The inclined plane, to tilt with your hands.

1. Step 1 · The forces

   First: the force diagram — which forces are really there

   Before any calculation you draw the forces acting on the body, all with their tail on the body. There are two. The weight P = m·g, vertical and pointing down: always vertical, never “tilted like the plane” — the Earth does not know there is a ramp underneath. And the normal reaction N, with which the plane pushes back, perpendicular to the surface and pointing outwards. Nothing else: in this first pass the plane is smooth, we add friction at step 5 and say so, instead of letting it disappear silently. If you cannot say WHO exerts a force, that force should not be drawn.

2. Step 2 · The axes

   Second: rotate the axes by θ — this is where it all happens

   The axes are usually horizontal and vertical. Here it is convenient to take them rotated by θ: x′ along the plane, pointing down the slope, and y′ perpendicular to the plane, pointing outwards. Watch the rotation: they are the same axes as always, turned by exactly the angle of the plane. Why is it convenient? Because in this system the motion happens along ONE axis only — along y′ the body never moves, it neither sinks in nor lifts off — and a two-dimensional problem becomes two one-dimensional problems. It is not a shortcut: it is the choice that turns the inclined plane into a ten-line exercise instead of a thirty-line one.

3. Step 3 · Components

   Third: resolve the weight along the new axes

   N is already perpendicular to the plane: it lies entirely on y′, there is nothing to resolve. The weight, on the other hand, is oblique with respect to the new axes, so we replace it with two forces that together have the same effect. Construction: from the tip of the weight draw the two perpendiculars to the axes, and read off the rectangle. What remains is to work out where the angle ends up, and it is the step almost everyone skips: the arc drawn on the body shows that the angle between the weight and the perpendicular to the plane is exactly θ, because they are angles whose sides are perpendicular in pairs. So θ is ATTACHED to y′: the component on y′ goes with the cosine and the one on x′ with the sine. P∥ = m·g·sin θ, P⊥ = m·g·cos θ.

4. Step 4 · The equilibria

   Fourth: write one equation for each axis

   Along y′ (perpendicular to the plane) the body does not move: the acceleration is zero and the sum of the forces must be zero. N − m·g·cos θ = 0, that is N = m·g·cos θ. This is a true equilibrium, and it always holds, whether the body is at rest or sliding. Along x′ (along the plane), on the other hand — careful — there is NO equilibrium: on the smooth plane nothing balances P∥, so the resultant is m·g·sin θ and the body accelerates. F = m·a gives m·g·sin θ = m·a, and the mass cancels: a = g·sin θ. Writing “the sum is zero along the plane too” is the costliest mistake of the chapter.

5. Step 5 · Friction

   Fifth (optional): with friction, x′ gets interesting

   So far the plane was smooth, by declaration. Switch friction on and a fourth force appears on the x′ axis, tangent to the surface and pointing up the plane. While the body is at rest, friction is exactly what is needed to hold it — m·g·sin θ − f = 0, a true equilibrium — but it cannot exceed μ·N. The body starts moving when m·g·sin θ > μ·m·g·cos θ, that is when tan θ > μ: the breakaway angle is arctan μ and, again, it does not depend on the mass. This figure stops there, at the limit of equilibrium: the motion that follows — net force, acceleration, a bit of kinematics — is another chapter, not yet drawn here.

6. Step 6 · Try it

   Now you move the ramp

   Drag the top of the plane (or use the keyboard arrows) and watch everything move together: the axes rotate with the plane, the weight arrow stays identical — it is the only one —, the two components swap importance and the normal reaction gets shorter. Switch friction on, raise μ and look for the exact angle at which the body breaks free. All the numbers and the two equations on the right are recalculated at every degree.

The inclined plane and its base

The surface the body rests on, the horizontal ground the angle is measured from and the little right-angle square at the other vertex: the triangle is right-angled, and θ sits between the hypotenuse and the base. In the drawing the ramp never changes length, only its slope — so the only thing that varies is θ.

The angle θ and the handle

The angle between the plane and the horizontal. It is the only quantity you decide: drag the top of the ramp, or reach it with Tab and use the arrow keys. Every other quantity is recalculated from it.

The body

A block of mass m = 2 kg. The mass is deliberately fixed: this way the weight arrow never changes length and you can see that the only thing varying is the way we resolve it. All the arrows start here, because this is where the forces act.

The weight P = m·g

The force with which the Earth pulls the body: vertical and downward, magnitude m·g. It does not tilt with the plane and does not change magnitude — it is the number one mistake of the chapter. Every other arrow comes from this one or answers it.

The normal reaction N

The push with which the plane answers: perpendicular to the surface, pointing outwards, so already entirely on the y′ axis. N = m·g·cos θ because along y′ the body does not move. N equals the weight only on a horizontal plane: at 60° it is already half.

The axes x′ and y′

The frame of reference chosen for this problem: the usual axes, rotated by θ. x′ runs along the plane downwards, y′ is perpendicular to the plane and points outwards. The original horizontal/vertical axes stay in grey, so you can see the angle between the two systems: it is exactly θ.

The parallel component P∥ = m·g·sin θ

The part of the weight directed along x′: it is the one that makes the body slide down, and nothing balances it on the smooth plane. It is zero on a horizontal plane and grows with the angle.

The perpendicular component P⊥ = m·g·cos θ

The part of the weight directed along y′, into the plane: it is the one the normal reaction has to balance. It is greatest on a horizontal plane and decreases as you tilt — and with it decreases the available friction, which is proportional to N.

The resultant along the plane

What is left after adding everything along x′: m·g·sin θ on the smooth plane, zero when friction manages to hold the body still. Divided by the mass it gives the acceleration — here always zero as soon as there is friction, because this figure does not go beyond the limit of equilibrium.

The friction force

Tangent to the surface and opposite to the direction in which the body would tend to move, so up the plane. With the body at rest it equals the parallel component exactly, up to a maximum of μ·N — μ here is static friction. Beyond that maximum equilibrium ends, which is exactly where this figure stops: the moving body, with kinetic friction slowing it down, is a separate case, not yet drawn here.

The values and the two equations

The column on the right shows every quantity in newtons and the two axis equations, recalculated at every degree you move: they are not fixed captions. Arrows and numbers come from the same source, so the length of one arrow can be compared with another.

1. y′ — perpendicular to the plane equilibrium

   N − m·g·cos θ = 0

   N = m·g·cos θ

   A true equilibrium: perpendicular to the plane the body does not move — it neither sinks in nor lifts off — so the acceleration along y′ is zero and the two forces cancel. It always holds, even while the body is sliding.

2. x′ — along the plane, no friction not an equilibrium

   m·g·sin θ = m·a

   a = g·sin θ

   Here there is NO equilibrium, and it is the point where most marks are lost: on the smooth plane nothing balances the parallel component, so the resultant is not zero and the body accelerates. The mass cancels, so the acceleration is the same for a heavy body and a light one.

3. x′ — along the plane, with friction (static equilibrium) equilibrium

   m·g·sin θ − f = 0, with f ≤ μ·N

   f = m·g·sin θ, as long as tan θ ≤ μ ⟹ a = 0

   As long as tan θ ≤ μ static friction exactly matches the parallel component and this axis too is an equilibrium: a = 0. This figure stays within that limit; beyond tan θ = μ friction is no longer enough and the body starts to move — the moving case, with kinetic friction and an acceleration different from zero, is another chapter.

Once it is established that θ is attached to y′, the usual rule applies: the component attached to the angle goes with the cosine, the opposite one with the sine. From there P⊥ = m·g·cos θ and P∥ = m·g·sin θ, and the dashed construction in the figure closes the rectangle that proves it.

• P = m·g — the weight, vertical, independent of θ
• P∥ = m·g·sin θ — along x′, down the slope
• P⊥ = m·g·cos θ — along y′, into the plane
• N = m·g·cos θ — the normal reaction, from the equilibrium along y′
• a = g·sin θ — smooth plane: the mass cancels
• a = g·(sin θ − μ·cos θ) — with friction, for a moving body
• θbreakaway = arctan μ — the angle beyond which the body starts to move

The check that saves the test is always the same: try θ = 0°. The plane is horizontal, nothing slides, so the component along the plane must be zero — and sin 0° = 0, while cos 0° = 1. If in your formula the component along the plane has a cosine, you have swapped the two factors, and you find out in two seconds instead of after handing in the paper.

• Dynamics and the inclined plane in Year 2 at a technical school — what the test really asks, with the lab report.
• Resolving forces without having done trigonometry — if sine and cosine have not yet arrived in maths.
• Practice test: the principles of dynamics — inclined plane, ropes and pulleys, with instant feedback.
• The unit circle, interactive — where sine and cosine come from, seen as the coordinates of a point.
• All the figures — the other figures in the section.