Why Friction Is Not Always μmg: HSC Physics Explained

A 2.0 kg box sits on a horizontal floor. You pull it to the right with a force of 6.0 N, but it doesn’t move.

Suppose the coefficient of static friction is 0.45. Since the normal force is 19.6 N, calculating \(\mu_s mg\) gives:

\[
f_{s,\max}=0.45\times 2.0\times 9.8
=8.82\text{ N}
\]

So is the friction force 8.82 N?

No. The actual friction force is 6.0 N, acting to the left. The value 8.82 N is the maximum static friction available, not the force automatically acting on the box.

This is why friction cannot always be calculated using \(\mu mg\). You need to know whether the object is stationary or sliding, and you need to calculate the normal force correctly before applying a friction model.

01Why the formula \(f=\mu mg\) can give the wrong answer

There are two different friction models to understand.

1. Static friction adjusts to the situation

Static friction acts between surfaces that are not sliding relative to each other.

Imagine pushing a stationary box horizontally. If you push with 2 N, the box stays still. Friction provides an opposing force of 2 N.

Increase your push to 4 N. If the box still doesn’t move, friction increases to 4 N.

It adjusts to match the force trying to make the surfaces slide, provided it has enough capacity to do so.

The maximum possible static friction is:

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

However, the actual static friction must satisfy:

\[
\boxed{0\leq f_s\leq\mu_s N}
\]

Here:

• \(f_s\) is the magnitude of the actual static friction force, measured in newtons (N).
• \(\mu_s\) is the coefficient of static friction, which has no units.
• \(N\) is the normal force, measured in newtons (N).

The important distinction is that \(\mu_s N\) gives an upper limit. It does not automatically give the actual friction force.

For the original 2.0 kg box, suppose the horizontal push is 6.0 N.

The maximum static friction is 8.82 N, so the box can remain stationary. Its actual static friction is 6.0 N.

If the push increases to 8.82 N, static friction reaches its maximum. Any further increase causes the box to start sliding under the assumed model.

This is the distinction explored in more detail in Why Static Friction Isn’t Always at Its Maximum.

2. Kinetic friction uses a different model

Once the surfaces are sliding relative to each other, kinetic friction acts.

For the standard HSC dry-sliding model, its magnitude is approximated by:

\[
\boxed{f_k=\mu_k N}
\]

Here:

• \(f_k\) is the kinetic friction force.
• \(\mu_k\) is the coefficient of kinetic friction.
• \(N\) is the normal force.

Unlike static friction, kinetic friction is modelled as having a magnitude proportional to the normal force while sliding occurs.

The coefficient of kinetic friction is often smaller than the coefficient of static friction:

\[
\mu_k<\mu_s
\]

This means it often takes more force to start an object moving than to keep it sliding.

For example, suppose a box has:

\[
\mu_s=0.50,\qquad \mu_k=0.40
\]

If its normal force is 20 N, then:

\[
f_{s,\max}=0.50\times20=10\text{ N}
\]

But once sliding:

\[
f_k=0.40\times20=8\text{ N}
\]

The maximum static friction and kinetic friction are different quantities.

One important limitation: \(f_k=\mu_kN\) is an idealised friction model. Real friction can depend on surface condition, lubrication, temperature, and sliding speed. For standard HSC problems, use the stated coefficients and the supplied model assumptions.

02Why the normal force is not always equal to the weight

Even when you choose the correct friction equation, there is another possible mistake.

Students often replace \(N\) with \(mg\) without checking the forces acting perpendicular to the surface.

The normal force is the contact force exerted perpendicular to the surface. It is not automatically equal to an object’s weight.

For a box on a horizontal floor, with no other vertical forces and no vertical acceleration:

\[
N=mg
\]

This is why the expression \(\mu mg\) sometimes works.

But change the direction of an applied force, and the normal force changes too.

Pulling upwards at an angle

Suppose you pull a box using a rope angled upwards.

The rope has an upward component that partially supports the box. The floor therefore needs to provide less normal force.

If the pulling force is \(P\), at an angle \(\theta\) above the horizontal:

\[
N+P\sin\theta-mg=0
\]

Rearranging:

\[
\boxed{N=mg-P\sin\theta}
\]

For a sliding box:

\[
f_k=\mu_k(mg-P\sin\theta)
\]

Using \(\mu_kmg\) would overestimate the friction force.

Pushing downwards at an angle

Now imagine pushing the same box downwards using an angled force.

The downward component adds to the weight, so the floor must provide a greater normal force.

The vertical equilibrium equation becomes:

\[
N-mg-P\sin\theta=0
\]

Therefore:

\[
\boxed{N=mg+P\sin\theta}
\]

For kinetic friction:

\[
f_k=\mu_k(mg+P\sin\theta)
\]

The friction force is now greater than it would be if you pushed horizontally with the same force magnitude.

What about an inclined plane?

For a block on an inclined plane, the weight has a component perpendicular to the surface.

If there are no other forces with components perpendicular to the plane, the normal force is:

\[
\boxed{N=mg\cos\theta}
\]

Consequently, the maximum static friction is:

\[
f_{s,\max}=\mu_smg\cos\theta
\]

And the kinetic friction model becomes:

\[
f_k=\mu_kmg\cos\theta
\]

Notice that neither expression is simply \(\mu mg\).

For a more detailed explanation of resolving forces on ramps, see Forces on Inclined Planes: Components, Normal Force and Friction.

03How an experiment demonstrates the difference

How could you experimentally determine whether friction is always equal to \(\mu mg\)?

A useful investigation is to measure the force required to pull a block across a horizontal bench.

Setting up the investigation

Place a block on a level bench. Connect it to a force sensor using a light string, with the string kept horizontal.

Connect the force sensor to a data logger so that the applied force can be recorded as it changes.

You can also place additional masses on the block to investigate how friction changes with normal force.

What should happen?

Start pulling very gently.

While the block remains stationary, the horizontal net force must be zero. Therefore, the static friction force balances the tension in the string.

As you gradually increase the pull, static friction increases too.

Eventually, the block is just about to slide. At this point, static friction reaches its maximum value.

Once the block starts sliding, the friction model changes to kinetic friction.

If you maintain approximately constant speed, the force sensor reading gives an estimate of the kinetic friction because the net horizontal force is approximately zero.

Illustrative experimental data

The following values are illustrative, not measurements from a specific experiment. Assume the string remains horizontal and the block is pulled at constant speed during the sliding measurements.

The normal force is calculated using \(N=mg\), since the string is horizontal and there is no vertical acceleration.

Consider the first trial.

The maximum static friction is 4.9 N, so:

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

\[
\mu_s=\frac{4.9}{9.8}=0.50
\]

For kinetic friction:

\[
\mu_k=\frac{f_k}{N}
\]

\[
\mu_k=\frac{3.9}{9.8}\approx0.40
\]

The data are consistent with the model that maximum static friction and kinetic friction increase approximately in proportion to normal force.

But here’s the important observation.

Suppose the 1.00 kg block is stationary while the force sensor reads 2.0 N.

The actual static friction is 2.0 N, not 4.9 N.

The experiment distinguishes three different measurements:

• The friction force while stationary below the threshold.
• The maximum static friction immediately before sliding.
• The kinetic friction while sliding.

What the original measurement cannot tell you

If you only record the largest force before the block starts moving, you can estimate the coefficient of static friction.

However, you cannot use that measurement to claim that static friction always equals this maximum value.

A better experimental design records the force continuously throughout the pull. This reveals how static friction adjusts before the block moves and how the measured force changes once sliding begins.

There is another limitation. The horizontal arrangement makes \(N=mg\) a valid assumption, but it cannot demonstrate what happens when the applied force has a vertical component.

To investigate that, repeat the experiment with an adjustable pulling angle and measure the normal force independently, or calculate it using vertical force equilibrium.

The important conclusion is that the friction model and the normal force must both be established before calculating friction.

04A reliable method for choosing the correct friction equation

When you encounter a friction problem in an HSC Physics exam, follow this sequence.

Then follow these steps:

1. Identify whether the surfaces are stationary relative to each other or sliding.
2. Draw a free-body diagram.
3. Calculate the normal force using Newton’s second law perpendicular to the surface.
4. If the object is stationary, calculate the friction needed for equilibrium.
5. Check whether the required static friction is less than or equal to \(\mu_sN\).
6. If the object is sliding, apply the kinetic friction model.

The key decision is not simply whether you know the coefficient of friction. It’s whether the object is sliding and what normal force is acting at the contact.

05Worked examples

06Questions and solutions

The following questions build from identifying actual static friction to combining normal-force changes, inclined planes, and multiple objects.

