Why Increasing Slope Angle Can Reduce Frictional Force

Raise one end of a board while a block slides down it. As the ramp gets steeper, the frictional force can actually become smaller.

That sounds odd. Surely a steeper ramp means the block is being pushed harder against the surface?

Not quite. Before calculating anything, consider what happens to the force pressing the block into the ramp as its angle increases. Does it stay equal to the block’s full weight, or does it decrease?

It decreases. On a straight, stationary incline with no additional forces perpendicular to the surface, the normal reaction is:

\[ N=mg\cos\theta \]

As the angle \(\theta\) increases, \(\cos\theta\) decreases. Since kinetic friction depends on the normal reaction, kinetic friction also decreases:

\[ \boxed{f_k=\mu_kmg\cos\theta} \]

There is an important qualification, though. If the block is stationary, the actual static friction needed to hold it in place increases as the ramp gets steeper. The maximum available static friction decreases.

Both behaviours are correct. The difference comes down to whether the block is sliding or remaining at rest.

01Resolve weight along the slope

Start by looking at the forces acting on a block resting on an inclined plane.

The block experiences its weight, \(mg\), vertically downwards. The ramp exerts a normal reaction, \(N\), perpendicular to its surface.

The normal reaction is not necessarily equal to the block’s weight. It only acts perpendicular to the ramp, so it cannot support the entire vertical weight when the surface is inclined.

To understand why, resolve the weight into two components.

Choosing the axes

For inclined-plane problems, choose axes that follow the surface:

• Parallel axis: along the slope, positive downwards.
• Perpendicular axis: at right angles to the slope, positive away from the surface.

This makes Newton’s second law much easier to apply.

The weight has two components:

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.

Here:

• \(m\) is mass, measured in kilograms (kg).
• \(g\) is gravitational acceleration, approximately \(9.8\ \mathrm{m\,s^{-2}}\).
• \(\theta\) is the angle of the slope above the horizontal.
• \(mg\) is the weight, measured in newtons (N).

Notice that the total weight does not change when the ramp is tilted. Only its components change.

Why does the normal reaction decrease?

Assume the ramp is straight and stationary, the block remains in contact with it, and there are no other forces with components perpendicular to the surface.

The block has no acceleration perpendicular to the ramp.

Therefore:

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

Taking away from the surface as positive:

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

Rearranging:

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

Now consider what happens as the angle increases.

At \(0^\circ\), the surface is horizontal, so the normal reaction equals the full weight.

At \(60^\circ\), the normal reaction is only half the block’s weight.

The steeper ramp supports a smaller component of the block’s weight perpendicular to its surface.

This is the physical reason friction can decrease.

02Why kinetic friction decreases as the slope gets steeper

Kinetic friction acts when two surfaces slide relative to each other.

Its magnitude is modelled by:

\[ f_k=\mu_kN \]

where:

• \(f_k\) is kinetic friction in newtons.
• \(\mu_k\) is the coefficient of kinetic friction, a dimensionless quantity.
• \(N\) is the normal reaction in newtons.

Substitute the expression for the normal reaction:

\[ N=mg\cos\theta \]

This gives:

\[ \boxed{f_k=\mu_kmg\cos\theta} \]

For the same block and surface, we assume that \(m\), \(g\), and \(\mu_k\) remain constant.

The only changing quantity is \(\cos\theta\).

As the slope gets steeper, \(\cos\theta\) decreases, so kinetic friction decreases.

Visualising the change

Imagine the same block sliding down the same ramp at three different angles.

The frictional force is smaller at the steeper angle because the block is pressed less firmly against the ramp.

However, something else is happening at the same time.

The component of gravity pulling the block down the slope is increasing:

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

So the block experiences:

• Less kinetic friction opposing its motion.
• A larger gravitational component pulling it downhill.

This is why a block generally accelerates more strongly down a steeper rough incline.

The two changes work together.

A graph of the competing forces

There are three useful quantities to compare as the angle increases:

1. The downslope component of weight, \(mg\sin\theta\).
2. The maximum available static friction, \(\mu_smg\cos\theta\).
3. The kinetic friction, \(\mu_kmg\cos\theta\).

The graph reveals an important distinction: the force required to prevent sliding and the maximum friction available are not the same thing.

03Static friction behaves differently

Suppose the block is not moving.

Static friction acts to prevent relative sliding between the block and the ramp.

