Why Is the Normal Force mg cos θ on a Simple Incline?

A 4.0 kg block sits on a ramp inclined at \(30^\circ\) to the horizontal. Its weight is \(39.2\text{ N}\). How hard does the ramp push back against it?

You might predict \(39.2\text{ N}\), because the block is supported by the ramp. But the actual normal reaction is only \(33.9\text{ N}\).

The reason is that the ramp only needs to balance the part of the block’s weight acting perpendicular to its surface. That component is \(mg\cos\theta\).

For a simple incline, this gives:

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

But why cosine rather than sine? And why does this equation sometimes stop working when another force is added?

Let’s build the result from the geometry and then test it in some less familiar situations.

01Why is the normal force \(mg\cos\theta\)?

The normal force, also called the normal reaction, is the contact force a surface exerts on an object. It always acts perpendicular to the surface.

This direction is important.

On a horizontal table, the normal force acts vertically upwards. Weight acts vertically downwards. These forces point in opposite directions, so if there is no vertical acceleration and no other vertical force, they balance:

\[ N=mg \]

An inclined plane changes the geometry. The normal force is no longer vertical, so it cannot directly balance the entire weight.

Instead, we need to resolve the weight into two components.

Choose axes along the slope

For an inclined-plane problem, choose:

• The parallel axis, running along the slope.
• The perpendicular axis, at \(90^\circ\) to the slope.

For consistency, we’ll take down the slope as positive along the parallel axis and away from the surface as positive along the perpendicular axis.

The weight \(mg\) acts vertically downwards. Its two components are:

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

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

Here:

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

Why does the perpendicular component use cosine?

Look at the right-angled triangle formed by the weight and its two components.

The full weight is the hypotenuse. The angle between the weight and the direction into the slope is \(\theta\).

The perpendicular component is therefore adjacent to this angle.

Using SOH CAH TOA:

\[ \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}} \]

Rearranging:

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

The parallel component is opposite the angle:

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

A useful way to picture this is to imagine the weight vector casting a shadow onto each of the two axes. The length of each shadow gives the corresponding component.

The components are not additional forces. They are two perpendicular descriptions of the same weight.

A common sine-versus-cosine mistake: If you accidentally measure the angle between the weight and the slope itself, rather than between the weight and the perpendicular direction, you’ll encounter the complementary angle. The same perpendicular component can then be written using sine. The geometry determines which ratio you use.

Apply Newton’s second law

We now know the component of weight pushing the block into the surface.

The normal reaction acts in the opposite direction.

Taking away from the surface as positive:

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

Therefore:

\[ N-mg\cos\theta=ma_{\perp} \]

For a block remaining in contact with a fixed, straight incline, there is no acceleration perpendicular to the surface.

This means:

\[ a_{\perp}=0 \]

So:

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

And finally:

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

That’s the complete reason.

The normal force equals \(mg\cos\theta\) because the perpendicular acceleration is zero, so the normal reaction balances the perpendicular component of weight.

Notice what we have not assumed: the block does not need to be stationary.

It could be sliding down the ramp and accelerating. Its acceleration is along the slope, not perpendicular to it.

This distinction is essential.

02What happens as the incline gets steeper?

Imagine gradually raising one end of a ramp while keeping the block’s mass unchanged.

Does the block’s weight increase?

No. Its weight remains \(mg\). What changes is how that weight is distributed between the two chosen directions.

The perpendicular component decreases as the ramp becomes steeper because \(\cos\theta\) decreases between \(0^\circ\) and \(90^\circ\).

The following values are theoretical predictions for a 2.0 kg block, using \(g=9.8\text{ m s}^{-2}\).

At \(0^\circ\), the ramp is horizontal. The normal reaction balances the entire weight.

At \(60^\circ\), only half the weight acts perpendicular to the ramp:

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

\[ N=0.5mg \]

At \(90^\circ\), the surface is vertical. Weight acts entirely parallel to the surface, so its perpendicular component is zero.

These are useful boundary checks. If your equation gives something different at \(0^\circ\) or \(90^\circ\), revisit your diagram.

