Game Development Reference
In-Depth Information
2.
Referring to the schematic diagram shown in Figure 3-2 earlier in this chapter, the
box will begin to slide when the force on the box parallel to the ramp is equal to the
maximum static friction force.
mg
sin
qm q
=
mg
cos
s
The angle at which the box begins to slide is therefore equal to the following:
tan
q
==
0.61
or
θ
=
31.4
s
An interesting thing to note about this problem is that the angle at which sliding begins
is independent of the mass of the box. A 10
kg
box will begin to slide at the same angle
as a 2
kg
box. This is an experiment you can easily try out on your own. If you don't have
a steel plate or aluminum box, use some other combination of materials listed in Table 3-1.
3.
As shown in Figure 3-14, the spring will reach its new equilibrium position when upward
force due to the extended spring balances the downward force due to the 1
kg
mass. If the
spring deflection is taken to be in the z-direction, the equation becomes
kz mg
Δ=
mg
9.81
Δ=
z
=
=
0.0302
k
325
-kΔz
mg
Figure 3-14.
The location of the spring is where the spring and gravity forces balance.
The spring will end up 3.02
cm
from its original equilibrium position.
4.
The radius of the moon's orbit will be at a value where the gravitational force between
the earth and moon equals the centripetal force required to keep the moon in that orbit.
2
Gm m
m v
me m
=
2
r
r
Gm
6.67
e
−
11*5.974
e
+
24
r
=
e
=
=
3.81
e
+
8
m
2
2
v
1023
