Problem 3 : A block of mass 10 kg is placed on a rough incline plane of inclination 30° with an external force “F” applied as shown in the figure. The coefficient of friction between block and incline is 0.5. Find the minimum force “F” so that block does not slide on the incline.
Solution : We first need to determine the direction of friction so that we could determine the minimum force required for block to remain stationary. Here, the gravitational force down the incline is :
m
g
sin
θ
=
10
X
10
X
sin
30
0
=
10
X
10
X
1
2
=
50
N
m
g
sin
θ
=
10
X
10
X
sin
30
0
=
10
X
10
X
1
2
=
50
N
The external force “F” has no component along the incline. As such, the component of weight along the incline is the net force parallel to incline. Friction, therefore, acts upward. If external force “F” is not applied, the limiting friction is given by :
F
S
=
μ
m
g
cos
θ
=
0.5
X
10
X
10
X
3
2
=
25
3
N
F
S
=
μ
m
g
cos
θ
=
0.5
X
10
X
10
X
3
2
=
25
3
N
As the downward gravitational force is greater than limiting friction, the block will slide down.
In order to restrict motion of block, the external force “F” should increase normal force, which in turn increases friction. Let the minimum force be “F” such that block does not side. In this limiting situation, the components of forces, parallel and perpendicular to incline, form balanced force systems. Free body diagram with external forces is shown in the figure. Considering "x" and "y" axes in parallel and perpendicular directions, we have :
∑
F
x
=
m
g
sin
θ
=
μ
N
∑
F
x
=
m
g
sin
θ
=
μ
N
∑
F
y
=
m
g
cos
θ
+
F
=
N
∑
F
y
=
m
g
cos
θ
+
F
=
N
Substituting for “N” in the first equation, we have :
⇒
m
g
sin
θ
=
μ
N
=
μ
m
g
cos
θ
+
μ
F
⇒
m
g
sin
θ
=
μ
N
=
μ
m
g
cos
θ
+
μ
F
⇒
F
=
m
g
sin
θ
−
μ
m
g
cos
θ
μ
⇒
F
=
m
g
sin
θ
−
μ
m
g
cos
θ
μ
⇒
F
=
50

25
3
0.5
=
13.4
N
⇒
F
=
50

25
3
0.5
=
13.4
N