slipping the map under the cap
TRANSCRIPT
Slipping the map under the cap
Dr Joseph Shapira
An abrupt pull of the map from the table leaves the table set (almost) intact. This is a familiar experience. An impressive demonstration of slipping the map under 24 table-sets over a 6 m long table, leaving the sets undisturbed (http://www.youtube.com/watch?v=vfnt8Sdj7cs#t=34) rose a vivid discussion in the physics-teachers community which deserves the following analysis.
Initial state: a cap ( marked A in Figure 1) is placed on the map covering a table, a distance d from its end. Its mass is mA and its coefficient of static friction is AS. The map is designated B, its mass is mB and its coefficient of static friction with the table is BS. A horizontal force F is applied to the map (pulling the map).
F
mBgmAgfB = (mA+mB)gBs
A
B
ʯʧ ʬʹdD
N
Figure 1: Initial state. Force diagram.
The following evolves in sequential phases. For sake of simplicity the cap is considered stable (assuming it is a point mass).
Phase 1: As the pull starts, the force F increases. The map does not move until F reaches the maximal static friction force between the map and the table.
F ≤ (mA+mB ) g μBs (1).
No horizontal force is applied to the cap, as the map does not move.
Phase 2: F surpasses the maximal static friction force between the map and the table:
1 Dr Joseph Shapira Slipping the map under the cap 1.4.2014
F
mBgmAgfB + fAB
A
שלחן d
fBA
F> (mA+mB ) g μBs.The map starts moving with acceleration aB. Two different phases ensue, depending on the coefficient of
static friction between the cap and the map:
1. The static friction between the cap and the map is high enough to move the cap with the map
without slipping aA=aB and mA aA <mA g μA s as portrayed in Figure 21.
Figure 2: Force diagram. The map accelerates and the cap moves with the map.
The horizontal forces on the cap:
f BA=mA aA
The horizontal forces on the map:
F−f AB−f B=mB aB=mB a A
⟹ F=mA aA+( mA +mB ) g μB+mB aB
or F=¿¿) (2)
which reaches its maximum for aA=gAS. The cap moves with the map as long as
F=(mA+mB ) (aB+g μB )<( mA+mB ) ( gμAs+g μB ) (3)Only when aA=gAS the static friction force is overcome and the cap slips on the map (see Figure 3).
1 The expression mAgAS defines the maximal static friction force between the cap and the map. The static friction force holds the cap to the map and is equal to mAaB.
2 Dr Joseph Shapira Slipping the map under the cap 1.4.2014
d
N
Figure 3: Balance of forces when the cap slips over the map
2. The maximal static friction between the cap and the map is small enough to allow the cap to slip over the map as the map starts moving(mA+mB ) ( gμ As+g μB ) ≤ (mA+mB ) g μBs (4)
The cap slips over the map with acceleration aA and against a dynamic friction force A. Then
F−mA aA−(mB+mA ) g μB=mB aB
F=mB (aB+g μB )+mA(g μA+g μB) ;(mA g μA=mA aA) . (5)The (dry) friction model assumes that the friction force mAgA is independent of the
value of cap velocity over the map, but only on the existence of a velocity difference between the map
and the cap. The initial condition for the slip is aB>aA , but once a difference in velocities is established
the slip continues, and the cap is accelerated, even when the map acceleration diminishes, as long as
there is a velocity difference.
These phases are portrayed in Figure 4. Equations (1)-(4) designate the value of the force for the
transitions between the phases.
3 Dr Joseph Shapira Slipping the map under the cap 1.4.2014
F
t
Static equilibrium
Case 1: The cap is accelerated with the map and slips once
the static friction is overcome
Case 2: The cap slips
Figure 4: Plot of the force along the phases
The distance that the cap travels before it slips off the map is readily calculated.
If the map accelerates with constant acceleration aB then the distance sA is
sA=da A
aB−aA= d
( aB
g μA)−1
Where d is the length of the “tail“ of the map as in Figures 1-3.
Once the cap slips off the map it moves on the bare table a distance sAT until it stops by the friction
between the cap and the table, according to its coefficient AT .
sAT/sA=ATA and the total distance is
sA+s AT=d
( aB
g μ A )−1(1+
μAT
μ A)
4 Dr Joseph Shapira Slipping the map under the cap 1.4.2014
The impulse acquired by the cap during the acceleration of the cap with the map (case 1) is neglected,
assuming that the rise of the force is fast and the time elapsed until the cap slips is negligible compared
to the time the cap slips.
