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Class 9 Newton's Laws of Motion (Advanced): cheat sheet
CBSE Science. Where Newton's laws need care: pseudo forces in accelerating buses and lifts, gravity above and below the surface, air resistance, and the turning effect of a force.
Key ideas
- Newton's laws hold as they are only in inertial frames; inside an accelerating frame we add a pseudo force.
- A pseudo force acts opposite to the frame's acceleration, has no physical source and no reaction pair.
- In a lift accelerating upward a scale reads more than your weight; accelerating downward, less.
- The Sun's pull provides the centripetal force that keeps the Earth in orbit; inertia keeps it moving along the curve.
- In air, the body with the smaller cross-section lands first; in a vacuum, all bodies fall together.
- g is greatest at the surface and falls both above it and below it, reaching zero at the centre.
- Torque is largest when the force is at right angles to the lever arm and zero when it points at the pivot.
Remember
| Pseudo force | F = −m × a(frame) |
| Scale reading in a lift | up: m(g + a); down: m(g − a) |
| g at a height h | g_h = g × R² ÷ (R + h)² |
| g at a depth d | g_d = g(1 − d/R) |
| Torque | τ = F × d × sin θ (N m); balance when the clockwise and anticlockwise torques are equal |
Watch out
- Giving the pseudo force the same direction as the frame's acceleration.
- Using R instead of (R + h) for the distance from the Earth's centre.
- Forgetting sin θ: a force at 30° gives only half the torque.
Worked example
A 35 kg person stands on a scale in a lift. Take g = 9.8 m/s². When the lift accelerates upward at 3.5 m/s², find (a) the pseudo force seen from inside the lift and (b) its direction (upward or downward), and (c) the scale reading. (d) What does the scale read when the lift accelerates downward at 3.5 m/s²?
Working: (a) m × a = 35 × 3.5; (b) opposite to the lift's acceleration; (c) m(g + a); (d) m(g − a)
Answer: (a) 122.5 N; (b) downward; (c) 465.5 N; (d) 220.5 N
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