Can a Block Stay at Rest on a Frictionless Slope?

Release a block on a shallow ramp and it may not move. Does that mean gravity has no force pulling it down the slope? No. The ramp might be supplying static friction to hold it in place.

But what happens if the ramp is perfectly frictionless?

A block cannot remain at rest on a non-horizontal frictionless slope under gravity alone. It accelerates down the slope because gravity has a component parallel to the surface, and a normal reaction cannot balance that component.

There is one important qualification: an additional force, such as tension in a string, can hold the block stationary even without friction.

Understanding why requires separating the weight into two components. It also requires knowing when the normal reaction changes, rather than automatically assuming it equals \(mg\).

01Why a frictionless slope cannot hold a block at rest

Imagine placing a block on a ramp inclined at \(30^\circ\) to the horizontal.

The block experiences two forces:

• Its weight, \(mg\), acting vertically downwards.
• The normal reaction, \(N\), exerted by the ramp perpendicular to its surface.

Because the surface is frictionless, there is no frictional force acting parallel to the ramp.

Before doing any calculations, consider this: could the normal reaction balance the entire weight?

It cannot. The normal reaction always acts perpendicular to the surface. On an inclined plane, that direction is not vertical.

The normal reaction can only balance the component of weight perpendicular to the ramp. The remaining component acts down the slope.

Resolve weight using axes along the slope

The easiest way to analyse an inclined-plane problem is to choose coordinate axes that match the surface.

Use:

• The parallel axis, along the slope, with positive direction downwards.
• The perpendicular axis, at right angles to the slope, with positive direction away from the surface.

The block’s weight acts vertically downwards, but we can resolve it into two perpendicular components.

These are:

1. The component parallel to the slope: \(mg\sin\theta\).
2. The component perpendicular to the slope: \(mg\cos\theta\).

Here:

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

Why does the sine component act along the slope?

Think of the weight and its two components as forming a right-angled triangle. The angle between the weight vector and the direction into the plane is \(\theta\).

The component opposite that angle is \(mg\sin\theta\), while the adjacent component is \(mg\cos\theta\).

This gives us the two equations needed to analyse the block.

What happens parallel to the slope?

Newton’s second law states:

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

There is no friction, and we are initially assuming there are no other applied forces.

Therefore, the only force parallel to the slope is the downslope component of weight.

\[
\sum F_{\parallel}=mg\sin\theta
\]

Applying Newton’s second law:

\[
mg\sin\theta=ma
\]

Cancel the mass:

\[
\boxed{a=g\sin\theta}
\]

This is the acceleration of a block sliding down a frictionless straight incline under gravity alone.

Notice something interesting: the mass disappears.

A 2 kg block and a 20 kg block released from rest on the same frictionless incline have the same acceleration.

The heavier block experiences a larger downslope gravitational force, but it also has proportionally more inertia. These effects cancel.

Most importantly, if \(\theta\) is greater than zero, then \(g\sin\theta\) is also greater than zero.

The block must accelerate down the slope. It cannot remain at rest.

What happens perpendicular to the slope?

Now consider the perpendicular direction.

The block does not accelerate through the surface or away from it. It remains in contact with the straight, stationary ramp.

Therefore, its perpendicular acceleration is zero.

The perpendicular forces must balance:

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

Taking away from the plane as positive:

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

Rearranging:

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

This is the normal reaction for a block on a straight incline when gravity is the only other force with a perpendicular component.

Notice that the normal reaction is not equal to the full weight.

It only balances the perpendicular component of weight.

The parallel component remains unbalanced and accelerates the block down the slope.

02When does the normal reaction change?

This is where students often make a mistake.

They learn that the normal reaction on a horizontal surface is \(N=mg\), then automatically use the same expression for every problem.

But \(N=mg\) is only correct in particular circumstances.

The normal reaction depends on the forces acting perpendicular to the surface, not simply on the block’s weight.

