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 $$tan$$ $$370$$ $$=34$$.

Question

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 $$tan$$ $$370$$ $$=34$$.

Infinity Learn · Accepted Answer

N $$=$$ Normal reaction of the inclined surface on the block $$=$$ (mg$$ $$cos$$ θ $$+$$ $$F)∴   $$fSmax=$$μ$$N=$$ μmg $$cos$$θ$$+F$$∴   μmg $$cos$$θ$$+F=mg$$  $$sin$$θ⇒  $$F=$$   mgsinθμ− $$cos$$θ$$=5$$ × $$10sin3700.2$$− $$cos370N=110N$$

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 $\tan 37^{0} = \frac{3}{4} .$

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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.

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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 $\tan 37^{0} = \frac{3}{4} .$