ΣF ÷ m = 40 ÷ 8 = 5.
Answer: Net force = 40 N right; a = 5 m/s² right

Example 2 · Worked application

A 5 kg object rests on a horizontal floor with g=10 N/kg and no other vertical forces. Find weight and normal force.
  1. Weight=mg=5×10=50 N down.
  2. Vertical acceleration is zero, so net vertical force is zero.
  3. Normal force must be 50 N up.
Answer: Weight 50 N down; normal 50 N up.
Check the reasoning

Normal equals weight only for the specified force and acceleration conditions.

Example 3

Two forces act on an object at a right angle: 9 N east and 12 N north. Find the magnitude of the net force.
  1. The forces are perpendicular, so use the Pythagorean theorem: net = √(Fₓ² + F_y²).
  2. net = √(9² + 12²) = √(81 + 144) = √225.
  3. = 15.
Answer: 15 N

Example 4 · Find and fix the mistake

Context: A 5 kg object rests on a horizontal floor with g=10 N/kg and no other vertical forces. Find weight and normal force. A student claims: “Normal force always equals mg in every situation.” Evaluate the claim and write a corrected explanation.
  1. Test the claim against the stated conditions: Normal equals weight only for the specified force and acceleration conditions.
  2. Inclines, vertical pushes and accelerating supports can change the normal force; use the force diagram and equations.
Answer: Inclines, vertical pushes and accelerating supports can change the normal force; use the force diagram and equations.
Check the reasoning

Normal equals weight only for the specified force and acceleration conditions.

Example 5 · Reason from the visual model

Write independent equations along the selected axesWrite independent equations along the selected axesAxisForce equationMotion constraintHorizontalTcosθ−f=maxAcceleration can benonzeroVerticalN+Tsinθ−mg=mayay=0 on a level surfaceTwo component equations share one physical diagram.
Original force equations for a level-table contact model with a single angled tension.
Why must N be found before μN in this pulled-block example?
  1. A vector equation becomes one scalar equation per perpendicular axis. Independent does not mean unrelated: the same tension angle enters both, and contact constraints can set a component of acceleration. For a block that remains on a horizontal table, ay=0 while ax may be nonzero. Assign signs to every component before substituting values; then solve the vertical support account before using a friction model that depends on N.
  2. The upward tension component changes N, so using mg directly in μN would use the wrong contact force.
Answer: The upward tension component changes N, so using mg directly in μN would use the wrong contact force.
Check the reasoning

Original force equations for a level-table contact model with a single angled tension.

Avoid these mistakes

Check your understanding

A 4 kg block has 20N right and 8N left. Find acceleration.

Discuss the answer

Net 12N right; a=12/4=3 m/s² right.

Return to the phenomenon

Return to the initial case. Why can an upward-angled pull change friction? Revise your drawing or explanation using two named ideas and one stated limitation.

Compare explanations after your attempt

The upward component reduces N; a friction model depending on N changes with it. Check contact remains possible.

What should we investigate next?

Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.

Investigate, calculate and revise

Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.

Activity 1 · Return to the opening case

Assigned reading and preparation

Use the opening phenomenon, unit definitions and explained Parts 1–4.

Your task

Return to the initial case. Why can an upward-angled pull change friction? Revise your drawing or explanation using two named ideas and one stated limitation.

Success criteria

  • Use the stated observations and appropriate units.
  • Connect a labeled model with the physical/environmental mechanism.
  • State assumptions and an evidence limitation.
  • Explain a substantive revision or next test.
Hint

Keep object/material identity, material transfer and energy account distinct.

Response guidance — discuss after your attempt

The upward component reduces N; a friction model depending on N changes with it. Check contact remains possible.

Activity 2 · Transfer to a changed case

Assigned reading and preparation

Cover the worked answer and use the explained models, relationships and notebook.

Your task

A 3 kg block is pulled by 15 N at 30° above horizontal. No vertical acceleration; resistance is 4 N. Find N and ax.

Success criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.
Hint

Name known quantities and conditions, show intermediate steps and check conservation.

Response guidance — discuss after your attempt

N=29.4−7.5=21.9 N. Horizontal net=15cos 30°−4≈8.99 N; ax≈3.00 m/s². The tension is counted once through its components.

Words to know

free-body diagram
A simplified sketch of a single object with an arrow drawn for every force acting on it, pointed in the correct direction and sized to show relative strength.
net force (ΣF)
The single overall force found by adding every force on an object as a vector, accounting for direction.
equilibrium
The condition in which the net force on an object is zero, so its velocity does not change.
normal force
The support force a surface pushes back with, always directed perpendicular to the surface.
vector
A quantity that has both magnitude and direction, such as force, velocity, or acceleration.

The free-body diagram: mechanics' most-used tool

Ask any physicist or engineer for the first step in a force problem and they will say the same thing: draw a free-body diagram. The technique is powerful precisely because it forces you to make three decisions explicit — which object you are analyzing, which forces act on it (and which do not), and which direction each one points. A frequent beginner mistake is to include forces the object exerts on other things; a free-body diagram shows only the forces acting on the chosen object itself. Done carefully, it converts a confusing real-world scene into a clean set of vectors ready for ΣF = ma.

Resolving forces into components

Because forces are vectors, the professional approach to any two-dimensional problem is to choose perpendicular axes (often x and y, sometimes tilted to match a ramp) and break every force into components along them. Then Newton's second law is applied separately in each direction: ΣFₓ = maₓ and ΣF_y = ma_y. A force F at angle θ splits into F cosθ and F sinθ, the same trigonometry used throughout geometry. This is why a rope pulled at an angle is less effective at dragging a sled than a horizontal pull — part of its force goes into lifting rather than pulling, and only the horizontal component contributes to forward motion.

