Weight Component Along an Incline (mg sin θ)

F∥=mgsin⁡θF_{\parallel} = m g \sin\theta

Worked example: 20 kg on a 30° ramp → 98.0665 N — press Try an example to run it live, then adjust anything.

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Weight Component Along an Incline (mg sin θ) explained

mmgF∥θ

Split the weight along the slope and you get mg sin θ — the force a rope, a brake or friction must hold to keep an object from sliding down. A 20 kg crate on a 30° ramp pulls down-slope with 20 × 9.80665 × sin 30° ≈ 98 N, exactly half its weight, because sin 30° = ½. Galileo built his entire kinematics programme on this: an incline "dilutes" gravity by sin θ, slowing a falling body enough to time it with a water clock in 1604, centuries before anything could time a free fall directly.

Highway grades use the same maths in disguise — a 6% grade means a rise of 6 m per 100 m, θ ≈ 3.43°, so a 40-tonne truck feels about 40,000 × 9.80665 × sin 3.43° ≈ 23 kN pushing it downhill, which is precisely why runaway-truck ramps exist. Pair this with N = mg cos θ and you have the complete free-body diagram for any slope; mixing up which one takes sine and which takes cosine is the standard exam trap, so check the limits: on level ground the down-slope force must vanish, and sin 0° = 0 does exactly that.

Weight Component Along an Incline (mg sin θ) formula

F∥=mgsin⁡θF_{\parallel} = m g \sin\theta
Where
  • F∥F_{\parallel}= Down-slope force (N)
  • mm= Mass (kg)
  • θ\theta= Incline angle (°)

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