Examples should include components of forces along and perpendicular to an…

The idea

On an inclined plane, choose axes parallel and perpendicular to the slope rather than horizontal and vertical. For a plane at angle θ to the horizontal, the weight W of an object on the slope resolves into W sin θ down the slope and W cos θ perpendicular into the slope. The angle between the weight and the perpendicular to the slope equals θ, which is why sine acts along the slope and cosine acts into it. For example, a 20 N weight on a 25° slope gives 20 sin 25° ≈ 8.5 N down the slope and 20 cos 25° ≈ 18 N perpendicular to the slope. The perpendicular component is balanced by the normal contact force when there is no acceleration perpendicular to the surface, while the parallel component drives motion or is balanced by friction.

What earns the marks

• Draw a clear free-body diagram showing the weight acting vertically downwards and mark the angle θ between the plane and the horizontal.
• Resolve the weight parallel to the plane as W sin θ and perpendicular to the plane as W cos θ, explaining that the angle between the weight and the perpendicular to the plane is θ.
• Identify the normal contact force as perpendicular to the surface and, in the absence of other perpendicular forces, equate it to W cos θ.
• Resolve any applied force or friction along the same parallel and perpendicular axes before combining components.
• Apply Newton's second law along each axis separately, using the parallel component to find acceleration or the balance of forces for equilibrium.

Common mistakes and how to put them right

• Using W cos θ down the slope and W sin θ into the slope.

  Put it right: Check by imagining θ = 0°, where the parallel component must be zero and the perpendicular component must equal the full weight.

• Assuming the normal contact force always equals the weight.

  Put it right: On a slope it equals W cos θ only when no other force acts perpendicular to the surface.

• Mixing horizontal and vertical axes with slope axes in the same calculation.

  Put it right: Choose one pair of perpendicular axes and resolve every force onto those axes consistently.

Exam tips

• Mark the angle θ at the slope and again where it appears between the weight and the perpendicular to the slope; this makes the sine and cosine assignment clear.
• Use the limiting cases θ = 0° and θ = 90° to check that your expressions behave sensibly.
• State the direction of each component, such as down the slope or perpendicular into the slope, so the physics remains clear.
• When friction is present, resolve it along the slope and include its sign relative to the chosen positive direction.