1. Changing the slope angle

For a block on a frictionless incline with no additional forces:

\[
N=mg\cos\theta
\]

As the slope becomes steeper, the cosine decreases.

Therefore, the normal reaction decreases.

Consider a 5 kg block. Its weight is:

\[
mg=5(9.8)=49\text{ N}
\]

The following table shows how its normal reaction changes as the slope angle increases.

These are theoretical values calculated using \(g=9.8\ \text{m s}^{-2}\), not experimental measurements.

There are two important trends:

• As the slope gets steeper, acceleration down the slope increases.
• As the slope gets steeper, the normal reaction decreases.

The block’s weight does not change. Only its components along the chosen axes change.

2. Adding a force perpendicular to the slope

Suppose someone pushes the block directly into the surface.

The ramp must now provide a larger normal reaction to prevent the block from moving through it.

If an additional perpendicular force \(P\) pushes into the plane:

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

The normal reaction increases.

Now imagine pulling the block away from the surface instead.

The perpendicular component of the pulling force acts in the same direction as the normal reaction, so the ramp does not need to push as hard.

The normal reaction decreases.

For a force \(P\) pulling at an angle \(\alpha\) above the slope:

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

This assumes the block remains in contact with a straight, stationary incline and has no perpendicular acceleration.

The important point is that a force can affect the normal reaction without directly changing the component of weight along the slope.

3. Adding a force parallel to the slope

What if a string pulls the block directly uphill, parallel to the surface?

This changes the forces along the slope, but it has no perpendicular component.

Therefore:

\[
N=mg\cos\theta
\]

The normal reaction remains unchanged.

If the tension is exactly equal to \(mg\sin\theta\), the block can remain stationary despite the absence of friction.

This is an important exception to the opening question.

A block can remain at rest on a frictionless slope, but it needs another force to balance the component of gravity pulling it downhill.

A useful comparison

There is one final restriction.

The normal reaction is a contact force. A surface can push an object away from itself, but it cannot pull the object towards itself.

Therefore, \(N\) cannot be negative.

If your calculation produces a negative normal reaction, the block has lost contact with the surface. You must reconsider the motion without assuming that the block remains on the incline.

03How could we test this experimentally?

So far, we have used Newton’s laws to predict what should happen.

But how could we distinguish a genuinely frictionless slope from an ordinary ramp that simply has very little friction?

Start with a wooden board.

Place a block on it and slowly increase the angle. At a small angle, the block might remain stationary.

A tempting conclusion would be that gravity has no component pulling the block downhill.

That conclusion is wrong.

The board may be providing static friction, which balances the downslope component of gravity. The block can remain stationary even though gravity is pulling it down the slope.

This is why an ordinary wooden ramp is not a good way to test the frictionless model.

For a more controlled experiment, we can use an air track.

The revised experimental design

An air track is a long, slightly inclined track with many small holes along its surface.

A compressor supplies air through these holes. The escaping air creates a thin cushion that supports a glider above the track, greatly reducing contact friction.

The experimental arrangement works as follows:

1. Raise one end of the air track to create a measured angle \(\theta\).
2. Place an air-track glider at the upper end.
3. Release the glider from rest without giving it an initial push.
4. Allow it to travel down the track.
5. Use two photogates, separated by a known distance, to measure its speed at two positions.

A small flag of known length attached to the glider passes through each photogate. The time taken to interrupt the light beam allows the speed at that position to be calculated.

What should the experiment show?

If the air track behaves approximately like a frictionless surface, the glider should accelerate down the track.

Our theoretical prediction is:

\[
a=g\sin\theta
\]

For example, suppose the track is inclined at \(10^\circ\).

The predicted acceleration is:

\[
a=9.8\sin10^\circ
\]

\[
\boxed{a=1.70\ \text{m s}^{-2}}
\]

We can use this prediction to estimate the glider’s speed after travelling 0.50 m from rest.

