The African Review of Physics (2012) 7:0009
67
Friction Controlled Three Stage Ladder Sliding Motion in a Non-conservative System:
From pre-detachment to post-detachment
Priyadarshi Majumdar1 and Sudipto Roy2
Jyotinagar Bidyasree Niketan Higher Secondary School, 41 Jyotinagar, Kolkata, India
2
Department of Physics, St. Xavier's College, 30, Mother Teresa Sarani, Kolkata, India
1
We have studied analytically and through numerical simulation a 2-dimensional ladder sliding through a rough wall and
attached floor. Our analysis mimics the actual situation successfully. We have shown that the initial external angle between
the ladder and the floor must be greater than some minimum value depending upon the parameters of the model (frictional
parameters) to initiate the constraint natural downfall of the ladder guided through the wall and floor. Also during the natural
down fall of the ladder there is a certain maximum value of the external angle mentioned earlier, at which the upper end of
the ladder detaches from the wall to avoid the paradoxical situation that its velocity circumvents the velocity of light.
Different physical situations like the effect of increasing or decreasing the length, mass, frictional effects, initial angle related
to the ladder are mimicked using simulation. The sliding ladder has been compared with a sliding bead through a rough
circular track. The situation after the detachment until the end of the motion is also analyzed. At each and every stage we
have also examined the mechanical energy lost from the system due to dissipative effect.
1.
Introduction
The problem of sliding dynamics of a uniform rigid
ladder under gravity with the additional constraint
that its upper and lower ends are in contact with the
wall and the floor (Fig. 1), respectively, is a
fundamental problem in classical mechanics [1].
While most of the related works on sliding ladders
are performed on conservative systems [2-8] there
is no such material in the literature, which tackles
the problem in a fully non-conservative system
where both the wall and the floor are rough ones. In
our present work, we consider a detailed analysis of
the sliding ladder in a complete non-conservative
system (rough wall and floor) having definite
frictional coefficients. In fact, for simplicity we
have taken the kinetic frictional coefficient same as
the coefficient of static friction, although the
former is slightly less than the latter. We briefly
discuss this problem below.
We choose a rectangular two-dimensional
Cartesian system of coordinates (although the
actual problem is clearly three dimensional but we
are allowed to choose its projection in 2-d without
loosing any valuable information at all). Here x and
y axes are taken along the floor and the wall,
respectively, (although the angle between the wall
and the floor can be taken as acute or obtuse but the
corresponding analysis will be more complicated).
The length and the mass of the ladder are taken as l
and m, respectively.
__________________
1
[email protected]
Y
A (0, y)
(X,Y)
π-θ
O
θ
B (x, 0)
X
Fig.1: A ladder supporting on a vertical wall and a
horizontal floor, (X ,Y) being its centre of mass.
At an instant t of its motion it makes an angle
θ (t ) with the positive direction of floor (X axis) as
shown in Fig.1. If it is assumed that the two ends
(A and B) are in contact with the wall and the floor,
respectively, during the entire process of its
downward fall, the constraint relation is clearly
x2 + y2 = l 2
(1)
y& = −( x / y ) x&
(2)
The African Review of Physics (2012) 7:0009
68
As the natural downward motion of the ladder
continues x increases and y decreases but x&
remains finite. Now, from Eqn. (2) as
y → 0 , we have y& → ∞
(3)
It implies that if the upper end (A) never looses
contact with the wall, its velocity ( y& ) will become
infinitely large as y → 0 . This phenomenon is
discussed in [2-8]. It is known that an obvious
solution to this problem is a detachment [2-5] of
the ladder from the wall much earlier than y → 0
(which can be observed even in a simple in-house
experiment). In this work, we consider the
following stages of ladder movement.
Stage 1: Starting from the initialization of its
motion and sliding phase through the wall and the
floor, the inputs to the first stage of motion are set
with the analyses of forces and torques.
Stage 2: After detachment from the wall and
simultaneous angular downfall of its upper end and
dragging motion of its lower end through the
ground, the inputs to this stage are the
corresponding simulation outputs of the first stage.
Stage 3: Finally, the dragging motion through the
floor of the entire ladder after the upper end
touches the ground and before it finally comes to
rest at the end, the inputs being the second stage
simulation outputs.
In the figure below (Fig. 2) we have shown the
schematic of the block diagram of the three stage
process we have discussed above.
Fig.2: Schematics of a three stage dynamics of the ladder sliding-dragging problem: the output of one stage being fed to the
input of next stage.
Kapranidis and Koo [2] in their work have
elaborately discussed the cases of sliding ladder in
a conservative system (both constrained as well as
unconstrained). In fact, they consider that initially
the ladder was inclined at a certain angle with the
wall and then starts falling. They derived the
equation of motion of the ladder from the
conservation of energy principle. Also computing
the reaction of the wall on the ladder they have
shown that when the ladder falls a vertical distance
equals to one third of its initial height along the
wall the said detachment occurs. They also
discussed the motion of the ladder after detachment
from the wall and shown that at that stage the
motion is governed only by the weight of the ladder
and the reaction at the point of contact of between
the ladder and the floor. Given an initial angle, the
velocity of the top, when the ladder hits the ground,
is proportional to the length of the ladder. The
cases of the constrained ladder were also discussed
by Kapranidis and Koo in which they have shown
that for a ladder which is compelled to be attached
to the wall while it is sliding down, x& → 0 as
y → 0 , hence not violating the law of nature.
Finally, they ends with the conclusion that no
external forces to the ladder can be adjusted so
that x& remains constant during the entire sliding
period of the ladder even its upper end hits the
ground. In a supporting work, Scholten and
Simoson [3] proved that in a conservative force
field it is indeed possible to make x& constant but
that will definitely leads towards detachment of the
ladder from the wall at some critical height. The
equation of motion derived by them when the
ladder is in contact with the wall and after it looses
contact from the wall resembles with those by the
others. They also determine the critical height
(which is equal to two third of the initial height of
the ladder along the vertical wall) and the
corresponding velocity in some specific cases to
mimic the real situation. The work of Freeman and
Palffy-Muhoray [4] also support the idea of critical
height in conservative system by determining an
analytical expression of it as a function of the
The African Review of Physics (2012) 7:0009
69
constant horizontal dragging velocity. In a different
work by McDonald [5] the ladder sliding problem
in a conservative system has been analyzed using
the Lagrange’s method as well as the torque
analysis. His main idea is to derive the equation of
motion of the ladder by torque analysis about
different identifiable points of the system and then
compute the angle of detachment again by setting
the expression regarding the normal reaction of the
wall to zero.
2.
+ 2m
l
X (t ) = − cos θ (t )
2
l
sin θ (t )
2
l2
4
(8a)
(8b)
Now from (6) and (8b)
mX&& = Rw − µ f R f
(9a)
mY&& = R f + µ w Rw − mg
(9b)
r d
 ml 2 && ˆ