Question 1

A 3.0 kg block rests on a horizontal floor. A horizontal force of 8.0 N is applied to the right.

The coefficient of static friction between the block and floor is 0.40.

Determine:

(a) The normal force acting on the block.

(b) The maximum possible static friction.

(c) Whether the block moves and the magnitude and direction of the actual friction force.

Take \(g=9.8\text{ m/s}^2\).

Solution 1

The block remains stationary, and the actual static friction is 8.0 N to the left.

(a) Normal force

There are no other vertical forces, so:

\[
N=mg
\]

\[
N=3.0\times9.8
\]

\[
\boxed{N=29.4\text{ N}}
\]

(b) Maximum static friction

\[
f_{s,\max}=\mu_sN
\]

\[
f_{s,\max}=0.40\times29.4
\]

\[
\boxed{f_{s,\max}=11.76\text{ N}}
\]

(c) Actual friction

Compare the applied force with the maximum static friction:

\[
8.0<11.76
\]

Static friction can therefore prevent the block from moving.

Since the block remains stationary, its horizontal net force must be zero.

\[
f_s=8.0\text{ N}
\]

The direction is opposite to the applied force.

\[
\boxed{f_s=8.0\text{ N to the left}}
\]

The trap is assuming that friction must equal 11.76 N simply because that is the maximum available value. The block only requires 8.0 N of friction to remain stationary.

Question 2

A student investigates friction using a block on a horizontal laboratory bench. The block is connected to a force sensor by a horizontal string.

The student records the following illustrative measurements:

Assume the string remains horizontal, the surfaces are unchanged between trials, and \(g=9.8\text{ m/s}^2\).

(a) Calculate the coefficient of static friction using Trial 1.

(b) Calculate the coefficient of kinetic friction using Trial 1.

(c) Repeat both calculations using Trial 2. Explain what the results suggest about the relationship between friction and normal force.

(d) During Trial 1, the student also applies a horizontal force of 3.0 N while the block remains stationary. Calculate the actual static friction. Explain why it is not 9.0 N.

Solution 2

The coefficients are approximately \(\mu_s=0.46\) and \(\mu_k=0.37\), while the actual static friction under the 3.0 N pull is 3.0 N.

(a) Coefficient of static friction in Trial 1

First, calculate the normal force:

\[
N=mg
\]

\[
N=2.0\times9.8=19.6\text{ N}
\]

The peak force immediately before sliding estimates the maximum static friction.

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

\[
\mu_s=\frac{9.0}{19.6}
\]

\[
\boxed{\mu_s\approx0.46}
\]

(b) Coefficient of kinetic friction in Trial 1

The block is sliding at constant speed, so its horizontal net force is zero.

The force sensor reading therefore equals the kinetic friction magnitude.

\[
\mu_k=\frac{f_k}{N}
\]

\[
\mu_k=\frac{7.2}{19.6}
\]

\[
\boxed{\mu_k\approx0.37}
\]

(c) Coefficients in Trial 2

Calculate the normal force:

\[
N=3.0\times9.8=29.4\text{ N}
\]

Static friction coefficient:

\[
\mu_s=\frac{13.5}{29.4}
\]

\[
\boxed{\mu_s\approx0.46}
\]

Kinetic friction coefficient:

\[
\mu_k=\frac{10.8}{29.4}
\]

\[
\boxed{\mu_k\approx0.37}
\]

Both coefficients are approximately unchanged between trials.

This supports the model that maximum static friction and kinetic friction are proportional to normal force for these surfaces under the experimental conditions.

It does not mean that every stationary friction measurement must equal \(\mu_sN\).

(d) Actual static friction under a 3.0 N pull

The block remains stationary, so its horizontal net force must be zero.

Therefore:

\[
\boxed{f_s=3.0\text{ N}}
\]

The 9.0 N measurement represents the maximum static friction from Trial 1, not the friction acting throughout the trial.

The student must distinguish between a measured threshold and an ordinary stationary reading.

Question 3

A 4.0 kg block rests on a rough inclined plane at an angle of 25° to the horizontal.

A rope runs parallel to the incline and pulls the block uphill.

The coefficients of friction are:

\[
\mu_s=0.45,\qquad \mu_k=0.30
\]

Initially, the rope tension is 12 N.

(a) Determine the magnitude and direction of static friction. Explain whether the block moves.

(b) The rope tension is increased to 35 N. Determine whether the block begins sliding.