Using the constant-acceleration equation:

\[
v^2=u^2+2as
\]

where \(u\) is initial velocity, \(v\) is final velocity, and \(s\) is displacement:

\[
v^2=0+2(1.70)(0.50)
\]

\[
v=\sqrt{1.70}
\]

\[
\boxed{v=1.30\ \text{m s}^{-1}}
\]

The glider should reach approximately \(1.30\ \text{m s}^{-1}\) after travelling 0.50 m.

In the actual experiment, the measured acceleration may be slightly lower because the air track does not eliminate every source of resistance.

We could calculate the acceleration between the photogates using:

\[
a=\frac{v^2-u^2}{2s}
\]

The measured acceleration can then be compared with the theoretical prediction.

If we instead record a velocity-time graph, its gradient gives the acceleration. This connects to the techniques covered in Velocity-Time Graphs: Slope, Area and Motion Explained.

What does this experiment tell us?

The key result is that the glider accelerates down the incline even when contact friction has been greatly reduced.

Increasing the angle should increase the acceleration, approximately following \(a=g\sin\theta\).

The normal reaction still exists, but it acts perpendicular to the track. It cannot cancel the downslope component of gravity.

The experiment also has limitations:

• Residual friction and air resistance may reduce the measured acceleration.
• An inaccurate incline angle produces an inaccurate theoretical prediction.
• The track must be properly aligned so that the glider does not experience additional sideways forces.

The air track therefore provides evidence consistent with the frictionless model. It does not create a perfectly frictionless surface.

04A reliable method for solving frictionless incline problems

When you encounter a block on a smooth slope in an HSC Physics question, follow this sequence.

Step 1: Draw the forces.

Include the weight \(mg\), the normal reaction \(N\), and any additional applied forces.

Do not draw friction on a frictionless surface.

Step 2: Choose axes along and perpendicular to the slope.

This avoids unnecessary trigonometry later.

Step 3: Resolve the weight.

Use:

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

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

Step 4: Apply Newton’s second law in both directions.

For a stationary straight incline with the block remaining in contact:

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

For motion along the slope:

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

Step 5: Check the normal reaction.

Make sure your calculated normal reaction is non-negative.

If it is negative, the block cannot remain in contact under the assumed conditions.

05Worked examples

06Questions and solutions

Question 1

A 2.0 kg block is released from rest on a frictionless incline at \(20^\circ\) to the horizontal.

Calculate:

(a) Its acceleration down the slope.

(b) The normal reaction exerted by the incline.

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

Solution 1

The block accelerates down the slope at \(3.35\ \text{m s}^{-2}\), and the normal reaction is \(18.4\text{ N}\).

Part (a): Acceleration

The only force parallel to the slope is the component of weight:

\[
mg\sin\theta=ma
\]

Cancel the mass:

\[
a=g\sin\theta
\]

Substitute:

\[
a=9.8\sin20^\circ
\]

\[
\boxed{a=3.35\ \text{m s}^{-2}}
\]

The acceleration is independent of the block’s mass.

Part (b): Normal reaction

There is no perpendicular acceleration, so:

\[
N=mg\cos\theta
\]

Substitute:

\[
N=2.0(9.8)\cos20^\circ
\]

\[
N=19.6(0.940)
\]

\[
\boxed{N=18.4\text{ N}}
\]

The normal reaction is smaller than the weight of 19.6 N because it balances only the perpendicular component of weight.

Question 2

A 5.0 kg block is stationary on a frictionless incline at \(30^\circ\) to the horizontal.

A light string runs parallel to the incline and holds the block in place.

A separate actuator also pushes the block directly into the surface with a perpendicular force of 8.0 N.

Calculate:

(a) The tension in the string.

(b) The normal reaction.

(c) The block’s initial acceleration if the string suddenly breaks, assuming the perpendicular 8.0 N force continues to act.

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

Solution 2

The tension is \(24.5\text{ N}\), the normal reaction is \(50.4\text{ N}\), and after the string breaks the block accelerates down the slope at \(4.90\ \text{m s}^{-2}\).