T =  ( Iθ&) kˆ = 
θ k
 dt

 3 
(10)
(4a)
Again,
(4b)
(5)
Apart from the gravitational force (mg) acting
vertically downwards through CM, there are
reaction forces from the wall and the floor acting
on the upper end (A) and on the lower end (B),
respectively. These two forces may be denoted by
Rw and Rf , acting along positive X axis and positive
Y axis, respectively. There are two frictional forces
(either limiting or during its motion), µ w Rw and
µ f R f acting on the upper and lower ends of the
rod along the positive Y and the negative X
direction, respectively. Here µ w and µ f are the
coefficients of kinetic friction (almost same as the
static frictional coefficients) of the wall and the
floor with the ladder, respectively.
r
Resultant force ( F ) acting on the ladder is
given by,
r
F = ( Rw − µ f R f )iˆ + ( R f + µ w Rw − mg ) ˆj
(7)
r
dP
= mX&&iˆ + mY&&ˆj
dt
i.e.,
X 2 +Y 2 =
l &&
X sin θ )kˆ
2
r
P = m[ X& (t )iˆ + Y& (t ) ˆj ]
Modeling
We ought to solve this paradoxical situation in this
work by considering the effects of all the forces
and torques acting on the system in details. Due to
uniform density and uniformity of cross section,
the CM of this rod is located at the middle (as
shown in Fig. 1), i.e., at a distance l/2 from its
either end. From Fig. 1 we have the following set
of constraint equations
Y (t ) =
r
T = (lµ w Rw cosθ − lRw sin θ − 0.5lmg cosθ
r
Also, the momentum ( P ) equations for CM are
Modeling and Simulation Before
Detachment
2.1.
Similarly, resultant torque acting on the ladder
along the axis perpendicular to the XY plane and
passing through its lower end B (Fig. 1) must
incorporate the effect of coordinate forces [5] (as B
is accelerating) and is given by,
From (7), (9a), (9b) and (10)
 mX&& + µ f (mY&& + mg ) 
ml &&

θ = ( µ w cos θ − sin θ )


3
1+ µ w µ f


−
mg
cos θ + mX&& sin θ
2
(11)
Now using (4a), (4b) and (11)
0.5lθ&&sin θ + 0.5lθ& 2 cos θ
 g cos θ
l &&
θ = f (θ ) 
−
2
&&
&
3
2
+ µ f g + 0.5l (θ cos θ − θ sin θ ) 
(
)
+ 0.5l sin θ (cos θθ& 2 + sin θθ&&).
(12)
with
 µ cos θ − sin θ
f (θ ) =  w
 1+ µ µ
w
f

