A body placed on an incline is subjected to downward pull due to gravity. The downward motion along the incline is, therefore, natural tendency of the body, which is opposed by friction acting in the opposite direction against the component of gravity along the incline. Depending on the relative strengths of two opposite forces, the body remains at rest or starts moving along the incline.
The translational motion of a body on an incline is along its surface. There is no motion in the direction perpendicular to the incline surface. The force components in the direction perpendicular to the incline form a balanced force system.
Let us examine the free body diagram as superimposed on the body systems and shown in the figure. In the direction of incline i.e. x – axis, we have :
∑
F
x
=
m
g
sin
θ

F
F
=
m
a
∑
F
x
=
m
g
sin
θ

F
F
=
m
a
where
F
F
F
F
is the force of friction, which can be static friction, maximum static friction or kinetic friction. In the ydirection,
∑
F
y
=
N

m
g
cos
θ
=
0
⇒
N
=
m
g
cos
θ
∑
F
y
=
N

m
g
cos
θ
=
0
⇒
N
=
m
g
cos
θ
Based on the angle of incline” θ”, there are following possibilities :
Case 1 : The friction is less than the maximum static friction and body is yet to start motion. Here,
θ
<
tan

1
(
μ
s
)
θ <
tan

1
(
μ
s
)
. In this case, friction is simply equal to component of gravity, parallel to the contact surface. Thus,
F
F
=
f
s
=
m
g
sin
θ
F
F
=
f
s
=
m
g
sin
θ
The net component of external forces in xdirection (including friction), therefore, is zero. As such, the block remains stationary.
∑
F
x
=
m
g
sin
θ

F
F
=
m
g
sin
θ

m
g
sin
θ
=
0
∑
F
x
=
m
g
sin
θ

F
F
=
m
g
sin
θ

m
g
sin
θ
=
0
Case 2 : The friction is equal to the maximum static friction and body is yet to start motion. Here,
θ
=
tan

1
(
μ
s
)
θ =
tan

1
(
μ
s
)
. In this case also, the component of net external force in xdirection (inclusive of friction) is zero and the block is stationary. Here,
⇒
tan
θ
=
tan
φ
=
μ
s
⇒
tan
θ
=
tan
φ
=
μ
s
Case 3 : The friction is equal to the kinetic friction and body is in motion. Initially,
θ
>
tan

1
(
μ
s
)
θ >
tan

1
(
μ
s
)
. In this case, friction is given by :
F
F
=
F
k
=
μ
k
N
F
F
=
F
k
=
μ
k
N
We must understand here that friction can be equal to kinetic friction only when body has started moving. This is possible when component of gravity, parallel to contact surface is greater than the maximum static friction. However, once this condition is fulfilled, the friction between the surfaces reduces slightly to become equal to kinetic friction. Subsequently, the motion of the body can be either without acceleration (uniform velocity) or with acceleration.
3a: Motion with uniform velocity
This is a case of balanced force system in xdirection. The component of net force in xdirection is zero and the block is moving with constant velocity. Here,
⇒
tan
θ
=
μ
k
⇒
tan
θ
=
μ
k
3b: Accelerated motion
This is a case of unbalanced force system in xdirection.
In x – direction,
∑
F
x
=
m
g
sin
θ

F
F
=
m
a
x
⇒
m
g
sin
θ

μ
k
N
=
m
a
∑
F
x
=
m
g
sin
θ

F
F
=
m
a
x
⇒
m
g
sin
θ

μ
k
N
=
m
a
Putting value of normal force, N, we have :
⇒
m
g
sin
θ

μ
k
m
g
cos
θ
=
m
a
⇒
a
=
g
sin
θ

μ
k
g
cos
θ
⇒
m
g
sin
θ

μ
k
m
g
cos
θ
=
m
a
⇒
a
=
g
sin
θ

μ
k
g
cos
θ
For the given value of “θ” and “
μ
k
μ
k
”, acceleration is constant and the block, therefore, moves down with uniform acceleration.
Problem : A block of mass “m” slides down an incline, which makes an angle of 30° with the horizontal. If the acceleration of the block is g/3, then find the coefficient of kinetic friction. Consider g = 10
m
/
s
2
m /
s
2
.
Solution : The coefficient of kinetic friction is given by :
μ
k
=
F
k
N
μ
k
=
F
k
N
In order to evaluate this ratio, we need to find normal and friction forces. The free body diagram of the block as superimposed on the system is shown in the figure.
In the xdirection,
∑
F
x
=
m
g
sin
30
0

F
k
=
m
a
x
⇒
m
g
sin
30
0

F
k
=
m
x
g
3
⇒
m
g
x
1
2

F
k
=
m
x
g
3
⇒
F
k
=
m
g
x
1
2

m
x
g
3
=
mg
6
∑
F
x
=
m
g
sin
30
0

F
k
=
m
a
x
⇒
m
g
sin
30
0

F
k
=
m
x
g
3
⇒
m
g
x
1
2

F
k
=
m
x
g
3
⇒
F
k
=
m
g
x
1
2

m
x
g
3
=
mg
6
In the ydirection, there is no acceleration,
∑
F
y
=
m
g
cos
30
0

N
=
0
⇒
N
=
m
g
cos
30
0
=
√
3
2
m
g
∑
F
y
=
m
g
cos
30
0

N
=
0
⇒
N
=
m
g
cos
30
0
=
√
3
2
m
g
Putting values in the expression of coefficient of friction, we have :
μ
k
=
2
6
√
3
=
1
3
√
3
=
0.192
μ
k
=
2
6
√
3
=
1
3
√
3
=
0.192