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Engineering mechanics – minimum external pull to initiate sliding on a rough horizontal plane\nA block of weight W rests on a rough, horizontal surface. The pull is applied at an angle θ to the horizontal. If θ equals the angle of friction, what is the minimum magnitude of force required to just start sliding the block?

Difficulty: Medium

Correct Answer: W sin θ

Explanation:


Introduction / Context:
This classical engineering mechanics question tests friction, impending motion, and the special case where the applied pull is inclined at the angle of friction. The goal is to find the least pull that will just overcome limiting friction and initiate sliding on a rough horizontal plane.



Given Data / Assumptions:

  • Block weight = W on a rough, horizontal plane.
  • Coefficient of friction = μ, so angle of friction φ satisfies tan φ = μ.
  • An external pull P is applied at an angle θ to the horizontal; for minimum P, θ = φ.
  • Impending motion condition (limiting friction) is used.


Concept / Approach:
At impending motion, the frictional force reaches its maximum value F = μN. For an inclined pull P at angle θ above the horizontal, the normal reaction reduces due to the vertical component of P. Minimizing P with respect to θ leads to θ = φ, the angle of friction. Substituting θ = φ yields a compact expression for the least pull.



Step-by-Step Solution:

Resolve forces vertically: N + P sin θ = W ⇒ N = W − P sin θ.At impending motion: horizontal equilibrium gives P cos θ = F = μN = μ(W − P sin θ).Rearrange: P(cos θ + μ sin θ) = μW ⇒ P = μW / (cos θ + μ sin θ).For minimum P, set θ = φ where tan φ = μ. Then cos φ + μ sin φ = cos φ + (tan φ) sin φ = cos φ + (sin φ / cos φ) sin φ = (cos^2 φ + sin^2 φ)/cos φ = 1 / cos φ.Hence P_min = μW / (1 / cos φ) = μW cos φ = W (μ cos φ) = W sin φ (since sin φ = μ / √(1 + μ^2) and cos φ = 1 / √(1 + μ^2)).


Verification / Alternative check:
Using the identity sin φ = μ / √(1 + μ^2) and cos φ = 1 / √(1 + μ^2), the reduced form P_min = μW / √(1 + μ^2) equals W sin φ, confirming consistency.



Why Other Options Are Wrong:

  • W cos θ: corresponds to a vertical-balance component, not the least-pull result.
  • W tan θ: dimensionally force, but not the friction-minimized pull.
  • None of these: incorrect because W sin θ is correct when θ = φ.


Common Pitfalls:
Forgetting that the normal reaction decreases due to the vertical component of the pull, or not using θ = φ for minimum P.


Final Answer:
W sin θ (with θ equal to the angle of friction)

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