(6)
and




(13)
The African Review of Physics (2012) 7:0009
Rw (t ) =
mX&& + µ f (mY&& + mg )
1 + µwµ f
mY&& − µ w mX&& + mg
R f (t ) =
1 + µwµ f
70
(15)
Because of its complexity it seems that the
analytical solution of Eqn. (12) is not possible
hence we have simulated it numerically using
MATLAB in intel platform to study the behavior of
the motion with the elapse of time. If in particular
we put µ f = µ w = 0 in Eqn. (12) we may obtain
the corresponding equation in conservative system
[2-5].
Since Eqn. (12) is a second order differential
equation in θ , hence to solve it we need two initial
conditions, one on θ and the other on θ& , namely
θ i and θ&i respectively. We have to be very careful
while choosing the initial conditions now. In order
to initiate both the translational (along X-axis and
Y-axis separately) and rotational motion about any
point, it must start simultaneously. All these can
only happen simultaneously if we have net forces
along +ve X-axis, -ve Y-axis and some net torque
along +ve Z-axis about any point in the system. As
a consequence of these we may have the following
set of inequalities:
2
cos θ iθ&i + sin θ iθ&&i ≥ 0
2
cos θ iθ&&i − sin θ iθ&i ≤ 0
tan θ i ≤
(14)
(16a)
(16b)
µ w Rw cosθ i − Rw sin θ i − 0.5mg cosθ i + mX&&i sin θ i ≥ 0
(16c)
Both, Eqns. (16a) and (16b), and the real life
situation suggest that if we supply suitable initial
angular velocity θ&i then even for a small initial
angle θ i (but obviously greater than 90 0 ) the
motion of the ladder can be initiated but those
motions are not due to natural initiation and hence
will not be discussed. We shall consider only the
self-initiated motion of the ladder starting from
zero initial angular velocity and at a definite
calculable initial angle, which perfectly mimics the
real situation that when θ i is greater than or equal
to a certain angle, depending upon the parameters
of the model, the ladder starts falling with zero
initial angular velocity. With the said assumption
the first two of the above set of inequalities become
obvious (irrespective of the system parameters)
while Eqn. (16c) yields
µ f µw −1
2µ f
(16d)
implying
θ i ≥ tan −1
µ f µw −1
2µ f
(16e)
Hence we may say that if the ladder starts from rest
(with only potential energy) then the starting
criteria depend only upon the friction of the floor
and attached wall and nothing else (not on the mass
and length of the ladder). The observational fact
also supports this. The corresponding result in the
conservative system [1-8] will obviously be
θ i = π / (since µ f = µ w = 0 ) [5].
We have presented the simulation results of
Eqn. (13) with the aid of Eqn. (16d) and the
numerical values of the parameters ( m, l , µ f , µ w )
involved in the dynamics. In the simulation study
of Eqn. (12), we observe that during the
constrained downfall of the ladder along the wall, a
situation arises when Rw becomes zero indicating a
detachment from the wall. Once any such
detachment occurs, Eqns. (1)-(4) become invalid,
as a result any further simulation leads negative Rw .
2.2.
Simulation
In the simulation process we have supplied the
initial condition on the angular position ( θ i ) of the
ladder. But to obtain a physical solution for θ and
of its two time derivatives ( θ& and θ&& ) we have to
choose the parameters and initial conditions
properly so as to mimic the real situations like:
(i) If, θ&i = 0 , the motion cannot be initialized
unless θ i becomes greater than or equal to some
definite θ , which again depends upon the
parameters of the model (we have already
mentioned this).
(ii) With the increasing friction the initial angle
increases and for very large friction the motion will
not initiate at all etc.
These are the phenomena, which can be
visualized easily in a simple in house experiment
even with a yard stick. If we are not aware of these
real life analogies while simulating equations like
Eqn. (12), we may obtain an output of the program
in which θ decreases from the very beginning,
which must not be the case. We have used the 4th
order Runge-Kutta simulation technique [9] in
MATLAB in a Windows PC (intel platform) to
obtain the nature of variations (Fig. 3) of different
The African Review of Physics (2012) 7:0009
71
t = t d , the time of detachment, the constraint
