BASIC DYNAMICS OF PACKAGES DOWN A CHUTE
[my name excluded for privacy purposes]
EME 3043 Dynamics
Dr. Vladimir Vantsevich
INTRODUCTION
This paper demonstrates a simple
approach of calculating the speed and normal
forces that act on a body as it travels down a
chute. This method can be generalized to
determine the motion and forces of a point-
mass as it travels along a curved surface.
PROBLEM: PACKAGES ALONG A CHUTE
50-lb boxes with an initial velocity of 3
ft/sec slide down a chute to a conveyor belt.
The chute has a height and radius of 5 feet, as
shown in Figure 1. The speed and normal force
of the chute must be determined at points B, C,
and D.
FIGURE 1. DIAGRAM OF BOXES TRAVELLING DOWN
CHUTE
In this problem, friction may be
neglected, and the boxes may be treated as
point masses.
DETERMINING VELOCITIES USING
CONSERVATION OF ENERGY
Since friction can be ignored and thus
the velocity of a box is path-independent, the
principle of conservation of energy [1] may be
used
T
2
−T
1
=−(V¿¿2−V
1
)¿
(1)
where T equals the potential energy of the box
and V equals its kinetic energy [1]. The
equation states that the change in potential
energy must be equal to the negative of the
change in kinetic energy. In other words, the
total energy of a box must remain constant
throughout its entire journey. Potential energy
and kinetic energy are respectively given by the
following equations [1]:
T=Wh
(2)
V=1
2mv
2
(3)
Since potential energy is relative, a
reference height must be chosen. In this
example, the bottom of the chute will be
assigned a height of h = 0 ft. By simple
trigonometry, points A, B, C, and D have
relative heights of 5, 4.33, 0.67, and 0 ft,
respectively.
Calculating T using equation (2) and W
= 50 lb, h = 5 ft yields a potential energy of T =
250 lb*ft at point A. By the same method, the
potential energies of a box and points B, C, and
D, are found to be 216.5, 33.5, and 0 lb*ft,
respectively. Thus the change in potential
energy from point A is Tnew - 250 lb*ft.
At point B, ∆T = -33.5 lb*ft, and by
equation (1) ∆V = +33.5 lb*ft. From (2),
∆ V =V
2
−V
1
=1
2m(v
2
2
−v
1
2
)
(4)
Using m = (50 lb)/(32.2 ft/sec), the velocity at
point B is calculated to be 7.22 ft/sec.
Employing the same technique for points C and
D yields 17.0 ft/sec and 18.2 ft/sec.
NORMAL FORCE CALCULATION
Two forces are acting on the body: the
normal force, which acts perpendicular to the
surface, and the weight which acts downward.
