How to Find Static Friction When a Block Stays at Rest

A toolbox sits motionless on a workshop floor. Someone pushes it 28 N to the right, while a cord pulls it 19 N to the left. How much friction is acting: 28 N, 19 N, or the difference between them?

The friction force is 9 N to the left. It balances the remaining 9 N of rightward force. When a block stays at rest, you find static friction by balancing the other forces parallel to the surface, then checking that the required friction does not exceed its maximum possible value.

01Find the friction required for equilibrium

If the toolbox remains stationary, its acceleration is zero. Newton’s second law tells us that the resultant force must also be zero.

Choose right as positive. The push contributes \(+28\ \mathrm{N}\), and the cord contributes \(-19\ \mathrm{N}\).

Before including friction, the net horizontal force is:

\[ 28-19=9\ \mathrm{N} \]

That is 9 N to the right.

For the toolbox to remain at rest, static friction must provide exactly 9 N to the left.

Writing \(f_s\) for the magnitude of static friction:

\[ \begin{aligned} \sum F_x &= 0\\ 28-19-f_s &= 0\\ f_s &= 9\ \mathrm{N} \end{aligned} \]

This is the key method: find the unbalanced force from everything except friction, then use friction to cancel it.

Why not calculate friction using the coefficient straight away?

You may have learnt the equation \(f=\mu F_N\). Applied without checking the situation, it can give the wrong answer.

Static friction is adjustable. The contact between the floor and the toolbox supplies as much friction as is needed to prevent sliding, up to a limit.

The correct relationship is:

\[ 0\leq f_s\leq\mu_s F_N \]

where:

• \(f_s\) is the actual static friction magnitude, measured in newtons (N).
• \(\mu_s\) is the coefficient of static friction, with no units.
• \(F_N\) is the normal contact force, perpendicular to the surface, measured in newtons (N).

The limiting value is:

\[ f_{s,\max}=\mu_s F_N \]

The equality applies at the threshold of slipping. It does not apply automatically whenever the block is stationary.

Imagine the toolbox could support up to 35 N of static friction. It would still experience only 9 N in our example, because 9 N is all that equilibrium requires.

That 35 N limit tells us what the contact could provide, not what it is providing.

02How could a spring-scale experiment demonstrate this?

The interesting measurement problem is that friction is difficult to measure directly. A spring scale measures the tension pulling on the block, not friction itself.

But while the block is stationary, we can infer friction from the force balance.

The first design: pull and watch for movement

Place a wooden block on a horizontal bench. Attach a string to one side of the block, connect the string to a spring scale, and pull slowly.

The string should ideally be horizontal. A ruler beside the block provides a fixed reference for detecting its first movement.

What would the scale show before the block moves? Would it immediately reach the maximum friction force?

No. As the scale reading increases, the opposing static friction increases by the same amount, provided the block remains stationary and the pull is the only other horizontal force.

Suppose we use the following illustrative model:

• Block mass: \(5.0\ \mathrm{kg}\)
• Gravitational acceleration: \(g=9.8\ \mathrm{m\,s^{-2}}\)
• Static friction coefficient: \(\mu_s=0.30\)
• Horizontal pulling force
• No other horizontal forces

Since the block has no vertical acceleration and the pull has no vertical component:

\[ \begin{aligned} F_N&=mg\\ &=5.0(9.8)\\ &=49.0\ \mathrm{N} \end{aligned} \]

The maximum static friction is:

\[ \begin{aligned} f_{s,\max}&=\mu_s F_N\\ &=0.30(49.0)\\ &=14.7\ \mathrm{N} \end{aligned} \]

Here is what the idealised model predicts as the horizontal pull increases.

These are calculated predictions, not recorded experimental readings.

The distinction in the final row is important. The contact cannot produce 15 N of static friction because its maximum is only 14.7 N. We must stop treating the block as stationary.

Why the original measurement can be misleading

A common practical problem is pulling the spring scale upwards at an angle instead of horizontally.

Suppose the string angles upwards from the block towards your hand.

The scale reading is now the total tension, but only part of that tension acts horizontally. The upward part also reduces the normal force.

Both effects change the reading at which slipping begins.

A better controlled test keeps the string horizontal, increases the pulling force gradually, and uses the fixed ruler to identify the first displacement. Repeating the measurement helps estimate a reproducible threshold rather than relying on one jerky pull.

If the scale must be angled, measure that angle and calculate its horizontal and vertical components.

There is another limitation. The greatest scale reading just before movement estimates the limiting static friction. A reading taken after the block starts accelerating may include effects of the motion and cannot automatically be treated as either the static-friction limit or kinetic friction.

For an ideal horizontal pull, measuring the force at impending slip gives:

\[ \mu_s=\frac{f_{s,\max}}{F_N} \]

This experiment explains why the coefficient is useful: it characterises a limiting contact force, not a friction force that acts constantly.

03A reliable method when the block stays at rest

Before substituting into a friction equation, work through these decisions.

Use a positive direction, such as right on a horizontal floor or uphill on an incline. A negative force result means the actual direction is opposite your chosen positive direction.

A useful warning: the magnitude and direction of static friction must come from the force balance. You cannot determine them from \(\mu_s\) and \(F_N\) alone.

04What changes when a pull is angled?

On a level floor, students often use \(F_N=mg\). That works only when no other vertical forces act and there is no vertical acceleration.

Suppose you pull a crate forwards and slightly upwards.

The upwards component of your pull supports some of the crate’s weight. The floor therefore exerts a smaller normal force.

That also reduces the maximum available static friction.

If the pulling force has magnitude \(P\) and acts at angle \(\theta\) above the horizontal:

\[ P_x=P\cos\theta \]

\[ P_y=P\sin\theta \]

The horizontal component \(P_x\) contributes to the force trying to slide the crate.

The vertical component \(P_y\) reduces the normal force:

\[ F_N=mg-P\sin\theta \]

This relationship assumes the crate remains in contact with the floor and has no vertical acceleration.

05Questions and solutions

Use \(g=9.8\ \mathrm{m\,s^{-2}}\) throughout. Unless otherwise stated, treat the blocks as rigid, the surfaces as fixed, the strings as light, and the friction coefficients as constant. Ignore air resistance.

The questions progress from a straightforward force balance to situations where friction changes direction, two surfaces impose separate limits, and even a block that stays fixed relative to a surface can be accelerating.

Question 1

A \(6.0\ \mathrm{kg}\) box rests on a horizontal floor. A horizontal force of 11 N acts to the right. The coefficient of static friction is 0.40.

Calculate:

(a) The normal force.

(b) The maximum possible static friction.

(c) The actual friction force and its direction, assuming the box was initially at rest.

Solution 1

The normal force is 58.8 N, the maximum static friction is 23.5 N, and the actual friction is 11 N to the left. The box can remain stationary because the required friction is smaller than the maximum.

For part (a), there are no other vertical forces:

\[ \begin{aligned} F_N&=mg\\ &=6.0(9.8)\\ &=58.8\ \mathrm{N} \end{aligned} \]

For part (b):

\[ \begin{aligned} f_{s,\max}&=\mu_sF_N\\ &=0.40(58.8)\\ &=23.52\ \mathrm{N} \end{aligned} \]

So the limiting static friction is approximately 23.5 N.

For part (c), only 11 N of friction is needed to balance the horizontal push.

\[ f_s=11\ \mathrm{N} \]

It acts to the left.

The tempting mistake is to report 23.5 N as the actual friction. That would produce an unbalanced force to the left, contradicting the stated stationary condition.

Question 2

A \(5.0\ \mathrm{kg}\) block sits on a plane inclined at \(20^\circ\) to the horizontal, rising towards the right. An adjustable cable pulls the block directly up the slope with tension \(T\).

The coefficients of friction are \(\mu_s=0.25\) and \(\mu_k=0.18\).

(a) Calculate the complete range of cable tensions for which the block can remain stationary.

(b) State the direction of static friction when \(T=12\ \mathrm{N}\) and when \(T=20\ \mathrm{N}\).

