How to Find the Angle at Which a Block Starts Sliding

Raise one end of a board with a block resting on it. At first, the block stays still. Keep increasing the angle, and eventually it starts sliding down. At what angle does that happen?

For a block about to slide down a rough slope, with no additional applied forces, the critical angle is:

\[
\boxed{\theta_c=\tan^{-1}(\mu_s)}
\]

Here, \(\mu_s\) is the coefficient of static friction. For example, if \(\mu_s=0.40\), the block starts sliding at:

\[
\theta_c=\tan^{-1}(0.40)=21.8^\circ
\]

Notice that the mass of the block does not appear in the equation. In the ideal friction model, doubling the block’s mass does not change the angle at which it starts sliding.

The key is understanding two forces: the component of weight pulling the block down the slope, and the maximum static friction resisting that movement. The normal reaction connects the two.

01Resolve the forces along the slope

Before using an equation, consider what happens as you gradually raise the board.

Does the normal reaction stay equal to the block’s weight because it supports the block?

No. The normal reaction acts perpendicular to the surface, not vertically upwards. As the slope gets steeper, the perpendicular component of the block’s weight gets smaller.

We need to resolve the weight into two components, using axes that follow the slope.

Choose the following sign convention:

• Positive parallel direction: down the slope.
• Positive perpendicular direction: away from the slope.

The block’s weight is \(mg\), where \(m\) is its mass in kilograms and \(g=9.8\text{ m s}^{-2}\) is gravitational acceleration.

Resolving weight gives:

Parallel to the slope:

\[
W_{\parallel}=mg\sin\theta
\]

This component pulls the block down the slope.

Perpendicular to the slope:

\[
W_{\perp}=mg\cos\theta
\]

This component pushes the block into the surface.

The normal reaction, \(N\), acts in the opposite perpendicular direction.

Finding the normal reaction

While the block remains in contact with the stationary slope, its acceleration perpendicular to the surface is zero.

Therefore:

\[
\sum F_{\perp}=0
\]

\[
N-mg\cos\theta=0
\]

So:

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

This is an important result.

The normal reaction is not generally equal to \(mg\). It equals the perpendicular component of the weight when gravity is the only force with a perpendicular component pushing the block into the slope.

For example:

As the slope gets steeper, the normal reaction decreases.

02Find the angle at which static friction reaches its maximum

Now consider the forces parallel to the slope.

The component of weight pulls the block downwards. Static friction acts upwards, preventing the block from sliding.

While the block remains stationary:

\[
\sum F_{\parallel}=0
\]

Therefore:

\[
mg\sin\theta-f_s=0
\]

Rearranging:

\[
f_s=mg\sin\theta
\]

Here, \(f_s\) is the actual static friction force.

There is an important distinction: static friction does not always equal its maximum value.

It adjusts to whatever value is needed to prevent movement, up to a maximum:

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

The symbol \(\mu_s\) represents the coefficient of static friction. It depends on the pair of contacting surfaces and their conditions.

For example, a block on a shallow slope might only need 2 N of friction to remain stationary, even if the maximum available static friction is 8 N.

As the slope becomes steeper, the required friction increases.

Eventually, the block reaches the point where static friction can no longer increase.

This is the instant just before sliding begins.

At this critical angle:

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

Substituting the maximum friction equation:

\[
mg\sin\theta_c=\mu_sN
\]

We already established that:

\[
N=mg\cos\theta_c
\]

Substitute this into the friction equation:

\[
mg\sin\theta_c
=
\mu_smg\cos\theta_c
\]

Cancel \(mg\):

\[
\sin\theta_c=\mu_s\cos\theta_c
\]

Divide by \(\cos\theta_c\):

\[
\tan\theta_c=\mu_s
\]

Therefore, the final equation is:

\[
\boxed{\theta_c=\tan^{-1}(\mu_s)}
\]

This is the equation you need when a question asks for the angle at which a block starts sliding down a rough incline.

What does the equation tell us?

• If \(\theta\) is smaller than the critical angle, static friction can prevent sliding.
• At the critical angle, static friction reaches its maximum.
• If the angle increases beyond the critical angle, the block starts sliding down, assuming no additional forces act.

The block’s mass cancels because both the downhill component of weight and the maximum static friction are proportional to mass.

This result assumes a constant coefficient of static friction, a rigid straight slope, and no additional applied forces.

If there is no friction at all, then \(\mu_s=0\), giving a critical angle of zero degrees. Any non-zero incline produces a downhill force that cannot be balanced by friction. See Can a Block Stay at Rest on a Frictionless Slope? for that contrasting situation.

03When does the normal reaction change?

This is where students often lose marks.

The normal reaction changes whenever the forces perpendicular to the surface change. Changing the slope angle is one example, but it is not the only one.

Consider these situations:

The distinction is important because maximum static friction depends directly on the normal reaction:

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

Increasing the normal reaction increases the maximum available static friction. Reducing it decreases the maximum available static friction.

A useful comparison

Imagine two identical blocks on identical \(25^\circ\) slopes.

One block has no additional force acting on it. The other is pushed directly into the slope.

The second block has a larger normal reaction, so it can have a larger maximum static friction force. It may therefore remain stationary even when the first block would slide.

The slope angle and the coefficient of friction have not changed. The normal reaction has.

There is another subtle point: the normal reaction does not automatically decrease when the block starts sliding.

If the slope angle and perpendicular applied forces remain unchanged, and the block accelerates only parallel to the surface, the normal reaction stays the same.

What changes at the transition is the type of friction:

• Immediately before movement, friction is static and at its maximum.
• Once movement begins, friction becomes kinetic.

Do not confuse a change in friction with a change in normal reaction.

04Finding the critical angle experimentally

You can also determine the critical angle without being given a coefficient of static friction.

Instead, measure the angle at which a block first starts moving.

The apparatus consists of:

• A rigid board that can be tilted.
• A block placed on the board.
• An adjustable support that raises one end of the board.
• An inclinometer or protractor to measure the angle.

Start with the board horizontal and the block stationary.

Gradually raise one end. As the angle increases, the component of weight acting down the slope increases, while the normal reaction decreases.

Eventually, the block begins to move.

What should the experiment show?

The prediction from our model is that movement begins when:

\[
mg\sin\theta_c=\mu_smg\cos\theta_c
\]

Therefore:

\[
\boxed{\mu_s=\tan\theta_c}
\]

The measured critical angle provides an estimate of the coefficient of static friction.

Notice what is actually measured: the angle at which movement begins. The coefficient of friction is calculated from that observation.

Improving the measurement

A rapid increase in slope angle can cause a problem.

If the board is raised too quickly, the block may already have started moving before the angle is recorded. At that point, kinetic friction is acting rather than maximum static friction.

A better method is to:

1. Increase the angle in small increments.
2. Record the angle at the first sustained movement.
3. Return the block to rest before repeating the measurement.
4. Repeat several trials and calculate an average critical angle.

Surface contamination, uneven contact, vibrations, and measurement uncertainty can also affect the result.

The experiment therefore gives an effective coefficient for the particular surfaces and conditions being tested, rather than a perfectly exact universal value.

05Worked examples

06Questions and solutions

Question 1

A \(1.2\text{ kg}\) block rests on a rough incline. The coefficient of static friction between the block and the surface is \(0.30\).

Calculate:

a) The angle at which the block begins sliding.

b) The normal reaction at this angle.

Use \(g=9.8\text{ m s}^{-2}\) and assume no other forces act on the block.

Solution 1

The block starts sliding at \(16.7^\circ\), and the normal reaction at this angle is approximately \(11.3\text{ N}\).

Step 1: Calculate the critical angle.