Constant velocity is not the same as 'no forces'

It is tempting to think that if an object's engine is running, it must be accelerating — but an airplane cruising at a steady altitude and speed is in perfect equilibrium. Lift balances weight, and the engines' thrust balances air drag; the net force is zero even though four large forces are acting. The engines run constantly not to speed the plane up, but simply to cancel the drag trying to slow it down. The same logic explains terminal velocity: a skydiver stops accelerating once air resistance grows to equal their weight, after which they fall at a constant speed with zero net force.

Static equilibrium: why bridges and buildings stand

Structural engineering is, at heart, the art of keeping enormous objects in static equilibrium — every net force (and every net rotational effect, called torque) balanced to zero. Engineers draw free-body diagrams of individual joints, beams, and cables to verify that the forces of traffic, wind, snow, and a structure's own weight are exactly opposed by the supports and tension members holding it up. When a designer gets a single force balance wrong, the consequence is not a lost point on a test but a structure that can sag, crack, or collapse — which is why the humble free-body diagram is a genuine matter of public safety.

In the real world

Structural engineers draw free-body diagrams of every joint in a bridge to confirm the forces of traffic, wind, and the bridge's own weight all cancel — an unbalanced joint is a bridge that could fail.

Exit check

A4 kg block has 20N right and 8N left. Find acceleration.

Check your explanation after answering

Net 12N right; a=12/4=3 m/s² right.

Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.

Problem 1

A 10 kg box on level ground is pushed right with 50 N while friction resists with 20 N. Find the net horizontal force.
Answer and explanation

30 N

Net force = 50 − 20 = 30 N to the right (the vertical weight and normal force cancel).

Problem 2

Two forces act on an object at a right angle: 6 N east and 8 N north. Find the magnitude of the net force.
Answer and explanation

10 N

Perpendicular forces combine by the Pythagorean theorem: √(6² + 8²) = √(36 + 64) = √100 = 10 N.

Problem 3

In a free-body diagram of a book resting on a flat table, which forces should be drawn?
  1. Only the normal force, upward
  2. An applied force and friction
  3. Weight downward and the normal force upward
  4. Only the weight, downward
Answer and explanation

Weight downward and the normal force upward

A resting book feels gravity pulling it down and the table's normal force pushing it up. The two are equal, so it stays in equilibrium.

Problem 4

An object moves in a straight line at constant velocity. Its free-body diagram must show forces that…
  1. include no forces at all
  2. grow larger over time
  3. point in the direction of motion
  4. add up to zero
Answer and explanation

add up to zero

Constant velocity means zero acceleration, so the net force is zero — the forces present must cancel out. (Forces can exist; they simply balance.)

Problem 5

List the forces you would draw on a free-body diagram of a sled being pulled across snow by a horizontal rope, and give the direction of each.
Answer and explanation

Weight pulls straight down; the normal force from the snow pushes straight up; the rope's tension pulls forward (horizontal); and kinetic friction from the snow pushes backward, opposing the sled's motion. If the sled moves at steady speed, the up/down pair cancels and tension equals friction.

Problem 6

A 6.5 kg object on frictionless ice feels two perpendicular forces: 12 N east and 5 N north. Find its acceleration.
Answer and explanation

2 m/s²

Net force = √(12² + 5²) = √(144 + 25) = √169 = 13 N. Then a = ΣF ÷ m = 13 ÷ 6.5 = 2 m/s².

Problem 7

A hanging mass is at rest on one vertical rope, with weight 30 N. Find tension. Show the reasoning, not just the conclusion.
Hints

Normal equals weight only for the specified force and acceleration conditions.

Answer and explanation

30 N upward, balancing weight in the given model.

Problem 8

Why draw only forces acting on the selected object?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

Newton's second law concerns that object's force sum; partner forces on other objects belong in other diagrams.

Problem 9

Write independent equations along the selected axesWrite independent equations along the selected axesAxisForce equationMotion constraintHorizontalTcosθ−f=maxAcceleration can benonzeroVerticalN+Tsinθ−mg=mayay=0 on a level surfaceTwo component equations share one physical diagram.
Original force equations for a level-table contact model with a single angled tension.
Interpret the supplied visual. Why must N be found before μN in this pulled-block example?
Answer and explanation

The upward tension component changes N, so using mg directly in μN would use the wrong contact force.

A vector equation becomes one scalar equation per perpendicular axis. Independent does not mean unrelated: the same tension angle enters both, and contact constraints can set a component of acceleration. For a block that remains on a horizontal table, ay=0 while ax may be nonzero. Assign signs to every component before substituting values; then solve the vertical support account before using a friction model that depends on N.

The upward tension component changes N, so using mg directly in μN would use the wrong contact force.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the physical or environmental relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 10

A 3 kg block is pulled by 15 N at 30° above horizontal. No vertical acceleration; resistance is 4 N. Find N and ax.
Answer and explanation

N=29.4−7.5=21.9 N. Horizontal net=15cos 30°−4≈8.99 N; ax≈3.00 m/s². The tension is counted once through its components.

N=29.4−7.5=21.9 N.

Horizontal net=15cos 30°−4≈8.99 N; ax≈3.00 m/s².

The tension is counted once through its components.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

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