A body of mass10kg rests on a rough inclined plane whose angle of tilt tan is variable.tan is gradually increased until the nody starts to slide down the plane at 30degrees.find coefficient of limiting friction.
Answers
Answer:
Explanation:
In first case when the body is in limiting equilibrium at angle of inclination of the inclined plane,
30
∘
then the downward force along the inclined plane is just balanced by the frictional force ,. So we have
m
g
sin
θ
=
μ
m
g
cos
θ
μ
=
sin
θ
cos
θ
=
tan
30
∘
=
1
√
3
,
where
m
→
mass of the body
θ
→
angle of inclination of the inclined plane
=
30
∘
μ
→
coefficient of friction
g
→
acceleration due to gravity
In the second case when the angle of inclination is
θ
=
60
∘
we have net downward force on the body along the inclined plane
F
=
m
g
sin
θ
−
μ
m
g
cos
θ
So acceleration of the body
a
=
F
m
=
g
(
sin
θ
−
μ
cos
θ
)
=
g
(
sin
60
∘
−
1
√
3
cos
60
∘
)
=
g
(
√
3
2
−
1
√
3
×
1
2
)
=
g
2
(
√
3
−
1
√
3
)
=
g
2
(
3
−
1
√
3
)
=
g
√
3