(c) If the cable tension is instead set to 4.0 N while the block is initially stationary, determine the direction of slipping and its acceleration immediately after sliding starts.

Solution 2

The block can remain stationary for cable tensions from approximately 5.25 N to 28.3 N, inclusive. Friction acts uphill at 12 N and downhill at 20 N. At a tension of 4.0 N, the block slips downhill, with an initial acceleration of approximately \(0.89\ \mathrm{m\,s^{-2}}\) under the kinetic-friction model.

Take uphill as positive.

For part (a), the component of the block’s weight downhill is:

\[ \begin{aligned} mg\sin20^\circ &=5.0(9.8)\sin20^\circ\\ &=16.76\ \mathrm{N} \end{aligned} \]

The normal force is:

\[ \begin{aligned} F_N&=mg\cos20^\circ\\ &=46.05\ \mathrm{N} \end{aligned} \]

Therefore:

\[ \begin{aligned} f_{s,\max}&=0.25(46.05)\\ &=11.51\ \mathrm{N} \end{aligned} \]

For equilibrium, friction must make up the difference between the cable tension and the downhill weight component.

The required friction magnitude is:

\[ f_s=|16.76-T| \]

The static-friction condition is:

\[ |16.76-T|\leq11.51 \]

Hence:

\[ 5.25\leq T\leq28.27 \]

All tensions are in newtons.

At the endpoints, static friction is at its limiting magnitude.

For part (b), at \(T=12\ \mathrm{N}\), the downhill gravitational component exceeds the uphill tension.

Friction must act uphill:

\[ f_s=16.76-12=4.76\ \mathrm{N} \]

At \(T=20\ \mathrm{N}\), the cable tension exceeds the downhill gravitational component.

Friction now acts downhill:

\[ f_s=20-16.76=3.24\ \mathrm{N} \]

This reversal is the key idea. Static friction opposes the tendency to slip relative to the surface, not necessarily the direction of the cable.

For part (c), a 4.0 N tension leaves a downhill tendency of:

\[ 16.76-4.0=12.76\ \mathrm{N} \]

That exceeds the maximum available static friction of 11.51 N, so the block starts slipping downhill.

Once it slides, kinetic friction acts uphill:

\[ \begin{aligned} f_k&=\mu_kF_N\\ &=0.18(46.05)\\ &=8.29\ \mathrm{N} \end{aligned} \]

The net downhill force is:

\[ \begin{aligned} F_{\mathrm{net}} &=16.76-4.0-8.29\\ &=4.47\ \mathrm{N} \end{aligned} \]

Therefore:

\[ \begin{aligned} a&=\frac{F_{\mathrm{net}}}{m}\\ &=\frac{4.47}{5.0}\\ &=0.89\ \mathrm{m\,s^{-2}} \end{aligned} \]

The block accelerates downhill.

A tempting shortcut is to assume friction must point uphill on any incline. That fails once the cable becomes strong enough to create a tendency to slip uphill.

Question 3

A student uses a spring scale to estimate the coefficient of static friction for a \(6.0\ \mathrm{kg}\) wooden block resting on a horizontal laboratory bench.

The scale pulls the block at \(30^\circ\) above the horizontal. The block is initially stationary, and the scale reads 24 N at the point when slipping is about to begin.

(a) Estimate the coefficient of static friction.

(b) The student then repeats the experiment with the scale held horizontally. Predict the pulling force required to reach the slipping threshold, assuming the same coefficient applies.

(c) Another student argues that a horizontal pull should need only \(24\cos30^\circ\) N because all of the force now points forwards. Explain why this prediction fails.

(d) Can the original measurement determine the kinetic-friction coefficient?

Solution 3

The estimated static-friction coefficient is 0.444, and the predicted horizontal pulling threshold is approximately 26.1 N. The horizontal test requires a larger scale reading because the original angled pull partly lifted the block, reducing its normal force and friction limit. The measurement does not determine kinetic friction.

