Answer:
Let
F
be horizontal force applied on the block as shown above.This force will have two components
(1)
F
cos
30
∘
that will act upward parallel to inclined plane.
(2)
F
sin
30
∘
that will act on inclined plane perpendicularly. So it will be added with the component of the weight of the block
5
g
cos
30
∘
to enhance the magnitude of Normal reaction
N
So the frictional force
f
fric
=
μ
N
=
0.2
(
F
sin
30
∘
+
5
g
cos
30
∘
)
,where acceleration due to gravity
(
g
)
=
9.8
m
/
s
2
Again the component of the weight of the block of mass 5kg acting downward parallel to the inclined plane is
5
g
sin
30
∘
(a) Considering that the equilibrium of forces when the body is on the verge of sliding up the incline we can write
F
cos
30
∘
=
5
g
sin
30
∘
+
0.2
(
F
sin
30
∘
+
5
g
cos
30
∘
)
⇒
F
=
5
g
(
sin
30
∘
+
0.2
cos
30
∘
)
cos
30
∘
−
0.2
sin
30
∘
≈
43
N
(b) Considering that the equilibrium of forces when the body is on the verge of sliding down the incline, we can write
F
cos
30
∘
=
5
g
sin
30
∘
−
0.2
(
F
sin
30
∘
+
5
g
cos
30
∘
)
⇒
F
=
5
g
(
sin
30
∘
−
0.2
cos
30
∘
)
cos
30
∘
+
0.2
sin
30
∘
≈
16.6
N
Explanation:
