Q.

In the arrangement shown, mass of the block is 5 kg and coefficient of static friction between the block and inclined surface is 0.2. For what minimum value of F, Will the block be on the verge of slipping down the plane ? Given tan 370 =34.

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a

60 N

b

110 N

c

80 N

d

40 N

answer is B.

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Detailed Solution

N = Normal reaction of the inclined surface on the block = (mg cos θ + F)∴   fSmax=μN= μmg cosθ+F∴   μmg cosθ+F=mg  sinθ⇒  F=   mgsinθμ− cosθ=5 × 10sin3700.2− cos370N=110N
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In the arrangement shown, mass of the block is 5 kg and coefficient of static friction between the block and inclined surface is 0.2. For what minimum value of F, Will the block be on the verge of slipping down the plane ? Given tan 370 =34.