For part (a), at impending slip, the horizontal component of the tension equals the maximum static friction.

\[ \begin{aligned} f_{s,\max}&=24\cos30^\circ\\ &=20.78\ \mathrm{N} \end{aligned} \]

However, the normal force is not simply the weight.

\[ \begin{aligned} F_N&=mg-24\sin30^\circ\\ &=58.8-12.0\\ &=46.8\ \mathrm{N} \end{aligned} \]

Therefore:

\[ \begin{aligned} \mu_s&=\frac{f_{s,\max}}{F_N}\\ &=\frac{20.78}{46.8}\\ &\approx0.444 \end{aligned} \]

For part (b), when the scale is horizontal, it supplies no upward component. The normal force returns to:

\[ F_N=mg=58.8\ \mathrm{N} \]

The predicted horizontal pulling threshold is:

\[ \begin{aligned} P_{\mathrm{limit}} &=\mu_sF_N\\ &=0.444(58.8)\\ &\approx26.1\ \mathrm{N} \end{aligned} \]

For part (c), the second student’s calculation of \(24\cos30^\circ=20.8\ \mathrm{N}\) identifies the horizontal force at the first threshold, but ignores the changed contact conditions.

In the angled test, the upward force reduces the normal force from 58.8 N to 46.8 N. In the horizontal test, the entire weight is supported by the bench, making the limiting friction larger.

The friction coefficient stays the same under the stated assumptions, but the normal force does not.

For part (d), the answer is no. The 24 N scale reading was taken at impending slip, so it provides information about limiting static friction.

It does not establish the friction force during sliding. To estimate \(\mu_k\), the student would need suitable measurements of the block’s motion and applied forces while it was sliding, or a controlled constant-speed sliding measurement.

The experimental trap is treating one scale reading as if it describes every friction regime.

Question 4

Two blocks are stacked on a horizontal floor.

The upper block A has mass \(2.0\ \mathrm{kg}\). The lower block B has mass \(5.0\ \mathrm{kg}\).

The coefficient of static friction between A and B is 0.45. Between B and the floor, it is 0.20.

A horizontal force \(P=8.0\ \mathrm{N}\) pushes A to the right. Separately, a horizontal cord pulls B to the left with tension \(T=18\ \mathrm{N}\). Both blocks are initially stationary.

(a) Find the magnitude and direction of the friction force on A.

(b) Find the magnitude and direction of the friction force exerted by the floor on B.

(c) Keeping \(P=8.0\ \mathrm{N}\), increase \(T\). Find the greatest tension for which both blocks can remain stationary.

(d) Instead, return to \(T=18\ \mathrm{N}\), but increase \(P\) to 10 N. Can both blocks remain stationary? Identify which contact condition fails.

Solution 4

Initially, friction on A is 8.0 N left, while floor friction on B is 10.0 N right. Both contacts can support these forces. The greatest allowable cable tension is 21.7 N. If the push on A instead increases to 10 N, the A-B contact cannot maintain static equilibrium, even though the floor contact could.

For part (a), consider A alone.

The 8.0 N push acts right. Its only other horizontal force is friction from B.

Since A is stationary, friction on A must be 8.0 N left.

Check the contact limit. The normal force between A and B is:

\[ F_{N,AB}=2.0(9.8)=19.6\ \mathrm{N} \]

So:

\[ \begin{aligned} f_{s,\max,AB}&=0.45(19.6)\\ &=8.82\ \mathrm{N} \end{aligned} \]

Since \(8.0<8.82\), this contact can remain static.

For part (b), Newton’s third law tells us that A exerts 8.0 N of friction to the right on B.

Block B therefore has an 18 N cord tension left and an 8.0 N friction force right from A.

To remain stationary, it needs an additional 10.0 N rightwards from the floor:

\[ 18-8.0=10.0\ \mathrm{N} \]

The floor supports the weight of both blocks:

\[ \begin{aligned} F_{N,\mathrm{floor}} &=(2.0+5.0)(9.8)\\ &=68.6\ \mathrm{N} \end{aligned} \]

Its limiting friction is:

\[ \begin{aligned} f_{s,\max,\mathrm{floor}} &=0.20(68.6)\\ &=13.72\ \mathrm{N} \end{aligned} \]

The required 10.0 N is within this limit.

