Problem 4 : A block “A” is placed over block “B”, whose base matches with the incline as shown in the figure. The incline is fixed on the horizontal plane. The friction between blocks is “μ”, whereas friction between block “B” and incline is negligible. Find the angle of incline for which block “A” is about to move over block “B”.
Solution : The block “A” is about to move on the block “B”. It means that the friction between the blocks is equal to limiting friction,
F
S
=
μ
N
F
S
=
μ
N
Now, we need to know the value of “N” in order to determine limiting friction. The block “A” is about to move relative to block “B”. As such, block “A” is not having relative motion with respect “B”. The two blocks move together with common acceleration, “a”, given by :
a
=
g
sin
θ
a
=
g
sin
θ
Since we require to find normal force "N", we shall consider forces in vertical direction. For this, we need to find acceleration in vertical direction. Hence, resolving acceleration in “x” and “y’ direction, we have :
a
x
=
a
cos
θ
=
g
sin
θ
cos
θ
a
x
=
a
cos
θ
=
g
sin
θ
cos
θ
a
y
=
a
sin
θ
=
g
sin
2
θ
a
y
=
a
sin
θ
=
g
sin
2
θ
Now, the forces on block “A” in vertical (“y”) direction are (i) its weight and (ii) normal force. Thus,
m
g
−
N
=
m
a
y
=
m
g
sin
2
θ
m
g
−
N
=
m
a
y
=
m
g
sin
2
θ
⇒
N
=
m
g
1
−
sin
2
θ
=
m
g
cos
2
θ
⇒
N
=
m
g
1
−
sin
2
θ
=
m
g
cos
2
θ
Putting expression of “N” in the equation of limiting friction, we have :
⇒
F
S
=
μ
m
g
cos
2
θ
⇒
F
S
=
μ
m
g
cos
2
θ
The limiting friction provides the acceleration in horizontal direction (note that blocks have no relative motion between them, but have accelerations with respect to ground),
F
S
=
μ
m
g
cos
2
θ
=
m
a
x
=
m
g
sin
θ
cos
θ
F
S
=
μ
m
g
cos
2
θ
=
m
a
x
=
m
g
sin
θ
cos
θ
⇒
tan
θ
=
μ
⇒
tan
θ
=
μ
θ
=
tan

1
μ
θ
=
tan

1
μ