Part (a): Tension

Because the block is stationary, the net force parallel to the slope must be zero.

The tension acts uphill, while the parallel component of weight acts downhill.

Therefore:

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

\[
T=5.0(9.8)(0.5)
\]

\[
\boxed{T=24.5\text{ N}}
\]

Part (b): Normal reaction

The perpendicular component of weight is:

\[
mg\cos30^\circ
\]

The actuator adds another 8.0 N into the surface.

Therefore, the normal reaction must balance both forces:

\[
N=mg\cos30^\circ+8.0
\]

\[
N=5.0(9.8)(0.866)+8.0
\]

\[
N=42.4+8.0
\]

\[
\boxed{N=50.4\text{ N}}
\]

Notice that the tension does not appear in this equation.

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

Part (c): When the string breaks

Once the string breaks, the tension disappears.

The remaining force parallel to the slope is:

\[
F_{\text{net}}=mg\sin30^\circ
\]

Using Newton’s second law:

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

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

\[
\boxed{a=4.90\ \text{m s}^{-2}}
\]

The 8.0 N perpendicular push remains, so the normal reaction stays at 50.4 N.

This is an important distinction: removing a force parallel to the surface changes the acceleration along the surface, but does not necessarily change the normal reaction.

Question 3

A 4.0 kg block sits on a frictionless incline at \(30^\circ\) to the horizontal. The incline rises towards the right.

A horizontal force of 10 N pushes the block towards the right, in the uphill direction.

Calculate:

(a) The block’s acceleration along the incline.

(b) The normal reaction.

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

Solution 3

The block accelerates down the incline at \(2.74\ \text{m s}^{-2}\), and the normal reaction is \(39.0\text{ N}\).

The important detail is that the applied force is horizontal, not parallel to the slope.

We must resolve it into components.

Step 1: Resolve the horizontal force.

The incline is at \(30^\circ\) to the horizontal.

The component of the applied force parallel to the slope is:

\[
P_{\parallel}=10\cos30^\circ
\]

\[
P_{\parallel}=8.66\text{ N}
\]

This acts uphill.

The perpendicular component is:

\[
P_{\perp}=10\sin30^\circ
\]

\[
P_{\perp}=5.0\text{ N}
\]

This component pushes the block into the surface.

Step 2: Calculate the acceleration.

The downslope component of weight is:

\[
mg\sin30^\circ
=4.0(9.8)(0.5)
=19.6\text{ N}
\]

The net force down the slope is:

\[
F_{\text{net}}=19.6-8.66
\]

\[
F_{\text{net}}=10.94\text{ N}
\]

Using Newton’s second law:

\[
10.94=4.0a
\]

\[
\boxed{a=2.74\ \text{m s}^{-2}}
\]

Step 3: Calculate the normal reaction.

The weight component perpendicular to the incline is:

\[
mg\cos30^\circ
=4.0(9.8)(0.866)
\]

\[
=33.95\text{ N}
\]

The horizontal push adds another 5.0 N into the surface.

Therefore:

\[
N=33.95+5.0
\]

\[
\boxed{N=38.95\text{ N}}
\]

Step 4: Explain the result.

A common mistake would be to calculate \(N=mg\cos30^\circ\) and stop there.

That would ignore the perpendicular component of the horizontal force.

The correct normal reaction is approximately 39.0 N because the applied force pushes the block harder against the surface.

Question 4

An air-track glider of mass 0.60 kg travels down a track inclined at \(12^\circ\) to the horizontal.

In an illustrative experimental trial, its measured acceleration is \(1.75\ \text{m s}^{-2}\).

Assume that the only departure from the ideal frictionless model is a small resistive force acting parallel to the track and uphill.

Calculate:

(a) The theoretical acceleration predicted by the frictionless model.

(b) The magnitude of the resistive force.

(c) The normal reaction.

A student argues that the normal reaction must be responsible for the difference between the theoretical and measured accelerations.