relations Eqn. (1) and Eqn. (2) will no longer be
valid. From Fig. 3, it is clear that both the angular
velocity and angular acceleration of the ladder
increase with time but the reaction forces as well as
the total energy of the ladder decrease as expected.
Some important simulation results of Eqn. (12)
are presented below in Table 1.
variables of the system defined earlier with time (t)
and also with angular position ( θ ).
All the graphs are plotted for a certain range of
θ , starting after the initiation of motion and just
before Rw becomes zero or negative imply
detachment from the wall and the corresponding
value of θ can be numerically identified as θ d , the
angle of detachment. It is that particular angle just
after which the simulation results of Eqn. (12)
become absurd, i.e., after an angle θ = θ d at
Table 1
Input
m = 5kg , l = 2m, µ f = 0.7, µ w = 0.1,
Output
θ d = 170.802 0 , θ&d = 96.647 0 s −1 ,
E i = 20.708 J ,θ i = 155 0 (> 146.402 0 ),
θ& = 0,θ&& = 170.547 0 s − 2 ;
i
θ&&d = 389.497 0 s − 2 , t d = 0.39s,
E d = 17.318 J , Y&d = −1.665ms −1
i
The initial height of the topmost point of the ladder
along the vertical wall is 0.8452 m while that at the
time of detachment is 0.3196 m. Hence unlike in
a
b
400
80
350
60
300
dθ /dt
170
θ
c
100
d2θ /dt2
175
165
the case of conservative system [3,4] the ladder has
fallen more than one third of its initial height.
40
160
20
155
200
0
0
0.2
t
0.4
150
0
35
30
250
0.2
t
0.4
30
21
20
20
10
19
0
0.2
t
0.4
0
0.2
t
0.4
20
0
15
10
E
Rf
Rw
25
0
0.2
t
0.4
-10
18
0
0.2
t
0.4
17
Fig.3: Simulation results of Eqn. (12) for m = 5kg , l = 2m, µ f = 0.7, µ w = 0.1,θi = 1550 , where θ i is greater than the
minimum required value 146.402 0 . All the results are plotted against t (sec.). In panel a (up and down) angle θ and R f are
plotted. In panel b (up and down) rates of variation of θ and Rw are plotted. In panel c (up and down) the rate of variation
of angular velocity and the energy of the ladder are plotted. The unit of θ is taken as degree and all other variables are in
S.I. The time at which Rw becomes zero is t d = 0.39 s. The loss of energy due to this entire process is 3.39J.
The African Review of Physics (2012) 7:0009
2.3.
72
(due to wall and floor) and energy of the system
decrease. Also at the instant of detachment we may
have
Real life mimicry
Fig. 3 shows that as the ladder leads towards
detachment from the wall both its angular velocity
and acceleration increase while both the reactions
[(
Rw
θ =θ d
)
)]
(
1
2
2
ml θ&&d sin θ d + θ&d cos θ d + µ f θ&&d cos θ d − θ&d sin θ d + µ f mg
2
=
=0
1+ µw µ f
Hence
(17a)
2µ g
2
− ( µ f sin θ d − cosθ d )θ&d + (µ f cosθ d + sin θ d )θ&&d = − f
l
(iii) Increase (decrease) in θ d and t d as a result of
increasing (decreasing) µ f and (or) µ w (see Figs.
6 and 7), other parameters kept constant.
(iv) Decrease in Y&d and t d , but increase in θ d with
the increase in θ i (see Fig. 8) keeping the other
parameters constant.
which can easily be verified using the data from
Table 1. Also θ d satisfies some demands of the
real situation that we may list below:
(i) Independence of θ d , Y&d and t d of mass (m) of
the ladder (see Fig. 4).
(ii) An increase in θ d , Y&d and t d with the increase
in length (l) of the ladder (see Fig. 5), other
parameters kept constant.
171
2.5
1.5
170.5
2
1
170
t
v
d
θd
2
d
3
1.5
0.5
169.5
1
0
169
0.5
-0.5
0
50
m
100
(17b)
168.5
0
50
m
100
0
50
m
100
Fig.4: Variations of the vertically downward speed vd = Y&d (ms −1 ) of the ladder along the wall at detachment t d (s) (left), that
of the time of detachment (middle) and finally that of the angle of detachment (degree) (right) with the mass of a ladder. The
parameters of the system are the length of the ladder (l) = 2m, µ f = 0.7, µ w = 0.1 . The motion of the ladder initiates at the
angle defined by the lower limit of Eqn. (16e). All these graphs show that the variables plotted against m (kg) are
independent of m.
The African Review of Physics (2012) 7:0009
73
5.5
4.5
5
4
4.5
169.5
169.4
3.5
169.3
4
θd
d
t
v
d
3
3.5
169.2
2.5
3
169.1
2
2.5
169
1.5
2
1.5
1
0
5
10
l
15
20
168.9
0
5
10
l
15
20
0
5
10
l
15
20
Fig.5: Variations of the vertically downward component of velocity in (ms −1 ) of the ladder along the wall at detachment
