Back to top

Inclined plane, Meaning (SPM Physics)

A flat surface set at an angle to the horizontal, on which an object's weight can be resolved into components along and perpendicular to the surface.

  • Specialist SPM Physics tutoring
  • 5,000+ students helped
  • Experienced Physics teachers
  • Fully online 1-to-1, nationwide
  • Real 1-hour paid trial, from RM50/hr
  • Built on the official SPM syllabus
EnglishInclined plane
Bahasa MelayuSatah condong
中文斜面

Definition

A flat surface set at an angle to the horizontal, on which an object's weight can be resolved into components along and perpendicular to the surface.

Force and Motion II

In exam terms

An inclined plane is a flat surface set at an angle θ to the horizontal. On it, the weight of an object is resolved into two components: one acting down the slope, mg sin θ, and one pressing into the slope, mg cos θ.

The component down the slope tends to make the object slide, while the perpendicular component sets the normal reaction and hence the friction available.

For example, a 2 kg block on a slope of 30° has a weight of mg = 2 kg × 9.81 m s⁻² = 19.6 N. The component pulling it down the slope is 19.6 N × sin 30° = 9.81 N, and the component pressing into the surface is 19.6 N × cos 30° = 17.0 N.

Don't confuse the two weight components

The common trap on an inclined plane is swapping the two weight components. The part along the slope uses sin θ and the part perpendicular to the slope uses cos θ; a quick check is that at θ = 0° (flat ground) the perpendicular part equals the full weight.

Do not confuse the normal reaction with weight either, on a slope the normal reaction equals mg cos θ, which is less than mg.

In Paper 2 this is usually asked with 'Resolve…', 'Calculate the acceleration…' or 'Determine the normal reaction…'. Keep newtons (N) on every line and state the angle you used.

Source: DSKP KSSM Physics Form 4 and 5 (Versi English) (Bahagian Pembangunan Kurikulum (BPK), KPM)

Frequently asked questions

Which weight component acts down the slope?
The component mg sin θ acts down the slope, and mg cos θ acts perpendicular to it, where θ is the angle of the incline to the horizontal.
Why is the normal reaction smaller on a slope?
Because only the perpendicular component of weight, mg cos θ, presses into the surface, and cos θ is less than one for any slope steeper than flat ground.
What is the weight component down a 30° slope for a 2 kg block?
It is mg sin θ = 2 kg × 9.81 m s⁻² × sin 30° = 9.81 N.

Book a Trial Class

One-hour paid trial · Same-day reply · from RM50/hr

Book a Trial Class

One-hour paid trial · Same-day reply