Topic 7 of 8 · Form 6
Moments and centres of mass
Moment = force × perpendicular distance. For a rigid body in equilibrium: resultant force zero and total moment about any point zero.
Notes for this topic 1 file, in order
Create a free account to download PDFs.
Moments
Moment = force × perpendicular distance. For a rigid body in equilibrium: resultant force zero and total moment about any point zero.
Worked example: A uniform beam AB of length 4 m and mass 20 kg rests on supports at A and B. A 30 kg child stands 1 m from A. Find the reactions (g = 9.8).
Moments about A: RB × 4 = 20g × 2 + 30g × 1 → RB = (392 + 294)/4 = 171.5 N
RA + RB = 50g = 490 → RA = 318.5 N
Ladders and rods
A ladder against a smooth wall on rough ground: resolve horizontally and vertically, take moments about the foot; limiting friction at the ground gives the critical angle.
Centre of mass
- Particles: x̄ = Σmx/Σm (and similarly ȳ).
- Standard laminas: rectangle (centre), triangle (⅔ along each median from the vertex), semicircle (4r/3π from the diameter).
- Composite bodies: add or subtract parts.
Worked example: Masses of 2 kg, 3 kg and 5 kg are at x = 0, 2 and 4 m. Find x̄.
x̄ = (0 + 6 + 20) ÷ 10 = 2.6 m
Suspension and toppling
A lamina suspended freely hangs with its centre of mass vertically below the point of suspension. A body on a slope topples when the vertical through its centre of mass falls outside its base.
Check yourself
- Where is the centre of mass of a uniform triangular lamina?
- A uniform rod 2 m long, weight 40 N, is supported at one end and held horizontal by a vertical force at the other. Find that force.
- When does a block on a slope topple?
Show answers
- At the intersection of the medians, ⅔ of the way from each vertex.
- 20 N
- When the vertical through its centre of mass passes outside its base.
Questions about this topic (0)
No questions about this topic yet.
Sign in or create a free account to join in.