Explain why this argument is incorrect.

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

Solution 4

The theoretical acceleration is \(2.04\ \text{m s}^{-2}\), the resistive force is approximately \(0.17\text{ N}\), and the normal reaction is \(5.75\text{ N}\).

The student’s explanation is incorrect because the normal reaction acts perpendicular to the track, not parallel to it.

Part (a): Theoretical acceleration

For the ideal frictionless model:

\[
a=g\sin\theta
\]

Substituting:

\[
a=9.8\sin12^\circ
\]

\[
\boxed{a=2.04\ \text{m s}^{-2}}
\]

The measured acceleration is lower than this prediction.

Part (b): Resistive force

Apply Newton’s second law parallel to the track.

Taking downhill as positive:

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

Rearrange:

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

Substitute:

\[
f=0.60(9.8)\sin12^\circ-0.60(1.75)
\]

\[
f=1.22-1.05
\]

\[
\boxed{f=0.17\text{ N}}
\]

The resistive force acts uphill and reduces the net force accelerating the glider downhill.

Part (c): Normal reaction

Assuming there is no acceleration perpendicular to the track:

\[
N=mg\cos\theta
\]

Substitute:

\[
N=0.60(9.8)\cos12^\circ
\]

\[
\boxed{N=5.75\text{ N}}
\]

Why the student’s argument is incorrect

The normal reaction acts perpendicular to the track.

It cannot explain a reduction in the acceleration along the track because it has no component in that direction.

The reduction is explained by the resistive force acting uphill.

This is a useful experimental distinction: a force can affect the motion in one direction without contributing to the force balance in another direction.

Question 5

A 2.0 kg block is on a frictionless incline at \(30^\circ\) to the horizontal.

A light string pulls the block uphill and away from the surface. The string makes an angle \(\alpha\) above the incline.

The tension is adjusted so that the block remains stationary while it is in contact with the incline.

Answer the following:

(a) Derive an expression for the tension \(T\) in terms of \(\alpha\).

(b) Derive an expression for the normal reaction \(N\) in terms of \(\alpha\).

(c) Calculate the angle \(\alpha\) at which the normal reaction becomes zero.

(d) A student claims that if \(\alpha\) increases beyond this angle, the incline can provide a negative normal reaction to keep the block stationary. Explain why this is impossible.

(e) At the exact angle found in part (c), does the block necessarily have to leave the surface immediately? Justify your answer.

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

Solution 5

The normal reaction becomes zero at \(\alpha=60^\circ\). At exactly this angle, the block can remain stationary with zero normal reaction. At any larger angle, the assumed equilibrium would require a negative normal reaction, so the block cannot remain in contact with the incline under the stated conditions.

This problem is more subtle than simply resolving weight into two components. We must consider how the tension affects both directions simultaneously.

Part (a): Derive the tension

Take uphill as the positive parallel direction.

The block is stationary, so its net force along the incline must be zero.

The uphill component of tension is:

\[
T\cos\alpha
\]

The downhill component of weight is:

\[
mg\sin30^\circ
\]

Therefore:

\[
T\cos\alpha-mg\sin30^\circ=0
\]

Rearrange:

\[
T\cos\alpha=mg\sin30^\circ
\]

\[
T=\frac{mg\sin30^\circ}{\cos\alpha}
\]

Substitute the mass and gravitational acceleration:

\[
\boxed{T=\frac{9.8}{\cos\alpha}\text{ N}}
\]

The required tension increases as the string becomes more perpendicular to the incline.

Why?

Because only the component \(T\cos\alpha\) acts uphill. As \(\alpha\) increases, this component becomes a smaller fraction of the total tension.

A larger tension is therefore needed to balance gravity along the slope.

Part (b): Derive the normal reaction

Now consider the perpendicular direction.

The normal reaction acts away from the surface.

The perpendicular component of tension also acts away from the surface.

The perpendicular component of weight acts into the surface.