03When does the normal reaction change?

The equation \(N=mg\cos\theta\) is not a universal formula for every object on every incline.

It relies on three conditions:

1. The incline is straight and stationary.
2. The object remains in contact with the surface, so its perpendicular acceleration is zero.
3. There are no additional forces with components perpendicular to the surface.

Change one of these conditions, and the normal reaction may change.

A force parallel to the slope

Suppose a rope pulls a block directly uphill, parallel to the incline.

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

Therefore:

\[ N=mg\cos\theta \]

The normal reaction is unchanged, even if the rope causes the block to accelerate uphill.

Friction also acts parallel to a straight incline, so it does not directly change the normal reaction.

A force pushing into the slope

Now suppose someone pushes the block directly into the surface with a force \(P\).

The surface must provide a larger normal reaction to balance both the perpendicular component of weight and the additional push.

Therefore:

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

The normal reaction increases.

A force pulling away from the slope

If a force pulls the block away from the surface, it helps support the block against gravity.

The normal reaction decreases:

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

This equation only applies while contact is maintained.

If the calculation gives a negative normal reaction, the assumed contact is impossible. A surface can push an object away, but it cannot pull it towards itself.

The object has lost contact, so \(N=0\), and its motion must be analysed again without the normal force.

What if the object accelerates perpendicular to the surface?

Return to Newton’s second law:

\[ N-mg\cos\theta=ma_{\perp} \]

If the perpendicular acceleration is not zero, then:

\[ N=mg\cos\theta+ma_{\perp} \]

For example, an object moving over a curved surface can have acceleration perpendicular to its instantaneous direction of motion. Its normal reaction must account for that acceleration.

This is why you should always check the acceleration before setting \(a_{\perp}=0\).

For a broader comparison involving lifts, angled forces, and other situations where the normal force differs from weight, see When Is the Normal Force Not Equal to Weight? HSC Physics.

04How could we test the relationship experimentally?

We have derived the prediction using Newton’s second law. But how could we measure the normal reaction and check whether it follows \(mg\cos\theta\)?

The important experimental decision is to measure the force perpendicular to the incline, rather than simply measuring the vertical support force.

The first experimental design

Imagine placing an entire inclined board, together with a block, on a bathroom scale.

The scale gives a reading, but does it directly measure the normal reaction on the block?

No.

The scale supports the board and its mounting as well as the forces transmitted through them. Its measurement is also along its own vertical sensing direction, not necessarily perpendicular to the inclined surface.

We need to isolate the force acting at the block’s contact with the ramp.

The revised experimental design

A better arrangement uses an adjustable inclined track and a single-axis force sensor.

The sensor is fitted underneath a small contact pad, flush with the inclined surface. Its sensitive axis is perpendicular to the ramp.

A block sits on the pad. A light cord runs from the block parallel to the slope and attaches to the upper end of the track. This holds the block in place without adding a perpendicular component of tension.

An angle gauge measures the incline angle, and the block’s mass is measured separately using a balance.

The measurement process is:

1. Measure the mass of the block.
2. Set the incline to a known angle.
3. Place the block on the force-sensor pad.
4. Use the parallel cord to prevent the block from sliding.
5. Record the normal force.
6. Increase the incline angle and repeat.

Because the cord is parallel to the incline, it does not contribute to the perpendicular force equation.

What should the experiment show?

For a 2.0 kg block, the theoretical normal reaction is:

\[ N=(2.0)(9.8)\cos\theta \]

The predicted readings are:

These are illustrative theoretical predictions, not recorded experimental measurements.

For example, at \(40^\circ\):

\[ N=19.6\cos40^\circ \]

\[ N=19.6(0.7660) \]

\[ N=15.0\text{ N} \]

The sensor should therefore register approximately \(15.0\text{ N}\).

If the readings follow the predicted cosine relationship within experimental uncertainty, they support the model.

There are some limitations:

• The force sensor must be calibrated correctly.
• Its sensitive axis must be perpendicular to the ramp.
• The contact pad must be flush with the surrounding surface.
• The angle must be measured accurately.
• Any sideways forces or misalignment can affect the reading.

