Friction: Static and Sliding Force Evidence

Friction: read the force before and after sliding starts

A block can remain still under several different pulling forces. That does not mean the forces have disappeared. It means the contact force is adjusting to balance the pull. The useful question is therefore not simply “What is the friction?” but “Are the surfaces still together, about to slip, or already sliding?” This lesson uses fictional supplied observations, so there is no equipment test to carry out.

Choose the object and state the model

Consider one 5.0 kg block on a horizontal surface. A pull acts horizontally to the right; the only vertical forces are its weight and the upward normal contact force. Use g = 10 N/kg. There is no vertical acceleration, so the normal force N equals mg = 5.0 × 10 = 50 N. The symbol N can denote the normal force, while N written after a number is the unit newton; context distinguishes them.

Draw only forces acting on the block. The force the block exerts on the table belongs on a different object’s diagram. For this dry-contact model, static friction has magnitude f_s ≤ μ_sN. Its maximum is μ_sN, reached at limiting rest. During sliding, approximate the magnitude by f_k = μ_kN. These are useful models rather than exact rules for every contact. The coefficients describe the pair of surfaces under stated conditions, not one material in isolation. OpenStax explains the static inequality and kinetic approximation.

Original schematic. A stationary block has a 12 N pull right and 12 N static friction left. At limiting rest these forces are each 20 N. In steady sliding they are each 15 N. In all three diagrams the upward normal force and downward weight are each 50 N. Vertical arrow lengths are schematic; horizontal arrows use one common scale.

Read a sequence without confusing state and force

The following invented record describes a gradually increased pull, followed by a separate interval of steady sliding. Force readings are rounded to 0.5 N. The observer also records motion, because a force number alone does not identify whether friction is static or sliding.

Fictional observation sequence for the same block and surfaces
Stage	Pull	Observed state	Inference about friction
1	0 N	Stationary	0 N in this horizontal model
2	6.0 N	Stationary	6.0 N left
3	12.0 N	Stationary	12.0 N left
4	19.5 N	Stationary	19.5 N left
5	20.0 N	Last stationary reading; slip follows	Near the limiting value
6	15.0 N	Already sliding at constant speed	15.0 N left

Three fictional runs give peak readings of 20.0, 19.5 and 20.5 N, and steady-sliding readings of 15.0, 15.5 and 14.5 N. Their means are 20.0 N and 15.0 N. The spread supports approximate values, not a perfectly repeatable threshold of exactly 20.000 N. Instrument resolution, small changes in the contact and judging the onset of slip can all matter. Repetition helps reveal variability; it cannot remove a shared calibration error.

Worked case 1: a small pull does not demand maximum friction

For calculation, take μ_s = 0.40 and μ_k = 0.30. The available maximum static friction is 0.40 × 50 = 20 N. At a 12 N pull, the block can remain stationary because only 12 N of opposing friction is needed. Thus the horizontal resultant is 12 − 12 = 0 N. Writing f = 20 N here would predict an 8 N resultant to the left, contradicting the stated stationary condition.

The inequality is a capacity check: first find the friction required for rest, then compare it with the available limit. A stationary block does not automatically use all the friction available to it.

Worked case 2: the limiting state

At limiting rest in the idealised calculation, the pull is 20 N and static friction is 20 N in the opposite direction. The forces still balance at that boundary. A larger pull cannot be balanced by static friction in this model. The phrase “just about to slide” describes the last static state; “has started sliding” calls for the sliding model instead. The supplied experimental readings identify the boundary only approximately.

Friction can be useful when the aim is to keep a package in position against a small sideways disturbance. The same resistance becomes a limitation when the aim is to slide the package. “Useful” describes the intended task; it is not another type of friction.

Worked case 3: steady sliding and speeding up

The approximate sliding friction is 0.30 × 50 = 15 N. A 15 N pull can maintain constant speed once the block is sliding, because 15 − 15 = 0 N. Constant speed does not mean that the individual forces vanish. If the pull rises to 24 N while it slides right, the resultant is 24 − 15 = 9 N right. The acceleration is F_net/m = 9/5.0 = 1.8 m/s² right.

Notice the role of history: a 15 N pull applied to a block initially at rest would be balanced by 15 N static friction. The same pull on this already sliding block balances 15 N sliding friction. The numerical equality does not make the two contact states identical.

Three practice tasks with complete solutions

Task: The original block remains still under a 9 N pull. Find friction and the horizontal resultant.

Solution: Required friction is 9 N left, below the 20 N limit. The resultant is 9 − 9 = 0 N. Using μ_kN would be inappropriate because the contact is not sliding.
Task: The block slides right while the pull is reduced to 11 N. Find its acceleration at that instant.

Solution: Sliding friction remains approximately 15 N left. Taking right as positive, F_net = 11 − 15 = −4 N and a = −4/5.0 = −0.8 m/s². It slows while moving right. Once it stops, reassess static friction; do not extend the sliding equation indefinitely.
Task: Another horizontal block has normal force 80 N and measured limiting friction 28 N. Find μ_s. What is its static friction under a 17 N pull?

Solution: μ_s = 28/80 = 0.35, with no unit. Since 17 N is below the 28 N limit, friction is 17 N opposite the pull, provided the block is initially at rest and no other horizontal forces act.

Frequently asked questions

Does friction always equal μN?: No. Static friction can be smaller than μ_sN. Equality gives its maximum; the sliding approximation uses μ_kN.
Is normal force always equal to weight?: No. It is equal here because the surface is horizontal, the pull has no vertical component and vertical acceleration is zero.
Does friction always oppose the object’s overall motion?: It opposes relative sliding, or the tendency to slide, at the contact. That is the direction test to use.
Can there be zero friction on a rough surface?: Yes. With no horizontal pull or tendency to slide in this model, no horizontal friction is needed.
Does a rougher-looking surface guarantee greater friction?: No. Materials, surface condition and the contact model matter. Appearance alone does not determine the coefficient.
Why is a steady-sliding reading useful?: Constant velocity implies zero resultant, so the measured horizontal pull balances sliding friction under the stated assumptions.
What should a good conclusion include?: State the motion evidence, infer the force balance, give direction and units, and acknowledge the resolution and variability of the supplied record.