SCIENCE · INVESTIGATE AND EXPLAIN

Recalculate friction when the pull changes the normal force

A 2 kg object is initially sliding to the right on a horizontal surface. A force of magnitude F pulls at angle θ above horizontal. Gravity is 10 m/s² and the kinetic coefficient is 0.3. This lesson evaluates the instantaneous state while rightward slipping is stipulated; it does not assume that the same direction of slip persists forever.

Learning outcome: Compare pulls of 10,20,30 N at 90°, then compare 0° and 30° at 15 N. Keep the conditional required normal and actual contact force separate. Explain what new event would require changing the assumed kinetic-friction direction.

Starting recap

Draw recipient-specific interactions, resolve signed components and identify the domain of each force/contact model before substituting.

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Try

At F=20 N and θ=90°, normal and friction are exactly zero and instantaneous vertical acceleration is zero. Raising F to 25 N at that angle produces upward acceleration 2.5 m/s², not a negative normal force.

Open the experiment

Predict

When a required floor normal is negative, the model should:

Your prediction and reason

Experiment

Compare pulls of 10,20,30 N at 90°, then compare 0° and 30° at 15 N. Keep the conditional required normal and actual contact force separate. Explain what new event would require changing the assumed kinetic-friction direction.

Audit the normal force before applying friction

Worked teaching model · synthetic values · A 2 kg point is initially sliding right on a horizontal support; g=10 m/s², kinetic coefficient .3. Evaluate instantaneous acceleration/contact, not an entire trajectory. Upward-only support cannot exert negative normal force.

Applied pull magnitude: 15 N Pull angle above horizontal: 30 degrees

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Normal force versus pull at selected angle

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: admissible normal (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Normal force versus pull at selected angle

admissible normal

pull (N)	normal force (N)
0	20
0.75	19.625
1.5	19.25
2.25	18.875
3	18.5
3.75	18.125
4.5	17.75
5.25	17.375
6	17
6.75	16.625
7.5	16.25
8.25	15.875
9	15.5
9.75	15.125
10.5	14.75
11.25	14.375
12	14
12.75	13.625
13.5	13.25
14.25	12.875
15	12.5
15.75	12.125
16.5	11.75
17.25	11.375
18	11
18.75	10.625
19.5	10.25
20.25	9.875
21	9.5
21.75	9.125
22.5	8.75
23.25	8.375
24	8
24.75	7.625
25.5	7.25
26.25	6.875
27	6.5
27.75	6.125
28.5	5.75
29.25	5.375
30	5

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
Quantity	Value	Unit
Required constrained normal	12.5	N
Actual normal	12.5	N
Kinetic friction magnitude	3.75	N
Instantaneous horizontal acceleration	4.62019	m/s²
Instantaneous vertical acceleration	0	m/s²
Floor constraint feasible	1	flag

Pulling upward changes normal force, so kinetic friction is not automatically mu times weight.

While contact holds, N=mg−F sinθ and ax=(F cosθ−mu_k N)/m for the stipulated rightward slip. At zero normal, friction is zero. Above lift-off, set contact force to zero and allow upward acceleration; do not retain vertical equilibrium with a negative normal. A later change of slip direction would require a new friction branch.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

Pulling upward changes normal force, so kinetic friction is not automatically mu times weight.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Define the recipient and model

Build the algebraic relationship

Draw the applied force as one interaction arrow, then resolve it algebraically into Fcosθ and Fsinθ. While floor contact imposes zero vertical acceleration, N+Fsinθ−mg=0, so N=20−Fsinθ. Kinetic friction is 0.3N leftward, not automatically 0.3mg. Horizontal acceleration is [Fcosθ−0.3N]/2.

Check the changed state

The surface cannot pull downward. If the required N is negative, floor contact is lost: actual N and friction are zero and vertical acceleration becomes (Fsinθ−20)/2 upward. At the exact zero-normal boundary, the instantaneous contact forces vanish. A negative number retained in μN would otherwise manufacture friction in the wrong direction.

Interpret evidence and limits

For F=15 N and θ=30°, upward applied force is 7.5 N, normal is 12.5 N and friction is 3.75 N. Horizontal acceleration is about 4.62019 m/s². At F=30 N and θ=90°, the object instead has vertical acceleration 5 m/s² and no floor friction. No real lifting, traction or equipment rating follows from this ideal point model.

Worked original packet

Bounded investigation

Changed-case numerical practice

Try the changed case before feedback

For a 2 kg object, g=10, a 10 N pull at 30° above horizontal and μk=0.2, find kinetic friction magnitude in newtons while rightward slip persists.

Numerical result in the requested units

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and boundaries

Original prose, data and diagrams checked 30 October 2026 against OpenStax College Physics 2e. This is an algebra-based virtual evidence exercise, not a physical experiment, equipment certification, transport recommendation or independent mastery result. Save assumptions and a competing explanation with the course evidence; supervised practical requirements remain external.

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Public practice checks a specific claim. It does not establish independent mastery.

Why are algebraic force components not additional interactions to add to the original force?

Your explanation and limitations

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Previous: Use system boundaries without inventing a pulling contact Next: Use an inequality before assigning a friction value

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