A box is lying on an inclined plane what is the coefficient of static friction if the box starts sliding when an angle of inclination is 60°
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Answered by
35
A box is lying on an inclined plane , if box starts sliding then , at this instant :
downward force along plane = upward force along plane.
downward force along plane = mgsin60° [ components of weight along plane]
upward force along plane = static friction
but we know,![f_s=\mu_s N f_s=\mu_s N](https://tex.z-dn.net/?f=f_s%3D%5Cmu_s+N)
where fs is static friction ,
is coefficient of static friction and N is normal reaction act between box and plane.
N = mgcos60° [ components of weight perpendicular to plane]
so, mgsin60° =
mgcos60°
![\mu_s=tan60^{\circ} \mu_s=tan60^{\circ}](https://tex.z-dn.net/?f=%5Cmu_s%3Dtan60%5E%7B%5Ccirc%7D)
hence, coefficient of static friction = tan60° = √3
downward force along plane = upward force along plane.
downward force along plane = mgsin60° [ components of weight along plane]
upward force along plane = static friction
but we know,
where fs is static friction ,
N = mgcos60° [ components of weight perpendicular to plane]
so, mgsin60° =
hence, coefficient of static friction = tan60° = √3
Answered by
21
Answer:
√3 = 1.73
Explanation:
as we know force on inclined plane will be mgsin60°
N on inclined plane will be = mgcos60°
applying above formula
mgsin60°=mu*mgcos60°
tan60°=√3
mu=1.73
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