At impending motion:

\[
\tan\theta_c=\mu_s
\]

Substitute:

\[
\theta_c=\tan^{-1}(0.30)
\]

\[
\boxed{\theta_c=16.7^\circ}
\]

Step 2: Calculate the normal reaction.

\[
N=mg\cos\theta_c
\]

Substitute:

\[
N=1.2\times9.8\times\cos(16.7^\circ)
\]

\[
\boxed{N=11.3\text{ N}}
\]

Step 3: Explain the physical meaning.

At the critical angle, maximum static friction is:

\[
f_{s,\max}=0.30\times11.3
\]

\[
f_{s,\max}=3.39\text{ N}
\]

The downhill component of weight is:

\[
W_{\parallel}=1.2\times9.8\times\sin(16.7^\circ)
\]

\[
W_{\parallel}\approx3.39\text{ N}
\]

The two forces are equal, so the block is at limiting equilibrium.

The normal reaction is smaller than the block’s weight because it only balances the perpendicular component of weight.

Question 2

A \(3.0\text{ kg}\) block rests on a rough incline at \(25^\circ\) to the horizontal. Its coefficient of static friction is \(0.40\).

An additional force of \(12\text{ N}\) is applied perpendicular to the surface, pushing the block into the incline.

Determine:

a) The normal reaction.

b) Whether the block remains stationary.

c) The actual static friction force acting on the block.

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

Solution 2

The block remains stationary. Its normal reaction is \(38.6\text{ N}\), and its actual static friction force is \(12.4\text{ N}\) up the slope.

Step 1: Resolve the weight perpendicular to the incline.

\[
W_{\perp}=mg\cos25^\circ
\]

\[
W_{\perp}=3.0\times9.8\times\cos25^\circ
\]

\[
W_{\perp}=26.6\text{ N}
\]

The additional \(12\text{ N}\) force also pushes the block into the surface.

Therefore:

\[
N=mg\cos25^\circ+12
\]

\[
N=26.6+12
\]

\[
\boxed{N=38.6\text{ N}}
\]

Step 2: Calculate the maximum available static friction.

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

\[
f_{s,\max}=0.40\times38.6
\]

\[
\boxed{f_{s,\max}=15.4\text{ N}}
\]

Step 3: Calculate the friction actually required.

The additional force is perpendicular to the surface, so it does not contribute to the force parallel to the slope.

The downhill component of weight is:

\[
W_{\parallel}=mg\sin25^\circ
\]

\[
W_{\parallel}=3.0\times9.8\times\sin25^\circ
\]

\[
\boxed{W_{\parallel}=12.4\text{ N}}
\]

For the block to remain stationary, static friction must balance this force.

Therefore, actual static friction is:

\[
\boxed{f_s=12.4\text{ N}}
\]

It acts up the slope.

Step 4: Determine whether the block moves.

Compare the required friction with the maximum available friction:

\[
12.4\text{ N}<15.4\text{ N}
\]

The block can generate enough static friction to remain stationary.

This question highlights an important difference between actual static friction and maximum static friction. The actual force is \(12.4\text{ N}\), not \(15.4\text{ N}\).

What if the additional perpendicular force were removed?

The normal reaction would decrease to \(26.6\text{ N}\).

Maximum static friction would then be:

\[
f_{s,\max}=0.40\times26.6
\]

\[
f_{s,\max}=10.6\text{ N}
\]

Since the downhill component of weight is \(12.4\text{ N}\), the block would slide.

The extra push prevents sliding by increasing the normal reaction and therefore increasing the maximum available friction.

Question 3

An adjustable ramp has a length of \(0.70\text{ m}\). A \(1.8\text{ kg}\) block is placed on the ramp.

The ramp is raised until its vertical height reaches \(0.26\text{ m}\), at which point the block just begins sliding.

The angle is then held constant. Once moving, the block accelerates down the ramp at \(0.90\text{ m s}^{-2}\).