The experiment tests the relationship directly, rather than relying on the block’s acceleration as an indirect measurement.

05Worked examples

06Questions and solutions

The following questions build from a straightforward normal-force calculation to situations where you must reconsider the assumptions behind \(N=mg\cos\theta\).

Question 1

A \(2.8\text{ kg}\) block rests on a fixed, smooth incline at \(35^\circ\) to the horizontal.

Calculate the normal reaction exerted by the incline.

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

Solution 1

The normal reaction is \(22.5\text{ N}\).

The block remains in contact with a fixed, straight incline, so it has no acceleration perpendicular to the surface.

The normal reaction must therefore balance the perpendicular component of weight.

\[ N=mg\cos\theta \]

Substituting:

\[ \begin{aligned} N&=(2.8)(9.8)\cos35^\circ\\ &=27.44(0.8192)\\ &=22.5\text{ N} \end{aligned} \]

The tempting answer is \(mg=27.44\text{ N}\), but that is the full weight, not its perpendicular component.

The incline only needs to balance the part of weight pushing directly into its surface.

Question 2

A \(4.0\text{ kg}\) block is pulled uphill along a fixed, smooth incline at \(25^\circ\) to the horizontal.

A light rope runs parallel to the incline. The block accelerates uphill at \(1.2\text{ m s}^{-2}\).

Calculate:

(a) The normal reaction.

(b) The tension in the rope.

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

Solution 2

The normal reaction is \(35.5\text{ N}\), and the rope tension is \(21.4\text{ N}\).

The key is that the rope pulls parallel to the incline. It changes the forces along the slope but has no perpendicular component.

Part (a): Normal reaction

Using the perpendicular direction:

\[ N-mg\cos25^\circ=0 \]

Therefore:

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

Substitute:

\[ \begin{aligned} N&=(4.0)(9.8)\cos25^\circ\\ &=39.2(0.9063)\\ &=35.5\text{ N} \end{aligned} \]

Part (b): Rope tension

Now take uphill as positive.

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

Newton’s second law gives:

\[ T-mg\sin25^\circ=ma \]

Rearranging:

\[ T=ma+mg\sin25^\circ \]

Substitute:

\[ \begin{aligned} T&=(4.0)(1.2)+39.2\sin25^\circ\\ &=4.8+16.6\\ &=21.4\text{ N} \end{aligned} \]

The block accelerates because the parallel forces are unbalanced.

However, its perpendicular acceleration remains zero. That is why its normal reaction is still \(mg\cos25^\circ\).

Question 3

A \(4.0\text{ kg}\) block slides down a rough, fixed incline.

A force sensor measures a normal reaction of \(31.3\text{ N}\). Motion analysis shows that the block accelerates down the incline at \(3.0\text{ m s}^{-2}\).

Calculate:

(a) The angle of the incline.

(b) The magnitude and direction of the kinetic frictional force.

Assume friction acts opposite to the block’s motion and take \(g=9.8\text{ m s}^{-2}\).

Solution 3

The incline angle is approximately \(37.0^\circ\), and the kinetic frictional force is \(11.6\text{ N}\) uphill.

This question requires us to use the normal reaction to determine the incline angle before analysing the motion along the slope.

Part (a): Find the incline angle

The block remains in contact with a fixed, straight incline, so:

\[ N=mg\cos\theta \]

Rearrange:

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

Substitute:

\[ \cos\theta=\frac{31.3}{(4.0)(9.8)} \]

\[ \cos\theta=\frac{31.3}{39.2} \]

\[ \theta=\cos^{-1}(0.7985) \]

\[ \theta\approx37.0^\circ \]

Part (b): Find the frictional force

The block slides downhill, so kinetic friction acts uphill.

Take downhill as positive.

The parallel forces are:

• \(mg\sin37^\circ\), acting downhill.
• \(f\), acting uphill.

