Problem 4 :
A block of mass "m" is placed on an incline of angle "θ" and mass "M", which is placed on a horizontal surface. All surfaces are friction - less. Find the acceleration of the block with respect to incline.

Solution :
Here, block and incline both have different accelerations. The block moves down along the incline. The incline, on the other hand, moves right as shown in the figure above. As such, we can not treat two elements as one body and find the their common accelerations with respect to ground.

Let "
a
B
a
B
"
, "
a
I
a
I
" and "
a
BI
a
BI
" respectively denote acceleration of block w.r.t ground, acceleration of the incline w.r.t ground and acceleration of the block w.r.t incline.

We first carry out force analysis on the "block" in the non-inertial frame of incline. Here, the forces on the block are (i) weight of the block, "mg", (ii) normal force due to underlying incline, "
N
1
N
1
", and (iii) Pseudo force, "
m
a
I
m
a
I
" applied in the direction opposite to the acceleration of the incline. The free body diagram of the incline in non-inertial reference of incline is shown here.

Force analysis on the block yields two relation,

∑
F
x
=
m
a
I
cos
θ
+
m
g
sin
θ
=
m
a
B
I
∑
F
x
=
m
a
I
cos
θ
+
m
g
sin
θ
=
m
a
B
I

⇒
a
B
I
=
a
I
cos
θ
+
g
sin
θ
⇒
a
B
I
=
a
I
cos
θ
+
g
sin
θ

∑
F
y
=
N
1
+
m
a
I
sin
θ
=
m
g
cos
θ
∑
F
y
=
N
1
+
m
a
I
sin
θ
=
m
g
cos
θ

It is clear from the first equation that we need to know "
a
I
a
I
" i.e. the acceleration of incline with respect to ground to determine the value of "
a
BI
a
BI
". It is imperative that we carry out force analysis on the "incline" in the inertial frame of reference of ground.

The forces on the incline are (i) weight of the incline, "Mg", (ii) normal force due to overlying block, "
N
1
N
1
", and (iii) Normal force due to underneath horizontal surface, "
N
2
N
2
". The free body diagram of the incline in inertial ground reference is shown here.

Force analysis of the incline yields two relations,

∑
F
x
=
N
1
sin
θ
=
m
a
I
∑
F
x
=
N
1
sin
θ
=
m
a
I

∑
F
y
=
N
2
=
N
1
cos
θ
+
M
g
∑
F
y
=
N
2
=
N
1
cos
θ
+
M
g

First of the equations is useful that can be combined with the analysis in non-inertial frame of reference. Substituting for "
N
1
N
1
" in the earlier equation, we have :

⇒
m
a
I
sin
θ
+
m
a
I
sin
θ
=
m
g
cos
θ
⇒
m
a
I
sin
θ
+
m
a
I
sin
θ
=
m
g
cos
θ

⇒
a
I
=
m
g
sin
θ
cos
θ
M
+
m
sin
2
θ
⇒
a
I
=
m
g
sin
θ
cos
θ
M
+
m
sin
2
θ

We can now determine the required acceleration of the block with respect to incline by substituting the value of "
a
I
a
I
" in the expression of "
a
BI
a
BI
" :

⇒
a
B
I
=
m
g
sin
θ
cos
2
θ
M
+
m
sin
2
θ
+
g
sin
θ
⇒
a
B
I
=
m
g
sin
θ
cos
2
θ
M
+
m
sin
2
θ
+
g
sin
θ

⇒
a
B
I
=
m
g
sin
θ
cos
2
θ
+
M
g
sin
θ
+
m
sin
3
θ
M
+
m
sin
2
θ
⇒
a
B
I
=
m
g
sin
θ
cos
2
θ
+
M
g
sin
θ
+
m
sin
3
θ
M
+
m
sin
2
θ

⇒
a
B
I
=
g
sin
θ
{
M
+
m
cos
2
θ
+
sin
2
θ
}
M
+
m
sin
2
θ
⇒
a
B
I
=
g
sin
θ
{
M
+
m
cos
2
θ
+
sin
2
θ
}
M
+
m
sin
2
θ

⇒
a
B
I
=
M
+
m
g
sin
θ
M
+
m
sin
2
θ
⇒
a
B
I
=
M
+
m
g
sin
θ
M
+
m
sin
2
θ