(c) If it begins sliding, calculate its initial acceleration.

Assume the rope remains parallel to the incline, the coefficients remain constant, and \(g=9.8\text{ m/s}^2\).

Solution 3

Initially, static friction acts uphill with a magnitude of 4.57 N, keeping the block stationary. When the tension increases to 35 N, the block slides uphill with an acceleration of approximately 1.94 m/s².

(a) Initial static friction

Resolve the weight into components parallel and perpendicular to the incline.

The component parallel to the incline is:

\[
mg\sin25^\circ
\]

\[
=4.0\times9.8\times\sin25^\circ
\]

\[
=16.57\text{ N}
\]

This component acts downhill.

The normal force is:

\[
N=mg\cos25^\circ
\]

\[
N=4.0\times9.8\times\cos25^\circ
\]

\[
\boxed{N=35.53\text{ N}}
\]

Calculate the maximum static friction:

\[
f_{s,\max}=\mu_sN
\]

\[
f_{s,\max}=0.45\times35.53
\]

\[
f_{s,\max}=15.99\text{ N}
\]

Now consider the forces parallel to the incline.

The weight component acts downhill with 16.57 N, while the rope pulls uphill with 12 N.

Without friction, the block would tend to slide downhill.

Therefore, static friction must act uphill.

For equilibrium:

\[
f_s+12-16.57=0
\]

\[
\boxed{f_s=4.57\text{ N}}
\]

Since:

\[
4.57<15.99
\]

the available static friction is sufficient.

The block remains stationary.

(b) Increasing the tension to 35 N

The rope now pulls uphill with 35 N.

The downhill component of gravity is still 16.57 N.

Without friction, the resultant force would be:

\[
F_{\text{resultant}}=35-16.57
\]

\[
F_{\text{resultant}}=18.43\text{ N uphill}
\]

To keep the block stationary, static friction would need to act downhill with a magnitude of 18.43 N.

However:

\[
18.43>15.99
\]

The required static friction exceeds its maximum possible value.

Therefore, the block begins sliding uphill.

(c) Initial acceleration

Once the block slides, use the kinetic friction model.

\[
f_k=\mu_kN
\]

\[
f_k=0.30\times35.53
\]

\[
f_k=10.66\text{ N}
\]

Kinetic friction acts downhill because the block is sliding uphill.

Take uphill as positive.

Newton’s second law gives:

\[
F_{\text{net}}=ma
\]

\[
35-16.57-10.66=4.0a
\]

\[
7.77=4.0a
\]

\[
\boxed{a=1.94\text{ m/s}^2}
\]

The block accelerates uphill.

The interesting result is that static friction changes direction between the two situations.

At 12 N, gravity is stronger than the rope’s uphill pull, so friction acts uphill.

At 35 N, the rope is strong enough to make the block slide uphill, so kinetic friction acts downhill.

Friction opposes the tendency for relative slipping, or the actual relative sliding. It does not always oppose gravity or the applied force.

Question 4

A 10 kg crate rests on a horizontal floor. A person pushes it with a force \(P\), directed 30° below the horizontal.

The coefficients of friction are:

\[
\mu_s=0.40,\qquad \mu_k=0.30
\]

Initially, the applied force is 50 N.

(a) Calculate the normal force.

(b) Determine whether the crate moves and calculate its actual friction force.

(c) The applied force increases to 60 N. Determine whether the crate now slides.

(d) Calculate the kinetic friction and acceleration immediately after sliding begins.

Assume the push remains at 30° below the horizontal, there is no vertical acceleration, and \(g=9.8\text{ m/s}^2\).

Solution 4

At 50 N, the crate remains stationary with 43.3 N of static friction acting to the left. At 60 N, it begins sliding to the right and accelerates at approximately 1.36 m/s².

(a) Normal force at 50 N

Resolve the applied force into components.

The horizontal component is:

\[
P_x=50\cos30^\circ
\]

\[
P_x=43.30\text{ N}
\]

The downward component is:

\[
P_y=50\sin30^\circ
\]

\[
P_y=25.0\text{ N}
\]

The normal force must balance both the weight and the additional downward push.

\[
N=mg+P_y
\]

\[
N=(10\times9.8)+25
\]

\[
\boxed{N=123\text{ N}}
\]

(b) Determine whether the crate moves

Calculate maximum static friction:

\[
f_{s,\max}=\mu_sN
\]

\[
f_{s,\max}=0.40\times123
\]

\[
f_{s,\max}=49.2\text{ N}
\]

The horizontal applied force is 43.30 N.

