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A block of mass M rests on a rough level floor with coefficient of friction μ. A person applies a horizontal pull, but the block remains at rest. If the total contact force exerted by the floor on the block is F, then

1. F = Mg
2. F = μMg
3. Mg ≤ F ≤ Mg√(1+μ²)
4. Mg ≥ F ≥ Mg√(1+μ²)

Correct answer: Mg ≤ F ≤ Mg√(1+μ²)

Solution

The floor's contact force combines normal N = Mg and friction f (0 to mu*Mg at impending motion): F = sqrt((Mg)^2 + f^2). Minimum (no pull) is Mg; maximum (impending slip) is sqrt((Mg)^2 + (mu Mg)^2) = Mg*sqrt(1+mu^2). So Mg <= F <= Mg*sqrt(1+mu^2).

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