If gravity is the only force acting parallel to the slope, static friction must balance the downslope component of weight.

Therefore:

\[ f_s=mg\sin\theta \]

As the angle increases, \(\sin\theta\) increases.

The actual static friction required to hold the block in place therefore increases.

But there is a limit.

The maximum static friction available is:

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

Using \(N=mg\cos\theta\):

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

This maximum decreases as the angle increases.

We now have three different behaviours:

This is the distinction students often miss.

Static friction does not automatically equal \(\mu_sN\). It adjusts to whatever value is needed to prevent slipping, provided that value does not exceed its maximum.

Eventually, the block reaches an angle where the required static friction equals the maximum available friction.

At that point, it is just about to slide.

Equating the two:

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

Cancel \(mg\) and rearrange:

\[ \tan\theta_c=\mu_s \]

Therefore:

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

Beyond this critical angle, static friction cannot hold the block stationary under these assumptions.

For a more detailed treatment of this threshold, see How to Find the Angle at Which a Block Starts Sliding.

04When does the normal reaction change?

So far, we have used:

\[ N=mg\cos\theta \]

This is correct under particular conditions. It is not a universal equation for every inclined-plane problem.

The normal reaction depends on all the forces acting perpendicular to the surface.

Consider three situations.

Situation 1: No additional perpendicular forces

A block rests on a stationary incline under gravity alone, with no perpendicular acceleration.

Then:

\[ N=mg\cos\theta \]

This is the standard situation.

Situation 2: A rope pulls the block away from the surface

Imagine attaching a rope to the block and pulling it uphill at an angle above the ramp.

The tension now has two components:

• A component parallel to the ramp.
• A component perpendicular to the ramp, directed away from the surface.

The perpendicular component reduces the normal reaction.

If the rope has tension \(T\) and makes an angle \(\alpha\) above the ramp:

\[ \boxed{N=mg\cos\theta-T\sin\alpha} \]

The normal reaction is smaller than it would be without the rope.

Consequently, kinetic friction also decreases.

Situation 3: An additional force pushes the block into the surface

Now imagine someone pushing the block perpendicular into the ramp.

This adds another force pressing the block against the surface.

If the additional perpendicular force is \(P\):

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

The normal reaction increases, so friction increases too.

The general rule is simple:

Calculate the normal reaction from the perpendicular force balance. Do not automatically substitute \(mg\cos\theta\).

If the block remains in contact with a fixed straight ramp, its perpendicular acceleration is zero. If it loses contact, the normal reaction becomes zero, and the usual friction equations no longer apply.

05How could we test whether friction decreases?

We can investigate this using an adjustable inclined plane, a block, and a spring balance.

The aim is to measure kinetic friction at different ramp angles while keeping the contacting surfaces the same.

Setting up the investigation

Imagine a board hinged at one end. The other end can be raised to change the angle.

Place a block on the board and attach it to a spring balance using a light string.

The string should run parallel to the surface of the ramp.

A protractor or angle sensor measures the incline.

We pull the block uphill at a constant speed.

Why uphill?

Because kinetic friction then acts down the slope, opposite to the block’s motion. Gravity’s parallel component also acts down the slope.

This makes the force balance straightforward.

What should the spring balance measure?

Let the spring balance reading be \(T\).

Choose uphill as the positive direction.

The block moves at constant speed, so its acceleration is zero.

Newton’s second law gives:

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

Therefore:

\[ T-mg\sin\theta-f_k=0 \]

Rearranging:

\[ \boxed{f_k=T-mg\sin\theta} \]

This is an important experimental detail.

The spring balance does not measure friction alone. It measures the tension needed to overcome both gravity’s downslope component and kinetic friction.

If we simply treated \(T\) as friction, our conclusion could be completely wrong.

Illustrative experimental data

Suppose we use a \(2.0\ \mathrm{kg}\) block and a surface with a kinetic friction coefficient of \(0.30\).

The following values are illustrative predictions from the friction model, not actual experimental measurements.

Take \(g=9.8\ \mathrm{m\,s^{-2}}\).

Look carefully at the last two columns.

As the ramp gets steeper:

• The actual kinetic friction decreases from \(5.88\ \mathrm N\) to \(2.94\ \mathrm N\).
• The spring balance reading increases from \(5.88\ \mathrm N\) to \(19.91\ \mathrm N\).