For part (c), A still requires 8.0 N of friction to the left. That does not change as the cord tension on B increases.

On B, the floor must supply the remaining rightward force:

\[ f_{\mathrm{floor}}=T-8.0 \]

At the limiting floor friction:

\[ \begin{aligned} T_{\max}-8.0&=13.72\\ T_{\max}&=21.72\ \mathrm{N} \end{aligned} \]

Therefore, the maximum allowable tension is approximately 21.7 N.

For part (d), increasing \(P\) to 10 N requires 10 N of friction on A to the left.

But the upper contact can provide at most 8.82 N.

\[ 10>8.82 \]

So both blocks cannot remain stationary.

Interestingly, the floor would need only:

\[ 18-10=8.0\ \mathrm{N} \]

to balance B if the upper contact could supply the required 10 N. That is well within the floor’s limit of 13.72 N.

The failure occurs at the A-B contact, not the floor.

The trap is treating the stacked blocks as a single object and checking only the floor friction. A whole-system force balance can hide an internal contact that has already exceeded its limit.

Question 5

A small \(2.0\ \mathrm{kg}\) block is placed on a ramp fixed to a trolley.

The ramp slopes upwards towards the left at \(30^\circ\) to the horizontal. The trolley accelerates horizontally to the right at a constant acceleration \(a\), where \(a\geq0\).

The coefficient of static friction between the block and ramp is 0.30.

The block is initially at rest relative to the trolley. Assume the ramp remains rigid and horizontal acceleration is maintained, the block stays in contact with the ramp, and there are no other applied forces.

(a) Find the complete range of trolley accelerations for which the block can remain at rest relative to the ramp.

(b) Determine the magnitude and direction of static friction when \(a=3.0\ \mathrm{m\,s^{-2}}\) and when \(a=7.0\ \mathrm{m\,s^{-2}}\).

(c) Find the trolley acceleration at which no friction is required.

(d) Explain why too little acceleration and too much acceleration can both cause the block to slip, even though the normal force increases with the trolley’s acceleration.

Solution 5

The block can remain fixed relative to the ramp only when the trolley accelerates right between approximately \(2.32\) and \(10.40\ \mathrm{m\,s^{-2}}\), inclusive. Friction acts uphill at \(3.0\ \mathrm{m\,s^{-2}}\), downhill at \(7.0\ \mathrm{m\,s^{-2}}\), and becomes zero at approximately \(5.66\ \mathrm{m\,s^{-2}}\).

The critical insight is that a block fixed to an accelerating ramp is not in equilibrium in the ground frame. It accelerates horizontally with the trolley.

We must use Newton’s second law, \(\sum F=ma\), rather than setting the total force to zero.

Take uphill, towards the left, as the positive direction parallel to the ramp.

For part (a), resolve the block’s horizontal acceleration along the slope.

Because uphill points left while acceleration points right, the acceleration component along the positive slope direction is:

\[ a_{\parallel}=-a\cos30^\circ \]

Let \(f_{\parallel}\) be the signed static-friction force, positive uphill.

Applying Newton’s second law parallel to the ramp:

\[ f_{\parallel}-mg\sin30^\circ =-ma\cos30^\circ \]

Rearranging:

\[ f_{\parallel} =m(g\sin30^\circ-a\cos30^\circ) \]

This equation determines both how much friction is required and which way it acts.

Now consider the direction perpendicular to the ramp, positive away from the surface.

The rightward acceleration has a component \(a\sin30^\circ\) away from the ramp.

