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Laws of dry friction; overturning and sliding

Short answer

Laws: friction opposes motion, self-adjusting up to limit, limiting friction F_max = μN, independent of contact area, depends on surface roughness, and kinetic friction μ_k < μ_s. For a block, sliding when P > μW, overturning when P > Wb/(2a). Example: 800 N cabinet, base 0.8 m, force at 1.2 m, μ = 0.5 gives P_tip = 266.7 N < P_slide = 400 N, so it tips first.

State laws of dry friction. How can we assume the condition of overturning and sliding of a block? Explain with suitable example.

Laws of dry friction (Coulomb's laws)

  1. The friction force acts along the contact surface and opposes the motion, or the tendency of motion, of the body.
  2. Up to the limit, the friction force is equal to the applied tangential force. It is a self-adjusting force.
  3. The limiting friction is directly proportional to the normal reaction, F_max = μₛN.
  4. The limiting friction does not depend on the area of contact (for the same normal reaction).
  5. The coefficient of friction depends on the nature and roughness of the surfaces in contact.
  6. For a body in motion, kinetic friction F = μₖN, with μₖ slightly less than μₛ and independent of the speed at low speeds.

Overturning and sliding of a block

A block of weight W, base width b, is pushed by a horizontal force P applied at height a above the base. With coefficient μ:

  • Sliding occurs when P > μW.
  • Overturning about the toe occurs when P a > W b/2, i.e. P > Wb/(2a).

The condition that needs the smaller P happens first.

Example: a cabinet W = 800 N, b = 0.8 m, pushed at a = 1.2 m, μ = 0.5. P_slide = 0.5 × 800 = 400 N. P_tip = 800 × 0.8/(2 × 1.2) = 266.7 N. Since 266.7 N < 400 N, the cabinet tips over before it slides. If P were applied lower, at a = 0.6 m, then P_tip = 533 N > 400 N, and it would slide first.

✓ Answer checked by Claude Opus 5.5 (high effort)

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