There is no contradiction.

The spring balance reading increases because the gravitational component pulling the block downhill increases substantially.

The frictional force itself still decreases.

What could go wrong in the experiment?

The first mistake would be to assume that the spring balance directly measures friction.

It does not. We must subtract \(mg\sin\theta\).

There is another important limitation: the block must move at constant speed.

If the block accelerates uphill, the equation becomes:

\[ T-mg\sin\theta-f_k=ma \]

Ignoring the acceleration would give an incorrect friction value.

A reliable investigation therefore needs to:

1. Keep the same block and contacting surfaces.
2. Keep the string parallel to the ramp.
3. Pull the block at constant speed.
4. Record the angle and spring balance reading.
5. Subtract \(mg\sin\theta\) to calculate kinetic friction.
6. Calculate the normal reaction using \(N=mg\cos\theta\).
7. Compare the calculated friction forces at different angles.

The experiment tests the relationship between friction and normal reaction rather than simply comparing spring balance readings.

06Worked examples

07A reliable method for HSC exam questions

When you see an inclined-plane question involving friction, work through these decisions in order.

First: Is the block sliding?

• If it is sliding, use kinetic friction: \(f_k=\mu_kN\).
• If it is stationary, calculate the static friction required for equilibrium.

Second: Resolve the weight.

Use:

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

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

Third: Calculate the normal reaction.

Do not automatically assume \(N=mg\cos\theta\).

Check whether additional forces have components perpendicular to the ramp.

Fourth: Apply the correct friction model.

For a sliding block:

\[ f_k=\mu_kN \]

For a stationary block:

\[ f_s\leq\mu_sN \]

Calculate the actual friction needed and check whether the surface can supply it.

Finally: Check your result.

Ask yourself whether the friction force should increase or decrease under the conditions given.

If your answer says the actual static friction decreases as an unsupported block remains stationary on a steeper incline, revisit your force resolution.

08Questions and solutions

Question 1

A \(3.0\ \mathrm{kg}\) block slides down a rough incline at \(35^\circ\) to the horizontal.

The coefficient of kinetic friction is \(0.25\).

Calculate:

a) The normal reaction acting on the block.

b) The magnitude of the kinetic friction.

Use \(g=9.8\ \mathrm{m\,s^{-2}}\). Assume no additional forces act on the block.

Solution 1

a) The normal reaction is approximately \(24.1\ \mathrm N\).

b) The kinetic friction is approximately \(6.02\ \mathrm N\), acting up the slope.

The block is sliding, so we use the kinetic friction equation.

Step 1: Calculate the normal reaction.

There are no additional perpendicular forces, so:

\[ N=mg\cos\theta \]

Substitute:

\[ N=3.0(9.8)\cos35^\circ \]

\[ \boxed{N=24.1\ \mathrm N} \]

Step 2: Calculate kinetic friction.

\[ f_k=\mu_kN \]

Substitute:

\[ f_k=0.25(24.1) \]

\[ \boxed{f_k=6.02\ \mathrm N} \]

The block is sliding down the incline, so kinetic friction acts up the incline to oppose the motion.

Question 2

A \(5.0\ \mathrm{kg}\) block slides down a rough, stationary incline.

The coefficient of kinetic friction is \(0.20\).

The incline is initially at \(40^\circ\). It is then increased to \(55^\circ\).

For both angles, calculate:

a) The normal reaction.

b) The kinetic friction.

c) The acceleration of the block down the slope.

d) Explain why the frictional force decreases even though the block accelerates more strongly.

Assume the coefficient of kinetic friction remains constant and no additional forces act on the block.

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

Solution 2

At \(40^\circ\), the block’s acceleration is approximately \(4.80\ \mathrm{m\,s^{-2}}\).

At \(55^\circ\), its acceleration increases to approximately \(6.90\ \mathrm{m\,s^{-2}}\), even though kinetic friction decreases.

Step 1: Calculate the normal reaction at \(40^\circ\).

\[ N=mg\cos\theta \]

\[ N=5.0(9.8)\cos40^\circ \]

\[ \boxed{N=37.5\ \mathrm N} \]

Step 2: Calculate kinetic friction at \(40^\circ\).

\[ f_k=\mu_kN \]

\[ f_k=0.20(37.5) \]

\[ \boxed{f_k=7.51\ \mathrm N} \]

Step 3: Calculate acceleration at \(40^\circ\).