Calculate:

a) The critical angle.

b) The coefficient of static friction.

c) The normal reaction at the critical angle.

d) The coefficient of kinetic friction.

Assume constant friction coefficients and no additional forces.

Solution 3

The critical angle is \(21.8^\circ\), the coefficient of static friction is \(0.400\), the normal reaction is \(16.4\text{ N}\), and the coefficient of kinetic friction is approximately \(0.301\).

Step 1: Find the critical angle.

Using the ramp geometry:

\[
\sin\theta_c=\frac{h}{L}
\]

Substitute:

\[
\sin\theta_c=\frac{0.26}{0.70}
\]

\[
\theta_c=\sin^{-1}(0.3714)
\]

\[
\boxed{\theta_c=21.8^\circ}
\]

Step 2: Calculate the coefficient of static friction.

At the instant sliding begins:

\[
\mu_s=\tan\theta_c
\]

\[
\mu_s=\tan(21.8^\circ)
\]

\[
\boxed{\mu_s=0.400}
\]

Step 3: Calculate the normal reaction.

\[
N=mg\cos\theta_c
\]

\[
N=1.8\times9.8\times\cos(21.8^\circ)
\]

\[
\boxed{N=16.4\text{ N}}
\]

Step 4: Calculate the coefficient of kinetic friction.

This is where the situation changes.

Before the block moves, static friction acts. Once the block is sliding, kinetic friction acts instead.

Take down the slope as positive.

Newton’s second law gives:

\[
mg\sin\theta_c-f_k=ma
\]

Since:

\[
f_k=\mu_kN
\]

Substitute:

\[
mg\sin\theta_c-\mu_kN=ma
\]

Rearrange:

\[
\mu_k=\frac{mg\sin\theta_c-ma}{N}
\]

Substitute the values:

\[
\mu_k=
\frac{
1.8(9.8)\sin(21.8^\circ)-1.8(0.90)
}{
16.4
}
\]

\[
\mu_k\approx\frac{4.93-1.62}{16.4}
\]

\[
\boxed{\mu_k=0.301}
\]

The kinetic friction coefficient is smaller than the static friction coefficient in this example.

This explains why the block can begin moving once the static friction limit is exceeded and then accelerate down the ramp.

Notice that the normal reaction does not change when the block starts sliding. The block has acceleration parallel to the slope, but no acceleration perpendicular to it.

Question 4

A \(2.0\text{ kg}\) block rests on a rough incline at \(30^\circ\) to the horizontal.

A rope attached to the block pulls it parallel to the slope, upwards. The coefficient of static friction is \(0.35\).

Calculate the range of rope tensions for which the block can remain stationary.

For each limiting tension, state the direction of static friction.

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

Solution 4

The block can remain stationary for rope tensions between approximately \(3.86\text{ N}\) and \(15.7\text{ N}\), inclusive.

The key is that static friction can act in either direction, depending on which way the block would otherwise move.

Step 1: Calculate the normal reaction.

The rope pulls parallel to the slope, so it has no perpendicular component.

Therefore:

\[
N=mg\cos30^\circ
\]

\[
N=2.0\times9.8\times\cos30^\circ
\]

\[
\boxed{N=17.0\text{ N}}
\]

Step 2: Calculate maximum static friction.

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

\[
f_{s,\max}=0.35\times16.97
\]

\[
\boxed{f_{s,\max}=5.94\text{ N}}
\]

Step 3: Calculate the downhill component of weight.

\[
W_{\parallel}=mg\sin30^\circ
\]

\[
W_{\parallel}=2.0\times9.8\times0.5
\]

\[
\boxed{W_{\parallel}=9.80\text{ N}}
\]

Step 4: Find the minimum tension.

If the rope tension is too small, the block tends to slide down the slope.

Static friction must therefore act up the slope.