(left), that of the time of detachment in (s) (middle) and finally that of the angle of detachment (degree) (right) with the
length of a ladder. The parameters of the system are the mass of the ladder (m) = 5kg, µ f = 0.7, µ w = 0.1 . Similar to the
pervious case the motion of the ladder initiates at the angle defined by the lower limit of Eqn. (16e). All the graphs show that
as the length of the ladder increases (decreases) all three y-variables increase (decrease) too.
1.75
0.82
176
0.81
175
1.7
0.8
174
0.79
1.65
θd
vd
173
td
0.78
172
1.6
0.77
171
0.76
1.55
170
0.75
1.5
0.1
0.2
0.3
µw
0.4
0.5
0.74
0.1
0.2
0.3
µw
0.4
0.5
169
0.1
0.2
0.3
0.4
0.5
µw
Fig.6: Variations of the vertically downward speed in ( ms −1 ) of the ladder along the wall at detachment (left), that of the
time of detachment in (s) (middle) and finally that of the angle of detachment (degree) (right) with µ w . The parameters of
the system are length of the ladder (l) = 2m, mass (m) = 5kg, µ f = 0.7 , the motion of the ladder initiates at the angle defined
by the lower limit of Eqn. (16e). The graphs indicate that as the detachment-speed decreases (increases) with increasing
(decreasing) wall roughness the other two parameters still increase (decrease).
The African Review of Physics (2012) 7:0009
74
As we can see from Fig. 5 (left) that the
vertically downward component of the velocity at
detachment increases with the length of the ladder.
Further, simulation shows that if we keep on
increasing the length with adjusted frictions then
although the said speed will increase, but in order
that speed to become comparable with that of light
the corresponding length of the ladder should be
enormous making the entire problem unrealistic,
hence the situation mentioned in Eqn. (3) can never
happen in reality, which discards any use of the
special theory of relativity in the context of this
problem.
However, an interesting situation arises in Fig.
8 (3rd sub-figure), where we can see that the upper
end of the ladder almost touches the ground just
before detachment but still there is no such
violation of the relativistic assumption because as
we can see from the graph that the corresponding
angle of initialization is very large making the
validity of the motion in a very short angular range
with a little amount of initial potential energy
imparted to the system. Consequently, x& → 0 as
y → 0 , keeping y& finite.
Also by varying the frictional coefficients (Figs.
6 and 7) and initial angle (Fig. 8) it is verified that
the angle of detachment varies with all of them
which nullifies any possibility of a definite critical
height [2-5] in non conservative system unlike in
the conservative system.
1.81
2
170
1.8
1.8
165
1.79
1.6
160
1.78
1.4
θd
155
td
vd
1.77
1.2
1.76
150
1
1.75
1.73
145
0.8
1.74
0
0.5
µf
1
0.6
0
0.5
µf
1
140
0
0.5
1
µf
Fig.7: Variations of the vertically downward speed of the ladder in ( ms −1 ) along the wall at detachment (left), that of the
time of detachment in (s) (middle) and finally that of the angle of detachment in degree (right) with µ f . The parameters of
the system are length of the ladder (l) = 2m, mass (m) = 5kg, µ f = 0.1 , the motion of the ladder initiates at the angle defined
by the lower limit of Eqn. (16e). The graphs indicate that both the detachment-speed and the corresponding time decrease
(increase) with increasing (decreasing) wall roughness but the corresponding angle still increases (decreases).
The African Review of Physics (2012) 7:0009
1.75
75
0.8
1.7
180
0.7
178
1.65
0.6
1.6
176
θ
t
V
d
d
d
0.5
1.55
1.5
174
0.4
1.45
172
0.3
1.4
170
0.2
1.35
1.3
140
160
180
0.1
140
160
θi
168
140
180
160
θi
180
θi
Fig.8: Variations of the vertically downward speed of the ladder in (ms −1 ) along the wall at detachment (left), that of the
time of detachment in (s) (middle) and finally that of the angle of detachment in degree (right) with the initial angle (also in
degree) of the ladder. The parameters of the system are length of the ladder (l) = 2m, mass (m) = 5kg, µ f = 0.7 , µ w = 0.1 .
The graphs indicate that both the detachment-speed and the corresponding time decrease (increase) with increasing
(decreasing) initial angle but the corresponding angle of detachment still increases (decreases).
2.4.
Energy analysis
30
Now at any instant the energy of the sliding ladder
is given by
PE
TKE
RKE
25
l