Take down the slope as positive.

The forces parallel to the incline are:

• Gravity’s component, \(mg\sin\theta\), down the slope.
• Kinetic friction, up the slope.

Newton’s second law gives:

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

Rearranging:

\[ a=\frac{mg\sin\theta-f_k}{m} \]

Substitute:

\[ a=9.8\sin40^\circ-0.20(9.8\cos40^\circ) \]

\[ \boxed{a=4.80\ \mathrm{m\,s^{-2}}} \]

Step 4: Calculate the normal reaction at \(55^\circ\).

\[ N=5.0(9.8)\cos55^\circ \]

\[ \boxed{N=28.1\ \mathrm N} \]

Step 5: Calculate kinetic friction at \(55^\circ\).

\[ f_k=0.20(28.1) \]

\[ \boxed{f_k=5.62\ \mathrm N} \]

Step 6: Calculate acceleration at \(55^\circ\).

\[ a=g(\sin\theta-\mu_k\cos\theta) \]

Substitute:

\[ a=9.8(\sin55^\circ-0.20\cos55^\circ) \]

\[ \boxed{a=6.90\ \mathrm{m\,s^{-2}}} \]

Step 7: Explain the changing forces.

As the incline becomes steeper, the normal reaction decreases because \(mg\cos\theta\) decreases.

This reduces kinetic friction.

At the same time, the downslope gravitational component increases because \(mg\sin\theta\) increases.

Both effects increase the resultant force down the slope.

Therefore, the block accelerates more strongly even though the frictional force becomes smaller.

Question 3

A student investigates kinetic friction by pulling a block uphill along an inclined plane using a force sensor.

The block has a mass of \(2.0\ \mathrm{kg}\). The incline is at \(30^\circ\), and the string is parallel to the surface.

The block is moving uphill and accelerating at \(0.50\ \mathrm{m\,s^{-2}}\). The force sensor reads \(13.0\ \mathrm N\).

Calculate:

a) The kinetic friction acting on the block.

b) The coefficient of kinetic friction.

c) Explain why using \(f_k=T-mg\sin\theta\) would produce an incorrect result in this situation.

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

Solution 3

The kinetic friction is \(2.20\ \mathrm N\), and the coefficient of kinetic friction is approximately \(0.13\).

The important detail is that the block is accelerating, so the resultant force is not zero.

Step 1: Choose a sign convention.

Take uphill as positive.

The forces parallel to the incline are:

• Tension, \(T\), acting uphill.
• The gravitational component, \(mg\sin\theta\), acting downhill.
• Kinetic friction, \(f_k\), acting downhill.

Step 2: Apply Newton’s second law.

\[ T-mg\sin\theta-f_k=ma \]

Rearrange:

\[ f_k=T-mg\sin\theta-ma \]

Substitute:

\[ f_k=13.0-2.0(9.8)\sin30^\circ -2.0(0.50) \]

\[ f_k=13.0-9.8-1.0 \]

\[ \boxed{f_k=2.20\ \mathrm N} \]

Step 3: Calculate the normal reaction.

The string is parallel to the incline, so it has no perpendicular component.

Therefore:

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

\[ N=2.0(9.8)\cos30^\circ \]

\[ N=16.97\ \mathrm N \]

Step 4: Calculate the coefficient of kinetic friction.

\[ f_k=\mu_kN \]

Rearrange:

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

Substitute:

\[ \mu_k=\frac{2.20}{16.97} \]

\[ \boxed{\mu_k=0.13} \]

Step 5: Explain the incorrect method.

The expression

\[ f_k=T-mg\sin\theta \]

assumes the block has zero acceleration.

Here, the block accelerates uphill, so some of the resultant force produces acceleration rather than balancing friction.

The student must include the \(ma\) term.

This is a common experimental-analysis mistake: assuming constant speed when the question explicitly gives a non-zero acceleration.

Question 4

A \(5.0\ \mathrm{kg}\) block rests on a rough incline at \(30^\circ\) to the horizontal.

A rope pulls the block uphill with a tension of \(20\ \mathrm N\). The rope makes an angle of \(30^\circ\) above the surface of the incline.

The coefficient of static friction is \(0.20\).

Determine whether the block can remain stationary.

If it cannot, explain which direction it will begin to move.