At the minimum tension, friction reaches its maximum value:

\[
T_{\min}+f_{s,\max}=mg\sin30^\circ
\]

Rearrange:

\[
T_{\min}=mg\sin30^\circ-f_{s,\max}
\]

\[
T_{\min}=9.80-5.94
\]

\[
\boxed{T_{\min}=3.86\text{ N}}
\]

At this limit, friction acts up the slope.

Step 5: Find the maximum tension.

If the rope pulls too strongly, the block tends to move up the slope.

Static friction must now act down the slope.

At the maximum tension:

\[
T_{\max}=mg\sin30^\circ+f_{s,\max}
\]

\[
T_{\max}=9.80+5.94
\]

\[
\boxed{T_{\max}=15.74\text{ N}}
\]

At this limit, friction acts down the slope.

Step 6: State the full range.

For the block to remain stationary:

\[
\boxed{3.86\leq T\leq15.74\text{ N}}
\]

The important lesson is that friction does not always act up the slope.

It opposes the tendency for relative motion. The rope can make the block tend to move upwards, in which case friction acts downwards.

Also notice that the normal reaction is unchanged by the rope tension because the rope is parallel to the surface.

Question 5

A \(2.0\text{ kg}\) block rests on a rough adjustable ramp. The ramp rises towards the right, and a constant horizontal force of \(4.0\text{ N}\) pushes the block towards the right.

The coefficient of static friction is \(0.40\), and the coefficient of kinetic friction is \(0.30\).

The ramp is initially shallow and is gradually raised. The block remains stationary until it eventually starts sliding down the ramp.

Once the block starts moving, the ramp angle and horizontal applied force remain constant.

Determine:

a) The angle at which the required static friction changes direction.

b) The angle at which the block first starts sliding down.

c) The normal reaction at the instant sliding begins.

d) The initial acceleration of the block immediately after sliding begins.

e) Whether the normal reaction suddenly decreases when static friction changes to kinetic friction.

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

Solution 5

The required friction changes direction at approximately \(11.5^\circ\). The block first slides down at approximately \(33.3^\circ\), with a normal reaction of \(18.6\text{ N}\). Its initial acceleration is approximately \(0.93\text{ m s}^{-2}\) down the slope. The normal reaction does not suddenly decrease when sliding begins.

This problem is more involved because the horizontal force has two components relative to the inclined surface.

One component acts along the slope, while the other pushes the block into the surface.

Step 1: Resolve the horizontal applied force.

The ramp rises towards the right, and the horizontal force pushes towards the right.

Its component parallel to the slope acts upwards:

\[
P_{\parallel}=P\cos\theta
\]

Its component perpendicular to the slope pushes into the surface:

\[
P_{\perp}=P\sin\theta
\]

The weight also has two components:

\[
W_{\parallel}=mg\sin\theta
\]

\[
W_{\perp}=mg\cos\theta
\]

The normal reaction must now balance both perpendicular components.

Therefore:

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

This is different from the simpler equation \(N=mg\cos\theta\).

Step 2: Find the angle at which friction changes direction.

Initially, the horizontal push has a stronger component up the slope than gravity has down the slope.

Therefore, the block would tend to move up the slope, and static friction acts down the slope.

As the ramp becomes steeper, the downhill component of weight increases.

At the angle where the required friction changes direction, the two parallel components balance each other:

\[
mg\sin\theta=P\cos\theta
\]

Rearrange:

\[
\tan\theta=\frac{P}{mg}
\]

Substitute:

\[
\tan\theta=\frac{4.0}{2.0\times9.8}
\]

\[
\theta=\tan^{-1}(0.2041)
\]

\[
\boxed{\theta=11.5^\circ}
\]

Below this angle, friction acts down the slope. Above it, friction acts up the slope.

The friction force is zero at the exact angle where the two parallel components balance.

Step 3: Find the angle at which the block starts sliding down.

At the critical angle, the block is about to slide downwards.

