Net force and friction: choose the right free-body diagram

MCAT content library / Chemistry & physics / Net force and friction: choose the right free-body diagram

A force diagram explains the chosen object. Follow two connected blocks from rest into sliding to see which forces balance, which are internal, and which change.

About 19 minutes, including the worked examples and checks.

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Useful foundations

Motion: translate between vectors, units, and graphs

On this page

A force diagram begins with a boundary
Name the interaction behind each arrow
Static friction adjusts; its maximum is a limit
Worked experiment: increase the pull on two blocks
Isolate one block to find the cord force
Third-law partners act on different objects
Check your reasoning
Continue learning
Sources

A force diagram begins with a boundary

Before adding forces, decide what object or group of objects you are analyzing. A force exerted by that object belongs on the other object’s diagram. A force exerted on it belongs on its own diagram. This boundary determines which interactions enter the net force. AAMC includes translational force balance, Newton’s laws, and friction in Content Category 4A.¹

For a constant-mass object in an inertial reference frame, the vector sum of external forces is ΣF = ma. In components, ΣF_x = ma_x and ΣF_y = ma_y. Zero net force means zero acceleration; it permits rest or constant velocity. A moving object does not require an unbalanced forward force merely to keep moving.²

Name the interaction behind each arrow

Weight is the gravitational force on an object, with magnitude mg near Earth’s surface. Mass measures inertia and is expressed in kilograms; weight is a force expressed in newtons. A normal force is a contact force perpendicular to a surface. A taut ideal cord supplies tension along the cord. These directions come from the interactions, not from the direction in which you hope the object will move.³

The normal force need not equal mg. On a horizontal surface, an upward component of an applied pull can reduce the normal force while vertical acceleration remains zero. If the pull has magnitude F at an angle θ above horizontal, the components are F cos θ horizontally and F sin θ upward. With no other vertical forces and maintained contact, N = mg − F sin θ. On an incline, the normal direction follows the surface instead of the vertical.

Static friction adjusts; its maximum is a limit

Static friction prevents relative slipping at a contact. Its magnitude satisfies 0 ≤ |f_s| ≤ μ_sN in the usual dry-contact model. The value μ_sN is the greatest static friction the model permits, not a force automatically present whenever two surfaces touch. Determine the friction needed for no slipping, then compare that requirement with the limit.⁴

During sliding, the simplified kinetic-friction model gives |f_k| = μ_kN, opposite the relative sliding direction. The direction of friction follows the contact’s actual or impending slip. It is not universally “opposite the object’s motion”: for example, static contact can help propel an object. Here, both blocks slide right relative to a stationary floor, so their kinetic friction points left.⁴

Worked experiment: increase the pull on two blocks

All quantities and force records below are original illustrative values. Blocks A and B have masses 2.0 kg and 3.0 kg. A light, taut, inextensible cord joins them on a level floor. A horizontal force pulls B right. Use g = 10 m/s², μ_s = 0.40 and μ_k = 0.20 at both contacts. Ignore air resistance and cord mass.

Because no other vertical forces act and neither block accelerates vertically, N_A = 20 N and N_B = 30 N. The largest available total static friction is 0.40(20 + 30) = 20 N. As the initially resting system is pulled with 8 N and then 18 N, total static friction matches the pull. The combined net horizontal force remains zero. These totals do not, by themselves, specify the division of static friction between the two contacts.

Applied pull	State specified	Total friction	Acceleration
8 N	Initially at rest	8 N left	0
18 N	Still at rest	18 N left	0
20 N	Limiting static balance	20 N left	0 remains possible
25 N	Sliding right	10 N left	3.0 m/s² right
12 N	Later, still sliding right	10 N left	0.40 m/s² right

An illustrative force ramp followed by a reduced pull

At exactly 20 N, the ideal static limit has been reached; equilibrium is still possible. A pull greater than 20 N cannot be balanced by the available static friction. Once both blocks slide right, total kinetic friction is 0.20 × 50 = 10 N. With a 25 N pull, the acceleration of the 5.0 kg system is (25 − 10)/5.0 = 3.0 m/s² right.

The last row changes the history. A 12 N pull could be balanced if the system were resting, but the stated system is already sliding. Its kinetic friction remains 10 N, leaving 2 N net force and 0.40 m/s² acceleration. Selecting a friction rule requires the contact state as well as the applied force.

Original two-block system, with the forces on A isolated below it. Arrow directions and numerical labels are specified; arrow lengths are not a magnitude scale.

Isolate one block to find the cord force

Choose A, B, and the connecting cord as the combined system. Cord interactions are internal, so they do not appear in its external-force sum. That choice was useful for finding the shared acceleration. To find the cord tension, now isolate block A.

On A, kinetic friction is 0.20 × 20 = 4 N leftward. Thus T − 4 = (2.0)(3.0), giving T = 10 N rightward. On B, the independent check is 25 − 10 − 6 = 9 N, which equals (3.0 kg)(3.0 m/s²). The same acceleration satisfies both object equations and the combined equation.

Third-law partners act on different objects

The partner of the cord’s 10 N rightward force on A is A’s 10 N leftward force on the cord. Those forces share an interaction and act on different objects. They are equal and opposite even if the objects have different masses or accelerations.⁵

The cord’s leftward force on B also has magnitude 10 N in this ideal cord model, but it is not the third-law partner of the force on A. Similarly, A’s weight and the floor’s normal force can balance on A without forming a third-law pair. The partner of Earth’s gravitational force on A acts on Earth.

For a changed experiment, identify the object, list its actual interactions, resolve the relevant components, and only then apply the net-force relation. Use a combined system to remove internal unknowns when helpful, and return to a single object when the question asks for an internal force.

Check your reasoning

Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.

Check 1

A block slides right on a level floor while a cord pulls it at 30° above horizontal toward the right. The block remains in contact with the floor and has zero vertical acceleration. Which list correctly describes the forces on the block?

A. Weight downward; normal force upward; cord force horizontally right; friction left.

Review this choice. The cord force acts along the actual angled cord. A horizontal arrow would omit its upward component and give the wrong vertical balance.

B. Weight downward; normal force upward; cord force up and right; friction left.

Correct. Gravity acts downward, the horizontal floor supplies an upward normal force, tension follows the cord, and friction opposes the rightward relative sliding. The upward tension component means the normal force is less than mg under the stated conditions.

C. Weight downward; normal force upward; cord force up and right; friction right.

Review this choice. The floor is stationary while the block slides right, so kinetic friction on the block points left. It does not add to this block’s rightward sliding.

D. Weight downward; normal force upward; cord force down and left; friction left.

Review this choice. The down-left arrow is the direction of the force the block exerts on the angled cord. The question asks for the force of the cord on the block, which points up and right.

The principle to carry forward

A free-body diagram contains forces on the chosen object, with directions fixed by the actual contacts and fields.

Check 2

In the worked two-block example, the cord pulls A right with 10 N. Which force is its Newton’s-third-law partner?

A. The 10 N force that A exerts leftward on the cord.

Correct. This is the same A–cord interaction viewed on the other object. Equal magnitude and opposite direction are accompanied by the required exchange of which object acts on which.

B. The 10 N force that the cord exerts leftward on B.

Review this choice. It happens to be equal and opposite for the ideal cord, but it belongs to the cord–B interaction. A third-law partner must reverse the actors in the same interaction.

C. The 4 N friction force that the floor exerts leftward on A.

Review this choice. This is a different interaction and acts on A again. It can oppose tension in A’s net force without being tension’s reaction partner.

D. The 20 N gravitational force that Earth exerts downward on A.

Review this choice. This is the Earth–A interaction, not the A–cord interaction. Its own partner is A’s gravitational force on Earth.

The principle to carry forward

Check the two actors before comparing arrows. Equal and opposite forces are not sufficient to identify a third-law pair.

Check 3

Both worked-example blocks are already sliding right under the same horizontal 25 N pull. A new surface gives μ_k = 0.10 at both contacts. Their masses remain 2.0 and 3.0 kg; g = 10 m/s², and all other model conditions are unchanged. What is their common acceleration?

A. 1.0 m/s² rightward

Review this choice. The total kinetic friction is 0.10 × 50 = 5 N, but friction alone divided by mass is not the net acceleration. The 25 N applied pull must also be included.

B. 3.0 m/s² rightward

Review this choice. This retains the old 10 N friction total. The lower coefficient reduces total kinetic friction to 5 N.

C. 4.0 m/s² rightward

Correct. The net force is 25 − 5 = 20 N on 5.0 kg, so a = 4.0 m/s². The stated ongoing sliding selects kinetic friction; no static threshold must be overcome anew.

D. 5.0 m/s² rightward

Review this choice. This divides the applied force by mass while neglecting the remaining friction. The new surface reduces friction but does not remove it.

The principle to carry forward

An altered contact changes a force term. Recompute the net external force for the chosen system rather than scaling the old acceleration alone.