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

Solution 4

The block cannot remain stationary. It will begin to slide down the incline because the required static friction exceeds the maximum available static friction.

The key is that the angled rope reduces the normal reaction.

Step 1: Resolve the rope tension.

The rope acts at \(30^\circ\) above the incline.

Its parallel component is:

\[ T_{\parallel}=20\cos30^\circ \]

\[ T_{\parallel}=17.32\ \mathrm N \]

Its perpendicular component, directed away from the surface, is:

\[ T_{\perp}=20\sin30^\circ \]

\[ T_{\perp}=10.0\ \mathrm N \]

Step 2: Calculate the normal reaction.

Take away from the surface as positive.

The normal reaction balances the inward gravitational component, reduced by the outward rope component:

\[ N=mg\cos30^\circ-T\sin30^\circ \]

Substitute:

\[ N=5.0(9.8)\cos30^\circ-20\sin30^\circ \]

\[ N=42.44-10.0 \]

\[ \boxed{N=32.4\ \mathrm N} \]

Step 3: Calculate the static friction required.

The downslope component of gravity is:

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

\[ W_{\parallel}=5.0(9.8)\sin30^\circ \]

\[ W_{\parallel}=24.5\ \mathrm N \]

The rope pulls uphill with \(17.32\ \mathrm N\).

Without friction, the resultant force would be:

\[ F_{\mathrm{net}}=24.5-17.32 \]

\[ F_{\mathrm{net}}=7.18\ \mathrm N \]

down the slope.

Therefore, static friction would need to act uphill with a magnitude of \(7.18\ \mathrm N\) to keep the block stationary.

Step 4: Calculate maximum static friction.

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

\[ f_{s,\max}=0.20(32.4) \]

\[ \boxed{f_{s,\max}=6.49\ \mathrm N} \]

Step 5: Compare the forces.

The required static friction is:

\[ 7.18\ \mathrm N \]

The maximum available static friction is:

\[ 6.49\ \mathrm N \]

Since:

\[ 7.18>6.49 \]

static friction cannot prevent the block from moving.

The block therefore begins sliding downhill.

Notice the tempting mistake here.

If you had assumed \(N=mg\cos30^\circ\), you would have calculated a maximum static friction of approximately \(8.49\ \mathrm N\).

That would incorrectly suggest the block could remain stationary.

The rope’s perpendicular component changes the normal reaction, so ignoring it changes the conclusion.

Question 5

A \(2.0\ \mathrm{kg}\) block slides down a rough incline.

Initially, the incline is at \(30^\circ\), and the block moves at constant speed. There are no additional forces acting on the block.

The incline is then raised to \(60^\circ\).

An adjustable clamp can apply a force perpendicular into the incline. This clamp is adjusted so that the normal reaction remains equal to its original value at \(30^\circ\).

The block continues sliding downhill.

Answer the following:

a) Calculate the coefficient of kinetic friction.

b) Calculate the perpendicular force the clamp must apply to maintain the original normal reaction at \(60^\circ\).

c) Calculate the kinetic friction and acceleration while the clamp is applied.

d) The clamp is suddenly removed, without changing the incline angle. Calculate the new kinetic friction and acceleration.

e) Explain why the frictional force does not decrease when the incline is first raised, but does decrease when the clamp is removed.

Assume the coefficient of kinetic friction remains constant, the block stays in contact with the ramp, and \(g=9.8\ \mathrm{m\,s^{-2}}\).

Solution 5

The coefficient of kinetic friction is approximately \(0.577\).

The clamp must apply approximately \(7.17\ \mathrm N\) into the surface.

While the clamp is applied, kinetic friction remains \(9.80\ \mathrm N\), and the block accelerates down the slope at \(3.59\ \mathrm{m\,s^{-2}}\).

After removing the clamp, friction decreases to \(5.66\ \mathrm N\), and acceleration increases to \(5.66\ \mathrm{m\,s^{-2}}\).

This question demonstrates that the slope angle does not directly determine the frictional force. The normal reaction does.

Step 1: Use the initial constant-speed condition to find the coefficient of kinetic friction.

At \(30^\circ\), the block moves at constant speed.