Newton’s second law gives:

\[ mg\sin37^\circ-f=ma \]

Rearranging:

\[ f=mg\sin37^\circ-ma \]

Substitute:

\[ \begin{aligned} f&=39.2\sin37^\circ-(4.0)(3.0)\\ &=23.6-12.0\\ &=11.6\text{ N} \end{aligned} \]

Therefore, friction acts uphill with a magnitude of \(11.6\text{ N}\).

Notice the important distinction: friction changes the block’s acceleration along the slope, but it does not change the normal reaction because friction acts parallel to the surface.

Question 4

A \(0.80\text{ kg}\) puck travels over the top of a smooth, convex curved track.

At the highest point, the track is locally horizontal. Its radius of curvature is \(2.0\text{ m}\), and the puck’s speed is \(4.0\text{ m s}^{-1}\).

Calculate:

(a) The normal reaction exerted by the track at the highest point.

(b) The maximum speed the puck can have at this point while maintaining contact with the track.

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

Solution 4

The normal reaction is \(1.44\text{ N}\), and the maximum speed before losing contact is \(4.43\text{ m s}^{-1}\).

This question exposes an important limitation of the simple-incline equation.

At the top of the curved track, the surface is locally horizontal. However, the puck has centripetal acceleration directed downwards, towards the centre of curvature.

Therefore, its perpendicular acceleration is not zero.

Part (a): Calculate the normal reaction

At the highest point:

• Weight acts downwards.
• The normal reaction acts upwards.
• Centripetal acceleration acts downwards.

Taking downwards as positive:

\[ mg-N=\frac{mv^2}{r} \]

Rearrange:

\[ N=mg-\frac{mv^2}{r} \]

Substitute:

\[ \begin{aligned} N&=(0.80)(9.8)-\frac{(0.80)(4.0)^2}{2.0}\\ &=7.84-6.40\\ &=1.44\text{ N} \end{aligned} \]

The normal reaction is \(1.44\text{ N}\).

Part (b): Find the maximum speed

The puck loses contact when the normal reaction becomes zero.

At this instant, gravity provides the entire centripetal force.

Therefore:

\[ mg=\frac{mv_{\max}^2}{r} \]

Cancel the mass:

\[ g=\frac{v_{\max}^2}{r} \]

Rearrange:

\[ v_{\max}=\sqrt{gr} \]

Substitute:

\[ \begin{aligned} v_{\max}&=\sqrt{(9.8)(2.0)}\\ &=\sqrt{19.6}\\ &=4.43\text{ m s}^{-1} \end{aligned} \]

The puck must travel at no more than approximately \(4.43\text{ m s}^{-1}\) at the crest to maintain contact in this ideal model.

Why can’t we simply use \(N=mg\) because the track is locally horizontal?

Because the puck is accelerating towards the centre of curvature. The zero-perpendicular-acceleration assumption used for a straight incline no longer applies.

Question 5

A \(4.0\text{ kg}\) block rests on a smooth, fixed incline. The incline can be adjusted to any angle \(\theta\) between \(0^\circ\) and \(90^\circ\).

The ramp rises towards the right. A constant horizontal force of \(39.2\text{ N}\) pushes the block towards the right.

The block is released from rest at each chosen incline angle.

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

(a) Derive an expression for the normal reaction \(N\) in terms of \(\theta\).

(b) Derive an expression for the block’s acceleration along the slope, taking uphill as positive.

(c) Find the incline angle at which the acceleration along the slope is zero.

(d) Prove that the normal reaction is greatest at this same angle. Calculate its maximum value and compare it with the normal reactions at \(0^\circ\) and \(90^\circ\).

Solution 5

The normal reaction is greatest at \(45^\circ\), where it reaches \(55.4\text{ N}\). This is also the angle at which the block has zero acceleration along the slope.

The surprising part is that the normal reaction does not simply decrease as the ramp becomes steeper.

The horizontal force changes its perpendicular component as the angle changes. We must account for both forces, not just weight.

Part (a): Derive the normal reaction

First resolve the weight:

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

Now resolve the horizontal force.