Static friction therefore acts up the slope and reaches its maximum:

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

The parallel force balance is:

\[
mg\sin\theta-P\cos\theta=\mu_sN
\]

Substitute the expression for the normal reaction:

\[
mg\sin\theta-P\cos\theta
=
\mu_s(mg\cos\theta+P\sin\theta)
\]

Expand:

\[
mg\sin\theta-P\cos\theta
=
\mu_smg\cos\theta+\mu_sP\sin\theta
\]

Collect the sine and cosine terms:

\[
(mg-\mu_sP)\sin\theta
=
(\mu_smg+P)\cos\theta
\]

Therefore:

\[
\tan\theta_c=
\frac{\mu_smg+P}{mg-\mu_sP}
\]

Substitute:

\[
\tan\theta_c=
\frac{(0.40)(2.0)(9.8)+4.0}
{(2.0)(9.8)-(0.40)(4.0)}
\]

\[
\tan\theta_c=\frac{11.84}{18.0}
\]

\[
\theta_c=\tan^{-1}(0.6578)
\]

\[
\boxed{\theta_c=33.3^\circ}
\]

The block first slides down at approximately \(33.3^\circ\).

Notice why simply using \(\tan^{-1}(0.40)\) would be incorrect. The horizontal force changes both the parallel force balance and the normal reaction.

Step 4: Calculate the normal reaction at the critical angle.

Use:

\[
N=mg\cos\theta_c+P\sin\theta_c
\]

Substitute:

\[
N=2.0(9.8)\cos(33.3^\circ)
+4.0\sin(33.3^\circ)
\]

\[
N\approx16.4+2.2
\]

\[
\boxed{N=18.6\text{ N}}
\]

The horizontal force increases the normal reaction because part of it pushes the block into the incline.

Step 5: Calculate the acceleration immediately after sliding begins.

The block is now moving, so kinetic friction applies.

The normal reaction remains approximately \(18.6\text{ N}\), provided the ramp angle and horizontal force remain unchanged and the block stays in contact with the surface.

Calculate kinetic friction:

\[
f_k=\mu_kN
\]

\[
f_k=0.30\times18.6
\]

\[
f_k=5.58\text{ N}
\]

The net force down the slope is:

\[
F_{\text{net}}=mg\sin\theta_c-P\cos\theta_c-f_k
\]

Substitute:

\[
F_{\text{net}}
=
19.6\sin(33.3^\circ)
-4.0\cos(33.3^\circ)-5.58
\]

\[
F_{\text{net}}\approx10.8-3.34-5.58
\]

\[
F_{\text{net}}\approx1.86\text{ N}
\]

Apply Newton’s second law:

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

\[
a=\frac{1.86}{2.0}
\]

\[
\boxed{a=0.93\text{ m s}^{-2}}
\]

The block accelerates down the slope.

Step 6: Evaluate the claim about the normal reaction.

A student argues that the normal reaction must decrease as soon as the block starts moving because static friction has been replaced by kinetic friction.

This is incorrect.

Static and kinetic friction both act parallel to the surface. Neither directly contributes to the perpendicular force balance.

The normal reaction is determined by:

\[
N=mg\cos\theta+P\sin\theta
\]

The angle and horizontal applied force have not changed, so neither has the normal reaction.

What changes is the friction force:

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

becomes:

\[
f_k=\mu_kN
\]

Because \(\mu_k=0.30\) is smaller than \(\mu_s=0.40\), friction decreases, allowing the block to accelerate.

This is the central distinction: a change in the friction model does not automatically mean a change in the normal reaction.

07Where to go next

Finding the critical angle tells you exactly when static friction can no longer keep a block stationary.

Once the block starts moving, the problem changes. You must use kinetic friction and Newton’s second law to calculate its acceleration.

The next useful method is How to Calculate Acceleration Down a Rough Incline, where the same force-resolution approach predicts what happens after the block begins sliding.