Therefore:

\[ a=0 \]

The forces parallel to the incline balance:

\[ mg\sin30^\circ=f_k \]

Using:

\[ f_k=\mu_kN \]

and:

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

we obtain:

\[ mg\sin30^\circ =\mu_kmg\cos30^\circ \]

Cancel \(mg\):

\[ \mu_k=\frac{\sin30^\circ}{\cos30^\circ} \]

Therefore:

\[ \boxed{\mu_k=\tan30^\circ=0.577} \]

Step 2: Calculate the original normal reaction.

At \(30^\circ\):

\[ N_1=mg\cos30^\circ \]

\[ N_1=2.0(9.8)\cos30^\circ \]

\[ \boxed{N_1=16.97\ \mathrm N} \]

This is the normal reaction that the clamp must maintain at \(60^\circ\).

Step 3: Calculate the clamp force at \(60^\circ\).

Without the clamp, the normal reaction would be:

\[ N=mg\cos60^\circ \]

\[ N=2.0(9.8)(0.5) \]

\[ N=9.80\ \mathrm N \]

The clamp pushes the block into the surface, increasing the normal reaction.

Therefore:

\[ N_2=mg\cos60^\circ+P \]

We want:

\[ N_2=16.97\ \mathrm N \]

Substitute:

\[ 16.97=9.80+P \]

\[ \boxed{P=7.17\ \mathrm N} \]

The clamp must push into the surface with a force of \(7.17\ \mathrm N\).

Step 4: Calculate kinetic friction while the clamp is applied.

The normal reaction is being maintained at \(16.97\ \mathrm N\).

Therefore:

\[ f_k=\mu_kN \]

\[ f_k=0.577(16.97) \]

\[ \boxed{f_k=9.80\ \mathrm N} \]

This is exactly the same kinetic friction as before the ramp was raised.

The angle has increased, but the normal reaction has not changed.

Step 5: Calculate the acceleration with the clamp applied.

The component of weight down the slope is:

\[ mg\sin60^\circ \]

\[ =2.0(9.8)\sin60^\circ \]

\[ =16.97\ \mathrm N \]

The clamp acts perpendicular to the surface, so it has no component along the slope.

Taking downhill as positive:

\[ F_{\mathrm{net}}=16.97-9.80 \]

\[ F_{\mathrm{net}}=7.17\ \mathrm N \]

Apply Newton’s second law:

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

\[ a=\frac{7.17}{2.0} \]

\[ \boxed{a=3.59\ \mathrm{m\,s^{-2}}} \]

The block accelerates downhill even though its frictional force has remained unchanged.

Step 6: Remove the clamp and recalculate the normal reaction.

The ramp remains at \(60^\circ\), but the additional inward force disappears.

Therefore:

\[ N=mg\cos60^\circ \]

\[ \boxed{N=9.80\ \mathrm N} \]

Step 7: Calculate the new kinetic friction.

\[ f_k=\mu_kN \]

\[ f_k=0.577(9.80) \]

\[ \boxed{f_k=5.66\ \mathrm N} \]

The frictional force has now decreased because the normal reaction has decreased.

Step 8: Calculate the new acceleration.

The downslope gravitational component is unchanged:

\[ mg\sin60^\circ=16.97\ \mathrm N \]

The new frictional force is \(5.66\ \mathrm N\).

Therefore:

\[ F_{\mathrm{net}}=16.97-5.66 \]

\[ F_{\mathrm{net}}=11.31\ \mathrm N \]

Using Newton’s second law:

\[ a=\frac{11.31}{2.0} \]

\[ \boxed{a=5.66\ \mathrm{m\,s^{-2}}} \]

Step 9: Explain the result.

There are two separate changes.

When the ramp was raised from \(30^\circ\) to \(60^\circ\), the clamp increased the normal reaction to keep it constant.

Because kinetic friction depends on the normal reaction, friction stayed at \(9.80\ \mathrm N\).

When the clamp was removed, the normal reaction fell from \(16.97\ \mathrm N\) to \(9.80\ \mathrm N\).

Kinetic friction consequently fell from \(9.80\ \mathrm N\) to \(5.66\ \mathrm N\).

The important conclusion is that increasing the angle does not automatically reduce friction. It reduces friction when the change in angle causes the normal reaction to decrease, assuming the coefficient of friction stays constant.

The next step in these problems is to combine the normal reaction and friction equations with Newton’s second law along the incline.

For a sliding block, that gives you the acceleration equation used in How to Calculate Acceleration Down a Rough Incline.

The key decision comes first: work out the normal reaction correctly, then decide whether you need static or kinetic friction.