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Forces on an inclined plane

G2.3 G2.4

By the end of this lesson you should be able to

  • Describe the forces acting on an object on an inclined plane through the use of free-body diagrams. G2.3
  • Determine the net force acting on an object on an inclined plane using vector analysis. G2.4

Key ideas

  • Tilt your axes: x along the slope, y perpendicular to it. Then only the weight needs resolving.
  • Weight splits into \(mg\sin\theta\) down the slope and \(mg\cos\theta\) into the slope.
  • The normal force balances the perpendicular part: \(F_N = mg\cos\theta\).
  • Frictionless slope: \(F_{net} = mg\sin\theta\) so \(a = g\sin\theta\), independent of mass. Friction acts up the slope against the motion.

Key formulas (QCAA formula book notation)

\[ F_{\parallel} = m g \sin\theta \]
\[ F_{\perp} = F_N = m g \cos\theta \]
\[ F_{net} = m g\sin\theta - F_f = m a \]
Worked example

A 5.0 kg crate slides down a 25° ramp with 8.0 N of friction. Find its acceleration.

  1. \(F_\parallel = 5.0 \times 9.8 \times \sin 25^\circ = 20.7\) N
  2. \(F_{net} = 20.7 - 8.0 = 12.7\) N
  3. \(a = 12.7/5.0 = 2.5\) m s⁻²

Common mistakes

  • Using cos for the down-slope component (it's sin when θ is the slope angle).
  • Setting F_N = mg on a slope.
  • Drawing friction in the direction of motion.

Do

  • Read: Textbook (Oxford Physics for Queensland Units 3 & 4) section 3.3
  • Do: CYL 3.3 Q2, 4, 5

Try it

Watch

Further reading


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