= 0.5ml  g sin θ (t ) + θ&(t ) 2 
3


(18)
Energy
20
E (t ) = mgY (t ) + 0.5m( X& (t ) 2 + Y& (t ) 2 ) + 0.5Iω (t ) 2
15
10
5
Hence the rate of loss of energy and the work done
by the ladder due to an elemental motion can be
expressed as
0
0
0.1
0.2
0.3
0.4
t
0.5
0.6
0.7
0.8
(19)
Fig.9: Various energy components (potential,
translational kinetic and rotational kinetic with their
abbreviated names mentioned in the legend) of the
sliding ladder (in J) in the non-conservative system have
been plotted against time. The parameters of the model
being m = 5 kg, l = 2 m, µ f = 0.7, µ w = 0.1 . As can be
Also we may obtain the elemental loss of energy of
the ladder in a time interval dt as
seen from the graph similar to conservative system the
potential energy decreases but both forms of kinetic
energy increases. But unlike the conservative system
here the sum total of all the energy decreases.
dE
= mgY& + mX&X&& + mY&Y&& + 1 / 12ml 2θ&θ&&
dt
2l 

= 0.5mlθ& g cosθ + θ&&
3 

E (t ) − E (t + dt ) = ( µ f R f − Rw )dX − ( R f + µ w Rw ) dY − Iθ&&dθ
(20)
In Fig. 9 we have plotted various energy
components of the constrained ladder against time.
It can be easily verified that the sum of the yheights of the energy graphs decrease with time
and the rate of decrease is actually the rate of work
against the frictional forces.
The African Review of Physics (2012) 7:0009
76
µR = − µ w Rw cosθ + µ f R f sin θ
In particular for a conservative system, Eqn. (20)
will get rid of µ f and µ w [2,5]. The resulting form
(23)
of Eqn. (20) will then merely indicate the work
done due to the constraint forces and torques (the
reaction applied by the wall and the floor and the
system torque). As we know that a conservative
system can never gain (loose) energy from (due to)
the constraint forces and torques hence the righthand side of Eqn. (20) must become zero implying
rightly that the system energy must remain
conserved. As an immediate implication of that,
right hand side of Eqn. (19) will also be zero,
which yields the equation of motion of the sliding
ladder and can easily be identified with Eqn. (12)
by putting µ f = µ w = 0 [2-5].
2.5.
Comparison with the sliding bead
We may put another interesting analysis here. We
know that during the constrained motion of the
ladder dragging through the wall and the floor its
CM always lies on a circle [1]. Hence the problem
of sliding ladder in a conservative system
( µ f = µ w = 0 ) can be identified to the slipping
motion (without rolling) of a point mass m along
the circumference of a frictionless circular path of
radius l/2. And the point of detachment of that
point mass will be exactly two third of its initial
height as in the case of the ladder, the problem
which is taught in the high school physics [1]. We
now want to check whether the same concept (i.e.,
the point mass m is traveling along the
circumference of a rough circular path of radius l/2,
obviously without rolling) can be applied in the
present non conservative case or not. If it is so then
the loss of energy by the ladder due to an elemental
motion (dx, dy) along the coordinate axes must be
the same as that by the corresponding point mass
during its identical elemental circular motion along
the imaginary rough track (the situation is
illustrated in Fig. 10) implying
µ w Rw dy + µ f R f dx = µR[(dx) 2 + (dy ) 2 ]
1/ 2
(21)
µ and R being the frictional coefficient and
normal reaction related to that imaginary rough
track. Now following Eqn. (1) and Eqn. (2) we may
write Eqn. (21) as
µ w Rw
x
µRl
+ µ f Rf =
y
y
yielding finally (following Fig. 1)
Fig.10: The motion of the point mass along the circular
track. It starts at P and detaches at R. The corresponding
polar
angles
are
and
∠POAi = φi = π − θ i
∠ROA f = φ f = π − θ f respectively. Ai Bi and A f B f are
the initial and final positions of the ladder respectively. Q
is an intermediate position of the point mass (the
corresponding ladder position being A(θ ) B (θ ) . The
radial reaction R (θ ) and transverse friction F (θ ) are
acting
on
it
in
that
position
(where
∠QOA(θ ) = φ = π − θ follows from Eqn. (26)).
Now for our assumption to be true the normal
reaction R must always act along the radial
direction while the frictional force Eqn. (23) should
act along the transverse direction implying
Rrˆ = R w iˆ + R f ˆj
(24a)
µRφˆ = − µ w Rw iˆ + µ f R f ˆj
(24b)
rˆ = (cos φ )iˆ + (sin φ ) ˆj
(25a)
φˆ = (− sin φ )iˆ + (cos φ ) ˆj
(25b)
where
are the radial and transverse unit vectors as usual
with the identification
φ = π −θ
(26)
(22)
from Fig. 4. Now all the results Eqns. (23)-(26) can
be verified to be consistent only when
The African Review of Physics (2012) 7:0009
77
µ = µ f = µw
initial angle of the ladder to a value less than the
minimum required value and then imparted an
initial angular velocity to the system. Similar to
real life situation our simulation result suggests that
the system will start moving as in the previous
cases. But unlike in the other cases here the initial
source of energy of the ladder is three-fold namely:
potential, kinetic as well as rotational. The entire
situation is been illustrated in Fig. 11 below. Fig.
11 shows that as the ladder leads towards
detachment from the wall its angular velocity,
acceleration increase while the reaction at the floor
decreases, but that due to the wall first increase and
then decreases. The energy of the system decrease
as usual.
(27)
i.e., only in the symmetric case. Hence we may
conclude that our present ladder sliding problem in
fully non-conservative system with perfect
symmetry (i.e., identical friction in both the wall
and floor) can be translated to a problem of
slipping of a mass point having the same mass as
that of the ladder along a rough circular track
having the similar roughness as the wall and floor.
Also the detachment for both the problems must
also happen at the same position of the mass point
and CM of the ladder.
2.6.
Effect of imparting initial angular velocity
We may now study the effect of imparting some
initial angular velocity to the ladder. We set the
a
b
150
130
145
125
θ
140
135
c
300
200
d2θ /dt2
135
dθ /dt
155
120
100
0
115
130
110
0
0.1
t
0.2
40
-100
0
0.1
t
0.2
15
0
0.1
t
0.2
0
0.1
t
0.2
50
48
10
30
E
Rf
Rw
46
5
44
20
0
10
42
-5
0
0.1
t
0.2
40
0
0.1
t
0.2
Fig.11: Simulation results of (12) with m = 5kg , l = 2m, µ f = 0.7, µ w = 0.1,θi = 131.760 , where θ i is less than the minimum
required value 146.402 0 , θ&i = 114.59 0 s −1 . All the results are plotted against t (sec.). In panel a (up and down) angle θ and
R f are plotted. In panel b (up and down) rates of variation of θ and Rw are plotted. In panel c (up and down) the rate of
variation of angular velocity and the energy of the ladder are plotted. The unit of θ is taken as degree and all other variables
are in S.I. The time at which Rw becomes zero is t d = 0.162 s. The loss of energy due to this entire process is 8.53J.
The African Review of Physics (2012) 7:0009
3.
78
Vanishing friction for smooth horizontal surface
( µ f = 0 ) clearly indicates (both from Eqn. (28a)
Modeling and Simulation After
Detachment
3.1.
and Eqn. (30)) that the ladder travels with a
constant horizontal velocity after detachment. On
the other hand, following (29) and (30) we finally
obtain
Modeling
Let us now discuss the situation after detachment
of the ladder from the wall. Just after detachment
the motion of the ladder can be described as
mX&& = −µ f R f
(28a)
mY&& = R f − mg
(28b)
(
 sin φ cos φ + µ f sin 2 φ
 2
&
&
φ&
φ =
2
 2 / 3 + µ sin φ cos φ − sin φ 
f


