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Calculating Speeds and Normal Forces of Packages Down a Chute: A Dynamics Problem, Study notes of Dynamics

A solution to calculate the velocities and normal forces of packages traveling down a chute using the principle of conservation of energy and inverse dynamics. The problem involves 50-lb boxes with an initial velocity of 3 ft/sec sliding down a chute with a height and radius of 5 feet. The document also includes a free body diagram and a force diagram to understand the forces acting on the packages.

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2009/2010

Uploaded on 12/09/2010

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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¿¿2V
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.
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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 =

mv 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 lbft 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 lbft, respectively. Thus the change in potential energy from point A is Tnew - 250 lbft. At point B, ∆ T = - 33.5 lbft, and by equation (1) ∆ V = + 33.5 lb*ft. From (2), ∆ V = V (^) 2 − V (^) 1 =

m ( 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.

The weight can be split into two components as shown in Figure 2 on the following page. FIGURE 2. FREE BODY DIAGRAM OF BOX Together, these forces contribute to the acceleration of the box along the curve of the chute. Because the velocity is parallel to the surface and the box is traveling along a curve, the acceleration is directed toward the center of curvature. Note that the acceleration is not completely perpendicular to the surface because the box is also accelerating down the chute. The observed acceleration and velocity is shown in Figure 3. FIGURE 3. OBSERVED MOTION Newton’s second law states [1]: Σ F (^) y = ma (^) y (5) If the y-axis is defined to be perpendicular to the surface of the chute, the acceleration in the y-direction must exactly equal the acceleration necessary to travel along a curve [1]. a (^) y = − v 2 r

Combing (5) and (6) results in: Σ F (^) y = NWcosθθ = − mv 2 r

Figure 1 shows that θB = θC = 30o. The previously obtained velocities and a radius 5 ft can be used to solve for the normal force N. Note that in position C, ay is positive. The normal forces at point B and C are calculated to be 27.1 lb and 133 lb. At point D, the radius of curvature becomes infinite and the slope becomes zero, resulting in a normal force equal to the weight, 50 lb. CONCLUSION The conservation of energy provides a quick and simple way of finding velocities after a change in height. Also, inverse dynamics (analyzing the motion to determine the forces) is shown to be a useful tool. REFERENCES [1] R. C. Hibbeler Engineering Mechanicsθ: Dynamicsθ. Prentice Hall, Upper Saddle River,