Game Development Reference
In-Depth Information
μ
μ
Surfaces
Aluminum on aluminum
1.40
1.10
Aluminum on steel
0.47
0.61
Copper on steel
0.36
0.53
Steel on steel
0.57
0.74
Nickel on nickel
0.53
1.10
Glass on glass
0.40
0.94
Copper on glass
0.53
0.68
Oak on oak (parallel to grain)
0.48
0.62
Oak on oak (perpendicular to grain)
0.32
0.54
Rubber on concrete (dry)
0.90
1.00
Rubber on concrete (wet)
0.25
0.30
Table 13.1. Typical values of the coefficient of kinetic friction
μ and coefficient of
static friction
μ .
Example 13.11. A block is resting on a horizontal plane for which the coeffi-
cient of static friction is given by
. Determine by what angle the plane
needs to be inclined before the block begins sliding under the influence of
gravity.
μ
=
0.5
S
Solution. We need to determine when the component of the gravitation force that
is parallel to the plane exceeds the static friction force. This occurs when
mg θμN μ mg θ
sin
=
=
cos
,
(13.108)
S
S
where θ is the angle of inclination. Solving for θ , we have
θ
=
tan
1
μ
26.6
°
.
(13.109)
S