Since:

\[
43.30<49.2
\]

the crate remains stationary.

Its actual static friction must balance the horizontal push:

\[
\boxed{f_s=43.30\text{ N to the left}}
\]

(c) Increase the applied force to 60 N

The horizontal component is now:

\[
P_x=60\cos30^\circ
\]

\[
P_x=51.96\text{ N}
\]

The downward component is:

\[
P_y=60\sin30^\circ
\]

\[
P_y=30.0\text{ N}
\]

Calculate the new normal force:

\[
N=98+30
\]

\[
\boxed{N=128\text{ N}}
\]

The new maximum static friction is:

\[
f_{s,\max}=0.40\times128
\]

\[
f_{s,\max}=51.2\text{ N}
\]

Compare the forces:

\[
51.96>51.2
\]

Static friction cannot balance the applied horizontal force.

The crate therefore begins sliding to the right.

(d) Calculate kinetic friction and acceleration

Once sliding begins:

\[
f_k=\mu_kN
\]

\[
f_k=0.30\times128
\]

\[
\boxed{f_k=38.4\text{ N}}
\]

Kinetic friction acts to the left.

Take right as positive.

The horizontal net force is:

\[
F_{\text{net}}=P_x-f_k
\]

\[
F_{\text{net}}=51.96-38.4
\]

\[
F_{\text{net}}=13.56\text{ N}
\]

Using Newton’s second law:

\[
F_{\text{net}}=ma
\]

\[
13.56=10a
\]

\[
\boxed{a=1.36\text{ m/s}^2}
\]

The trap here is assuming the normal force remains equal to the weight.

If you had used \(N=mg=98\text{ N}\), you would have calculated a maximum static friction of only 39.2 N. You might incorrectly conclude that the crate moves even when the applied force is 50 N.

The downward push increases the normal force, which increases the available friction.

This is why resolving the applied force before calculating friction is essential.

Question 5

Two blocks are stacked on top of each other.

• Block A has a mass of 4.0 kg.
• Block B has a mass of 2.0 kg.
• Block A rests directly on block B.
• Block B rests on a smooth, horizontal floor.
• The coefficient of static friction between A and B is 0.50.

A force \(F\) is applied to block A, directed horizontally to the right.

Assume both blocks initially start from rest, remain in contact, and the floor provides no horizontal friction.

(a) Calculate the maximum acceleration of the two blocks before A begins slipping relative to B.

(b) Determine the corresponding applied force \(F\).

(c) Now suppose the applied force is instead directed 30° above the horizontal. Derive the maximum applied force that allows both blocks to move together without slipping. Calculate the corresponding acceleration.

Use \(g=9.8\text{ m/s}^2\).

Solution 5

With a horizontal pull, the maximum common acceleration is 9.8 m/s². With a pull at 30° above the horizontal, the maximum common acceleration decreases to approximately 5.25 m/s², and the maximum applied force is approximately 36.4 N.

This question exposes a particularly important limitation of treating \(\mu g\) as a universal maximum acceleration.

The friction force between the blocks is determined by the normal force at their contact. However, that friction force is responsible for accelerating block B, whose mass is different from the mass producing the contact normal force.

(a) Maximum acceleration with a horizontal pull

First, calculate the normal force between the blocks.

For block A, the only vertical forces are its weight and the normal force from block B.

There is no vertical acceleration, so:

\[
N=m_Ag
\]

\[
N=4.0\times9.8
\]

\[
N=39.2\text{ N}
\]

The maximum static friction is:

\[
f_{s,\max}=\mu_sN
\]

\[
f_{s,\max}=0.50\times39.2
\]

\[
\boxed{f_{s,\max}=19.6\text{ N}}
\]

Now consider block B.

The only horizontal force acting on B is the friction exerted by block A.

Therefore, using Newton’s second law:

\[
f_s=m_Ba
\]

At the point where slipping is just about to occur:

\[
f_s=f_{s,\max}
\]

Therefore:

\[
19.6=2.0a
\]

\[
\boxed{a_{\max}=9.8\text{ m/s}^2}
\]

This is the maximum acceleration before the two blocks begin slipping relative to each other.

(b) Corresponding applied force

Since the blocks move together, they have the same horizontal acceleration.