Because the force acts towards the right while the ramp rises towards the right, its components are:

\[ F_{\perp}=F\sin\theta \]

directed into the surface, and:

\[ F_{\parallel}=F\cos\theta \]

directed uphill.

Taking away from the surface as positive:

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

Therefore:

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

We know:

\[ mg=(4.0)(9.8)=39.2\text{ N} \]

and \(F=39.2\text{ N}\).

So:

\[ \boxed{N=39.2(\cos\theta+\sin\theta)} \]

Part (b): Derive the acceleration

Take uphill as positive.

The horizontal force contributes \(F\cos\theta\) uphill, while gravity contributes \(mg\sin\theta\) downhill.

Applying Newton’s second law:

\[ F\cos\theta-mg\sin\theta=ma \]

Substitute \(F=mg=39.2\text{ N}\):

\[ 39.2\cos\theta-39.2\sin\theta=4.0a \]

Therefore:

\[ \boxed{a=9.8(\cos\theta-\sin\theta)} \]

This expression tells us the direction of acceleration at every incline angle.

Part (c): Find the angle at which acceleration is zero

For zero acceleration:

\[ 9.8(\cos\theta-\sin\theta)=0 \]

Therefore:

\[ \cos\theta=\sin\theta \]

Dividing by \(\cos\theta\):

\[ \tan\theta=1 \]

Hence:

\[ \boxed{\theta=45^\circ} \]

At this angle, the uphill component of the horizontal force exactly balances the downhill component of weight.

If released from rest, the block remains stationary along the incline.

Part (d): Prove that the normal reaction is greatest at this angle

We found:

\[ N=39.2(\cos\theta+\sin\theta) \]

We need to show that this expression is largest at \(45^\circ\).

Use the trigonometric identity:

\[ \sin^2\theta+\cos^2\theta=1 \]

Squaring the expression:

\[ \begin{aligned} (\cos\theta+\sin\theta)^2 &=\cos^2\theta+\sin^2\theta\\ &\quad+2\sin\theta\cos\theta\\ &=1+\sin2\theta \end{aligned} \]

Since the maximum value of \(\sin2\theta\) is 1:

\[ (\cos\theta+\sin\theta)^2\leq2 \]

Therefore:

\[ \cos\theta+\sin\theta\leq\sqrt{2} \]

The maximum normal reaction is:

\[ N_{\max}=39.2\sqrt{2} \]

\[ \boxed{N_{\max}=55.4\text{ N}} \]

Equality occurs when:

\[ \sin2\theta=1 \]

which gives:

\[ 2\theta=90^\circ \]

Therefore:

\[ \boxed{\theta=45^\circ} \]

The maximum normal reaction occurs at exactly the same angle as zero acceleration along the slope.

Compare the normal reaction at the endpoints.

At \(0^\circ\):

\[ N=39.2(\cos0^\circ+\sin0^\circ) \]

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

At \(90^\circ\):

\[ N=39.2(\cos90^\circ+\sin90^\circ) \]

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

So the normal reaction is \(39.2\text{ N}\) at both endpoints, but reaches \(55.4\text{ N}\) at \(45^\circ\).

This is the key conceptual result.

A student who looks only at \(mg\cos\theta\) might confidently predict that the normal reaction must decrease as the ramp gets steeper.

That prediction ignores the horizontal force.

As the incline becomes steeper, the perpendicular component of the horizontal push increases. At \(45^\circ\), both forces have equal parallel components in opposite directions, while their perpendicular components add together.

The combined effect is a larger normal reaction than the block’s weight.

07Where to go next

The most useful habit to take from this topic is to separate the forces into two directions before deciding what the normal reaction should be.

For a simple straight incline, the perpendicular acceleration is zero, giving \(N=mg\cos\theta\). But an additional perpendicular force or acceleration changes that result.

Once the perpendicular equation is established, use the parallel equation to investigate acceleration, tension, and friction.

For the next step into curved motion, Banked Tracks: Normal Force and Design Speed in HSC Physics explores what happens when the normal force also provides the acceleration needed to change an object’s direction.