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Displacement of COM:Application

A force of 10...

Question

A force of 100 N is applied on a block of mass 3 kg as shown in figure. The coefficient of friction between the surface of the block is 0.25. The friction force acting on the block is :

15 N downwards

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25 N upwards

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20 N downwards

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20 N upwards

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Solution

The correct option is B 20 N downwards
Net force in upward direction $f_{a} = F s i n 30 - m g$
$= 100 \times \frac{1}{2} - 3 \times 10 = 20 N$
Limiting friction $F_{l} = μ F c o s 30^{o}$
$= \frac{1}{4} \times 100 \frac{\sqrt{3}}{2} = 21.15 N$
Hence block will not slip. Hence
$f_{r} = f_{a} = 20 N$

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A force of 100 N is applied on a block of mass 3 kg as shown in figure. The coefficient of friction between the surface of the block is 0.25. The friction force acting on the block is :

The correct option is B 20 N downwardsNet force in upward direction fa=Fsin30−mg=100×12−3×10=20NLimiting friction Fl=μFcos30o=14×100√32=21.15NHence block will not slip. Hencefr=fa=20N