First slide

Static friction

Question

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

Moderate

A

9.8 N
B

0.7 × 9.8 × √3 N
C

9.8 × √3 N
D

0.7 × 9.8 N

Solution

$A n g l e o f f r i c t i o n φ = \tan^{- 1} (0.7) > 30^{0} . T h e r e f o r e t h e b l o c k w i l l b e i n e q i l i b r i u m .$
The actual frictional force will be equal to $mgsin θ$
$= 2 \times 9 \cdot 8 \times \frac{1}{2} = 9 \cdot 8 N$ .

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