A block of mass $2 \,kg$ rests on a rough inclined plane making an angle of $30°$ with the horizontal. The coefficient of static friction between the block and the plane is $ 0.7$. The frictional force on the block is ....... $N$.
A block of mass $2 \,kg$ rests on a rough inclined plane making an angle of $30°$ with the horizontal. The coefficient of static friction between the block and the plane is $ 0.7$. The frictional force on the block is ....... $N$.
$9.8$
$0.7 \times 9.8 \times \sqrt 3$
$9.8 \times \sqrt 3$
$0.8 \times 9.8$
A block of mass $2 \,kg$ rests on a rough inclined plane making an angle of $30°$ with the horizontal. The coefficient of static friction between the block and the plane is $ 0.7$. The frictional force on the block is ....... $N$.
Limiting friction ${F_l} = \mu \;mg\cos \theta $
${F_l} = 0.7 \times 2 \times 10 \times \cos 30^\circ = 12\;N$(approximately)
But when the block is lying on the inclined plane then component of weight down the plane $ = mg\sin \theta $ $ = 2 \times 9.8 \times \sin 30^\circ = 9.8\;N$
It means the body is stationary, so static friction will work on it
$\therefore $ Static friction $=$ Applied force $= 9.8 \,N$
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