The Falling Chain: - College of the Redwoods

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Transcript The Falling Chain: - College of the Redwoods

The Force of a Falling
Chain
By
Gina Giacone and Linda Lindsley
The Set-up:
• “A very flexible
uniform chain of
length L and mass M
is suspended from one
end so that it hangs
vertically,” the other
end is hooked on a
force sensor, though it
exerts no force at the
beginning.
The Problem:
• To calculate the
force that the chain
exerts on the sensor
as it falls to a
position hanging
from the hook of
the sensor
The force exerted by the chain on the sensor is a
combination of two forces:
•The impulse force(momentum) of each link as it
is stopped by the link above it (F1), and
•the weight force of the links already hanging
from the hook(F2).
The total force (FT) exerted on the sensor is:
FT = F1 + F2
• The chain itself is
made up of links that
are individual mass
elements (dm), which
are associated with
their length
increments (dx).
These individual masses and lengths are related to the
mass and length of the entire chain by:
dm M

dx
L
Rearranging this we get:
M
dm  dx
L
Center of Mass:
1 L
xc 
xdm

M 0
Velocity of the center of mass:
dxc
vc 
dt
dxi
1
vc 
m i
M
dt
m i v i
vc 
M
Rearranging the equation:
Mv c  m i v i  p i  P
The acceleration of the center of mass is:
1 dvc
ac 
M dt
dvi
1
a c  m i
M
dt
m i a i
ac 
M
Rearranging this equation we arrive at:
Mac  mi ai  Fi  F1
Previously defined is:
P  Mv c
So we have a more basic form of Newton’s Second Law:
dP
dm
F1 
v
dt
dt
Substituting dm into the previous force equation:
M dx
F1  v
L dt
M 2

v
L
Using Newton’s Second Law:
v  v  2ax
2
2
0
Where,
a  acce le rati
on due to gravity g
x  2x
v0  0
Then,
v  2 g(2 x )  4 gx
2
Substituting this into the previous equation,
M
x
F1 
(4 gx )  4Mg
L
L
In addition to the force of the falling link, the
chain already hanging on the sensor exerts a
force equal to its weight force,
x
F2  Mg
L
So, the total instantaneous force exerted on the
sensor as the final link falls is,
FT  F1  F2
x
x
 4M g  M g
L
L
x
 5M g
L
Since the sensor measures only force
versus time, a substitution of x=0.5gt2
is made,
2 2
5Mg t
FT 
2L
As the final link comes to rest:
F  Mg
Graph of Theoretical Values:
0
-0.05
____ Falling Force
-0.1
____ Weight Force
-0.15
____ Maximum Force
-0.2
-0.25
-0.3
-0.35
-0.4
-0.45
0
0.1
0.2
0.3
0.4
0.5
Experimental Data:
Length: 0.7645 m
Mass:
0.008479 Kg
Weight: 0.0832 N
Five times the weight force:
0.4159 N