Since the block has no perpendicular acceleration:

\[
N+T\sin\alpha-mg\cos30^\circ=0
\]

Rearrange:

\[
N=mg\cos30^\circ-T\sin\alpha
\]

Substitute the expression for tension from part (a):

\[
N=mg\cos30^\circ-
\frac{mg\sin30^\circ}{\cos\alpha}
\sin\alpha
\]

Simplify:

\[
N=mg\cos30^\circ-mg\sin30^\circ\tan\alpha
\]

Factor out \(mg\):

\[
\boxed{
N=mg(\cos30^\circ-\sin30^\circ\tan\alpha)
}
\]

For this 2.0 kg block:

\[
\boxed{
N=19.6(0.866-0.5\tan\alpha)
}
\]

This equation describes how the normal reaction changes as the string’s angle increases.

Part (c): Find the angle at which the normal reaction becomes zero

At the limiting angle:

\[
N=0
\]

Therefore:

\[
mg(\cos30^\circ-\sin30^\circ\tan\alpha)=0
\]

Divide by \(mg\):

\[
\cos30^\circ-\sin30^\circ\tan\alpha=0
\]

Rearrange:

\[
\sin30^\circ\tan\alpha=\cos30^\circ
\]

\[
\tan\alpha=
\frac{\cos30^\circ}{\sin30^\circ}
\]

\[
\tan\alpha=\cot30^\circ
\]

Therefore:

\[
\tan\alpha=\sqrt{3}
\]

\[
\boxed{\alpha=60^\circ}
\]

The normal reaction reaches zero when the string is at \(60^\circ\) above the incline.

At this angle, the tension is:

\[
T=\frac{9.8}{\cos60^\circ}
\]

\[
T=\frac{9.8}{0.5}
\]

\[
\boxed{T=19.6\text{ N}}
\]

Interestingly, the string is now vertical because the incline is at \(30^\circ\) to the horizontal.

The tension is exactly equal to the block’s weight.

Part (d): Can the normal reaction become negative?

No. A negative normal reaction is physically impossible for an ordinary contact surface.

A normal reaction is a pushing force. The incline cannot pull the block towards itself.

For angles greater than \(60^\circ\), the equation derived in part (b) predicts:

\[
N<0
\]

This tells us that our assumption of continued contact is no longer valid.

The tension now has such a large perpendicular component that the incline would need to pull the block towards itself to maintain equilibrium.

Since it cannot do that, the block loses contact.

With the tension adjusted to balance the parallel component of gravity, the remaining net force is directed away from the surface.

This is the key physical meaning of a negative normal reaction in a calculation: it signals that the assumed contact condition has failed.

Part (e): Must the block leave the surface at exactly \(60^\circ\)?

No. At exactly \(60^\circ\), the block can remain stationary with zero normal reaction.

At this limiting angle, the string is vertical and its tension equals the block’s weight:

\[
T=mg
\]

The tension balances gravity completely.

There is no net force, and the incline does not need to exert any normal reaction.

The block can therefore remain stationary while just touching the surface.

This is different from angles greater than \(60^\circ\), where the required equilibrium would demand a negative normal reaction.

The distinction is subtle but important: zero normal reaction does not automatically mean that an object must immediately move away from a surface.

It means that the surface is exerting no contact force. Whether the object separates depends on the other forces acting on it.

07Where to go next

The central idea is simple: a frictionless incline still exerts a normal reaction, but that reaction only acts perpendicular to the surface.

Under gravity alone:

\[
\boxed{a=g\sin\theta}
\]

and

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

An additional force parallel to the slope can hold a block stationary without changing the normal reaction. A force perpendicular to the slope changes the normal reaction, and a sufficiently large outward force can cause the block to lose contact.

The natural next step is to introduce static friction and compare the force required to prevent sliding with the maximum friction available.

That is the difference between an ideal frictionless incline and a real ramp that can actually hold a block in place. See Static and Kinetic Friction Explained for HSC Physics for the next part of that analysis.