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A minimum force F is applied to a block of mass 102 kg to prevent it from sliding on a plane with an inclination angle $30^{o}$ with the horizontal. If the coefficients of static and kinetic friction between the block and the plane are 0.4 and 0.3 respectively, then the force F is

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224 N

zero

157 N

315 N

answer is A.

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Detailed Solution

$m g \sin θ = (102) (10) \sin 30^{o} = 510 N μ_{s} m g \cos θ = (0.4) (102) (10) \cos 30^{o} = 353 N F = 510 - 353 = 157 N$

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