−
)
0.5mgl cos φ − ml X&& − 0.5l cos φφ& 2 − 0.5l sin φφ&& sin φ
= −1 / 3ml 2φ&&
(28c)
Similar to Eqn. (12), the above equation also can
not be solved analytically rather we may simulate
it, the inputs being the outputs of Eqn. (12). The
results are shown in Fig. 12 below.
the first two of the above equations are due to the
motion of the centre of mass while the last one is
due to the motion about the centre of mass with the
consideration of the effect of coordinate forces [5].
The angle φ is the internal angle between the
ladder and the ground which can be identified from
(26). As Y = 0.5l sin φ , hence
R f = 0.5ml (cos φφ&& − sin φφ& 2 ) + mg
3.2.
(30)
a
b
10
c
-90
-405
-100
0
-410
-110
d2φ/dt2
φ
dφ/dt
5
-120
0.4
0.45
0.5
-140
0.35
-415
-420
-130
-5
0.35
Simulation
Some important simulation results of Eqn. (31) are
presented below in Table 2. Also Fig. 12 shows
that as the ladder tends to be horizontal after
detaching from the wall both its angular velocity
and the normal reaction at the ground increases (in
fact quite clearly it increases towards the weight of
the ladder) while the energy of the system
decreases as it should be. On the contrary, the
angular acceleration first decreases up to a
minimum value then increases.
(29)
X&& = −0.5lµ f (cos φφ&& − sin φφ& 2 ) − µ f g

0.5 cos φ + µ f sin φ
2 g 
 (31)
2
l  2 / 3 + µ f sin φ cos φ − sin φ 
0.4
t
0.45
0.5
-425
0.35
0.4
t
1.005
0.45
0.5
0.45
0.5
t
12.5
17.35
1
17.3
E
0.995
Rf
X
12
17.25
11.5
0.99
0.985
0.35
17.2
0.4
0.45
t
0.5
11
0.35
0.4
0.45
t
0.5
17.15
0.35
0.4
t
Fig.12: Simulation results for Eqn. (31) with m = 5kg , l = 2m, µ f = 0.7, φ i = 9.198 0 , φ&i = −96.647 0 s −1 . All the results are
plotted against t (s) measured from just after detachment (i.e., form t = 0.362 s). In panel a (up and down) rates of variation
of angle φ and X with time are plotted. In panel b (up and down) rates of variation of φ and R f with time are plotted. In
panel c (up and down) the rate of variation of angular velocity and energy of the ladder are plotted. The unit of φ is taken as
degree and all other variables are plotted in S.I. The initial values of
The energy loss during this process is 0.109J.
φ
and its time derivative are the outputs of Eqn. (13).
The African Review of Physics (2012) 7:0009
79
Table 2
Input
φ&&i = −389.497 0 s − 2 , t d = 0.39s,
Output
X g = 1.004m, t g = 0.472s, φ g = 0 0
Y& = −2.27ms −1 , φ& = −129.92 0 s −1 ,
E d = 17.318 J
X& g = 0.134ms −1 , E g = 17.209 J ,
φ i = 9.198 0 , φ&i = −96.647 0 s −1 ,
3.3.
Energy analysis and bi-dimensional
equation of motion
Similar to our analysis in the last section we may
find the energy of the sliding ladder at any instant t
to be
E (t ) = mgY + 0.5m( X& 2 + Y& 2 ) + 0.5Iω ′ 2
(
)
1
1

= 0.5m X& 2 + lg sin φ + ml 2φ& 2  cos 2 φ +  (32)
8
3

Hence, similar to Eqn. (19) and Eqn. (20), the rate
of loss of energy and the work done by the ladder
due to its elemental displacement can be expressed
as
dE
= mgY& + mX&X&& + mY&Y&& + 1 / 12ml 2φ&φ&&
dt
(33)
E (t ) − E (t + dt ) = µ f R f dX − R f dY − Iφ&&dφ
(34)
X&&
µf
= Y&& + g
(35)
Integrating Eqn. (35) with proper initial conditions
to the system we obtain (noting that t d being the
time of detachment)
−
−
X&
µf
= Y& + g (t − t d ) −
X& d
µf
− Y&d
(36a)