Therefore:

\[ F_N-mg\cos30^\circ =ma\sin30^\circ \]

So:

\[ F_N=m(g\cos30^\circ+a\sin30^\circ) \]

The static-friction condition is:

\[ |f_{\parallel}|\leq\mu_sF_N \]

Substituting both force expressions and cancelling the nonzero mass gives:

\[ |g\sin30^\circ-a\cos30^\circ| \leq 0.30(g\cos30^\circ+a\sin30^\circ) \]

Using \(g=9.8\ \mathrm{m\,s^{-2}}\):

\[ |4.90-0.866a| \leq2.55+0.150a \]

For the required friction to be within its limit, both sides of the absolute-value condition must be satisfied.

First:

\[ \begin{aligned} 4.90-0.866a&\leq2.55+0.150a\\ 2.35&\leq1.016a\\ a&\geq2.32 \end{aligned} \]

Second:

\[ \begin{aligned} 0.866a-4.90&\leq2.55+0.150a\\ 0.716a&\leq7.45\\ a&\leq10.40 \end{aligned} \]

Using unrounded values in the calculations, the range is:

\[ 2.32\leq a\leq10.40 \]

with \(a\) measured in \(\mathrm{m\,s^{-2}}\).

For part (b), at \(a=3.0\ \mathrm{m\,s^{-2}}\):

\[ \begin{aligned} f_{\parallel} &=2.0(4.90-0.866(3.0))\\ &\approx+4.60\ \mathrm{N} \end{aligned} \]

The positive result means 4.60 N uphill.

The corresponding limiting friction is approximately 5.99 N, so the contact can supply the required force.

At \(a=7.0\ \mathrm{m\,s^{-2}}\):

\[ \begin{aligned} f_{\parallel} &=2.0(4.90-0.866(7.0))\\ &\approx-2.12\ \mathrm{N} \end{aligned} \]

The negative result means 2.12 N downhill.

The limiting friction is approximately 7.19 N, again sufficient.

For part (c), no friction is required when \(f_{\parallel}=0\):

\[ g\sin30^\circ=a\cos30^\circ \]

Hence:

\[ \begin{aligned} a&=g\tan30^\circ\\ &=9.8\tan30^\circ\\ &=5.66\ \mathrm{m\,s^{-2}} \end{aligned} \]

At this acceleration, the components of gravity and the normal force alone produce the required rightward acceleration. Static friction is zero despite the rough contact.

For part (d), consider the two ends of the allowable range.

At low trolley accelerations, gravity tends to make the block slide downhill towards the right. Static friction must act uphill, but below \(2.32\ \mathrm{m\,s^{-2}}\), its maximum is insufficient.

As the trolley accelerates faster, the required uphill friction decreases. At \(5.66\ \mathrm{m\,s^{-2}}\), it reaches zero.

Above that acceleration, the block needs friction downhill. Although this seems strange, the surface is trying to carry the block rightwards so rapidly that the block would tend to move uphill-left relative to the trolley without enough downhill friction.

The normal force grows with acceleration, increasing the available static friction. But the downhill friction required also grows.

Beyond \(10.40\ \mathrm{m\,s^{-2}}\), the required friction grows too large for the contact to supply. The block tends to slip uphill relative to the ramp.

The tempting approach is to assume that because the block does not move relative to the trolley, all its forces must balance.

That would be true only in a non-accelerating reference frame.

Here, the block shares the trolley’s horizontal acceleration. The resultant force cannot be zero, and the direction of static friction depends on the acceleration required.

This also shows why the static-friction inequality is more useful than automatically setting \(f_s=\mu_sF_N\). It captures an entire range of possible stationary-relative-to-surface states, including one in which no friction acts at all.

06The distinction to carry into your next forces problem

When a block stays at rest on a fixed surface, friction is the force required to make the resultant force zero, provided it lies within the available static-friction limit.

When the block stays fixed relative to an accelerating surface, it may still have acceleration in the ground frame. Find its required acceleration first, then use Newton’s second law to determine the friction force.

In both cases, the method is the same in spirit: calculate what friction must do, then check whether the contact can do it.

For a broader explanation of how the static and sliding models relate, see Static and Kinetic Friction Explained for HSC Physics.

The next useful skill is drawing a free-body diagram before choosing an equation. Once the forces, contact surfaces, and acceleration are correctly identified, the friction calculation is usually the easy part.