A brick of mass $2\, kg$ begins to slide down on a plane inclined at an angle of $45^° $ with the horizontal. The force of friction will be
A brick of mass $2\, kg$ begins to slide down on a plane inclined at an angle of $45^° $ with the horizontal. The force of friction will be
$19.6 \,sin \,45^°$
$19.6\, cos\, 45^°$
$9.8\, sin\, 45^°$
$9.8\, cos \,45^°$
A brick of mass $2\, kg$ begins to slide down on a plane inclined at an angle of $45^° $ with the horizontal. The force of friction will be
For angle of repose,
Friction = Component of weight along the plane
=$mg\sin \theta $ $ = 2 \times 9.8 \times \sin {45^o}$ $ = 19.6\sin {45^o}$
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