Game Development Reference
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component of gravity perpendicular to the surface, and from Table 12.1, the coe
cient of
static friction between concrete and wood is 0.62. Thus, we have
f
s
= g
,
(
s
n) = g
,
(0.62g
⊥
) = g
.
The normal and lateral components of gravity can be expressed in terms of the angle of
inclination θ as
g
⊥
= gcos θ,
g
= gsin θ,
where g is the total magnitude of the force of gravity on the block. Plugging in these
values and solving this for θ, we have
0.62g
⊥
= g
,
0.62gcos θ = gsin θ,
gsin θ
gcos θ
,
0.62 =
sin θ
0.62 =
cos θ
,
0.62 = tan θ,
arctan 0.62 = θ,
32
o
≈ θ.
Notice that neither the weight of the block nor the acceleration due to gravity was relevant
in this experiment. Thus, if you were to conduct this experiment on the moon, you would
get the same critical angle.
6.
(a) Hooke's law tells us f = kx
0
. The force in this case is gravity, which is proportional
to the mass and given by f = mg. Thus, the relation is
mg = kx
0
.
(b) Substituting into the equation obtained in part (a), we have
mg = kx
0
,
(5.00 kg)(9.8 m/s
2
) = k(10.0 cm),
49 N
0.100 m
= k,
4.9×10
2
N/m = k.
(c) mg = kx
0
m(9.8 m/s
2
) = (4.9×10
2
N/m)(17.0 cm)
m =
(4.9×10
2
N/m)(0.170 m)
9.8 m/s
2
= 8.5 kg
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