A more realistic case will be addressed now, where the force is high at the start (the map is pulled
abruptly) up to time tm , and then it reduces to a value that allows the map to move at constant speed,
as portrayed in Figure 5.
t
v
tm td t0
Map’s speedCap’s speedForce
aA
aB
F
Figure 5: Sequence of events for the cap and for the map
The initial force accelerates the map with acceleration aB ≥ g μB+gμ AS up to time tm and
consequently maintains a fixed speed, by just overcoming the friction forces
F=(mA+mB ) g μB+mA gμ A
While the cap maintains a constant acceleration aA.
Two conditions are required for the cap to stay on the table:
1. The cap slips off the “tail” of the map (if not – it is dragged with the map from the table)
2. The path that the cap travels (sA+sAT) is shorter than the initial distance of the cap from
the end of the table D (see Figure 1) : D>sA+sAT.
The cap will slip off the “tail” of the map at a time td only as long as its speed is lower than that of
the map, td<t0. This is translated to a condition on the minimum length of time that the map is
accelerated tm (derived in appendix 1):
5 Dr Joseph Shapira Slipping the map under the cap 1.4.2014
tm
√ 2dg μ A
>
a A
aB
√1−aA
aB
The minimum duration of acceleration tm that ensures that the cap slips off the map is plotted in
Figure 6.
0.00
0.50
1.00
1.50
2.00
2.50
3.00
3.50
4.00
4.50
0.95 0.9 0.85 0.8 0.75 0.7 0.65 0.6 0.55 0.5 0.45 0.4 0.35 0.3 0.25
td>t0The cap is dragged with the map
td t0The cap slips off the map
Figure 6: minimum duration of acceleration
Slipping off the “tail” of the map is not enough. The point where the cap slips has to be far
enough from the end of the table to ensure that it stops before the end - sA+sAT<D. This
distance bounds the time of slippage td and depends only on the friction coefficients (refer to
appendix 2): t
d<¿√ 2 D
g μA(1+μ A
μ AT )¿
A few comments
1. Static friction is a shear force between surfaces adhering to each other by electrical, chemical or
surface roughness. The shear stress is elastic to the point of breakdown, which is very fast
avalanche process ( of the order of fractions of a second), and is therefore considered by the
model as instantaneous.
6 Dr Joseph Shapira Slipping the map under the cap 1.4.2014
2. Typical values for static friction coefficient between various materials are provided in
http://ffden-2.phys.uaf.edu/211_fall2002.web.dir/ben_townsend/staticandkineticfriction.htm
3. A successful demonstration of the cap slippage off the map depends on an initial acceleration of
the map exceeding the static friction force of the cap with the map. Once applied, the process of
tearing the static friction is almost instantaneous and does not cause any measurable shift of the
cap (it may cause the cap to fall over, however, due to the moment between its inertia, acting at
its center of mass, and the friction that acts at the base). The cap then accelerates steadily till its
speed matches that of the map or it slips off the “tail” of the map. Such demonstrations are
therefore successful when the “tail” of the map is short enough.
Conclusions
The mystery of the bikes pulling the map, that initiated this discussion, is unveiled in the following links
http://www.youtube.com/watch?v=4Uzc8wrKCrQ
http://www.youtube.com/watch?v=p9JrTUbtDNU
Acknowledgement
Dr. Yaron Lehavi and Dr. Hana Berger are acknowledged for reviewing, commenting and correcting.
Appendix 1
The distance that the “tail” of the map travelled is sB=12
aB tm2+aB tm (t d−tm )=aB tm( td−
12
tm).
The distance that the cap shifted is sA=12
aA t d2.
The difference between these when the cap slips off is d.
aB tm( td−12
tm)−12
aA t d2−d=0
But tm=aA
aBt 0
Then
t d2−2 t 0t d+
aA
aBt 0
2−2 da A
=0
7 Dr Joseph Shapira Slipping the map under the cap 1.4.2014
td 1,2
t 0=1±√1−( a A
aB)−2 d
aA( aA
aB)
21
tm2
But t d<t 0 (see figure 4). Only the second solution is applicable, and the expression under the square-
root sign has to be positive. Then
tm
√ 2dg μ A
>
a A
aB
√1−aA
aB
Appendix 2
The distance that the cap shifted until slipping off the map is sA=12
gμ A td2
The distance it slips over the bare table until stop is sAT=12
gμAT tAT2 where tAT is the duration of
stopping.
But v=g μ A t d=g μAT tAT and thereforetAT
t d=
μ A
μ AT.
The distance of the cap from the edge of the table at the time td has to exceed sAT
D−s A=D−12
g μA td2> 1
2g μAT tAT
2
D> 12
g μ A t d2(1+
μA
μ AT)
td<¿√ 2 D
g μA(1+μ A
μ AT )¿
8 Dr Joseph Shapira Slipping the map under the cap 1.4.2014