Consider both blocks as one system.

The only external horizontal force is the applied force \(F\), because the floor is smooth.

The total mass is:

\[
m_{\text{total}}=4.0+2.0=6.0\text{ kg}
\]

Using Newton’s second law:

\[
F=m_{\text{total}}a
\]

\[
F=6.0\times9.8
\]

\[
\boxed{F=58.8\text{ N}}
\]

At this force, the friction between the blocks reaches its maximum of 19.6 N.

Any further increase in the horizontal force causes block A to slip relative to block B.

(c) Force applied at 30° above the horizontal

This is where the problem becomes more interesting.

The upward component of the applied force reduces the normal force between the blocks.

Therefore, the maximum static friction is no longer 19.6 N.

Let the applied force be \(F\).

First, calculate the normal force between the blocks.

For block A, vertical equilibrium gives:

\[
N+F\sin30^\circ-m_Ag=0
\]

Therefore:

\[
N=m_Ag-F\sin30^\circ
\]

Substituting the known values:

\[
\boxed{N=39.2-0.5F}
\]

The maximum static friction becomes:

\[
f_{s,\max}=\mu_sN
\]

\[
f_{s,\max}=0.50(39.2-0.5F)
\]

\[
\boxed{f_{s,\max}=19.6-0.25F}
\]

Notice that the maximum available friction decreases as the applied force increases.

Now consider the horizontal acceleration of both blocks, assuming they move together.

The total horizontal applied force is \(F\cos30^\circ\).

Using Newton’s second law for the combined system:

\[
F\cos30^\circ=(m_A+m_B)a
\]

Therefore:

\[
a=\frac{F\cos30^\circ}{6.0}
\]

The friction force required to accelerate block B is:

\[
f_s=m_Ba
\]

Substituting the expression for acceleration:

\[
f_s=\frac{2.0F\cos30^\circ}{6.0}
\]

\[
\boxed{f_s=\frac{F\cos30^\circ}{3}}
\]

At the threshold of slipping, the required static friction equals the maximum available static friction.

Therefore:

\[
\frac{F\cos30^\circ}{3}
=
0.50(39.2-F\sin30^\circ)
\]

Substituting the trigonometric values:

\[
0.2887F=19.6-0.25F
\]

Collecting terms:

\[
0.5387F=19.6
\]

\[
\boxed{F_{\max}\approx36.4\text{ N}}
\]

Now calculate the corresponding acceleration:

\[
a=\frac{F\cos30^\circ}{6.0}
\]

\[
a=\frac{36.4\times0.866}{6.0}
\]

\[
\boxed{a_{\max}\approx5.25\text{ m/s}^2}
\]

At this threshold, the normal force is approximately:

\[
N=39.2-(36.4\times0.5)
\]

\[
N\approx21.0\text{ N}
\]

The maximum static friction is therefore approximately:

\[
f_{s,\max}=0.50\times21.0
\]

\[
f_{s,\max}\approx10.5\text{ N}
\]

This is also the friction required to accelerate block B at 5.25 m/s².

Why this result is surprising

A tempting approach is to assume the maximum acceleration must be:

\[
a_{\max}=\mu_sg
\]

This would give:

\[
a_{\max}=0.50\times9.8
\]

\[
a_{\max}=4.9\text{ m/s}^2
\]

But that is not a general rule.

In this system, friction is accelerating block B, while the normal force at the contact is determined by block A.

These are different masses.

With a horizontal pull, the maximum friction is 19.6 N, and it accelerates the 2.0 kg lower block. This allows a maximum acceleration of 9.8 m/s².

With an upward-angled pull, the normal force decreases as the applied force increases. The available friction decreases at the same time that more friction is required to accelerate block B.

Both effects must be considered together.

This is the deeper reason why you cannot treat \(\mu mg\) as a universal friction formula.

07The main decision to remember

Before calculating friction, establish two things:

1. Is the object stationary or sliding relative to the surface?
2. What is the actual normal force at the contact?

For stationary objects, calculate the friction required by the force balance and check that it does not exceed \(\mu_sN\).

For sliding objects, use the kinetic friction model \(f_k=\mu_kN\).

Only substitute \(N=mg\) when the vertical force balance actually justifies it.

Once you can find the normal force correctly, the next important skill is resolving forces perpendicular and parallel to an inclined plane, particularly when an applied force changes the normal reaction.