X&
= Y + 0.5 gt 2 −  gt d + d + Y&d t


µf
µf


&

 X

X
2
−  − 0.5 gt d + d + Yd . +  d + Y&d t d (36b)




µf

  µf

X
g
With
X d = −l / 2 cosθ d ,
Yd = l / 2 sin θ d ,
&
&
&
&
X d = l / 2 sin θ dθ d , Yd = l / 2 cos θ d θ d . Eqn. (36b)
also suggests that even after the detachment from
the wall the motion of the ladder is still guided by
some constraint relation between the abscissa and
ordinate of its CM as we have seen before (see
Eqn. (5)) but unlike the situation before
detachment, now the constraint relation between X
and Y are time dependent, i.e., at a particular instant
t and for a definite X we can have only a definite Y
depending upon the initial conditions and
parameters of the model. It can also be verified that
in Eqn. (36b) as t increases Y decreases but X
increases, as expected from real observation. In
Eqn. (36a) and Eqn. (36b) X d = 0.987mt ,
Y = 0.159mt
and
X& = 0.269ms −1 ,
d
On the other hand, we may find the equation of
motion of the ladder using Eqn. (28a) and Eqn.
(28b) as
−
g
d
Y&d = −1.665ms −1
are the coordinates and
corresponding velocity components (in S.I.) of the
CM of the ladder and we are measuring the time
from just after the detachment and using the
simulation results of Eqn. (12).
Now at the very moment the entire ladder
touches the ground ( φ = 0 , in our simulation result
of Eqn. (31)) the y-component of its CM becomes
zero and let the corresponding time is t g . Hence
the relevant X coordinate will be



X&
2
X g = − µ f  0.5 gt g −  gtd + d + Y&d t g



µf




  X&
 
X
2
−  − 0.5 gtd + d + Yd  +  d + Y&d td 



 
µf

  µf
 
(37)
We have estimated t g from the simulation result of
Eqn. (24) to be 0.472 s, it is that particular value of
time at which φ just becomes zero. Using Eqn.
(37) we then estimate X g = 1.0017m and when it
is compared with the simulation result in Eqn. (31)
we can observe the order of agreement therein. The
loss of energy of the ladder at the very moment
The African Review of Physics (2012) 7:0009
80
when the entire ladder just becomes horizontal is
given by
4.
Now just after the entire ladder becomes horizontal
and lies on the ground
2
2
∆E = 0.5mgl cos θ i − 0.5mX& g − 0.5mY&g
− 1 / 24ml 2φ&g = 3.499 J
2
Modeling and Simulation of the Horizontal
Ladder System
(38)
X&& = − µ f g
Now as both the ladder and the ground are
considered to be perfectly horizontal, hence either
there is a single point of contact between them
(which is any one of the terminal point of the
ladder) or all the points of the ladder are in contact
with the ground. So just at t = t g , the point of
action of R f shifts from one end of the ladder to its
CM. At the same instant after the collision with the
ground the ladder may jump of the ground to make
the further motion much complicated. To avoid
those kinds of situations, we consider the collision
of the ladder with the ground to be perfectly
inelastic, i.e., after the collision takes place ladder
will stick to the ground and does not jump at all. In
the process the translational kinetic energy of the
ladder corresponding to its vertical component of
velocity (downwards) and also the entire rotational
energy will be transferred to the ground and
surroundings. The only energy remains in the
system will be that due to the further possible
horizontal dragging motion of the ladder through
the ground.
(39a)
integrating and using initial conditions (the inputs
being the outputs of Eqn. (31)),
X& = − µ f g (t − t g ) + X& g
(39b)
X = −0.5µ f gt 2 + µ f gtt g + X& g t + X g
2
− 0.5µ f gt g − X& g t g
(39c)
Motion will cease at
t f = t g + X& g / µ f g
(40a)
and the corresponding displacement of the CM will
be
2
X f = 0.5 X& g / µ f g + X g
(40b)
In Table 3, we have presented the input and output
data for the final dragging phase of the ladder.
Table 3
Input
t g = 0.472s, X g = 1.004m, X& g = 0.134ms −1 ,
Output
t f = 0.492s, X f = 1.0053m
E ghorizontal = 0.0449 J
5.
Conclusion
In the above manner and with the help of analytical
understanding and numerical simulation we may
analyze the entire sliding and dragging 2dimensional motion of the ladder under certain
constraints. All our results can be verified to be
true (already discussed a few times) in conservative
systems by putting µ f = µ w = 0 [1-8], e.g. the
energy loss equations will disappear in the
conservative system. Also the comparison of the
sliding ladder with that of a sliding point mass
seems to be of certain importance both from the
standpoints of reaction and detachment. This kind
of problem may find their importance in the
understanding of the basic laws of nature, such as
translational and rotational kinematics and
dynamics, non-violation of special theory of
relativity and obviously mechanics of nonconservative systems. The data output and
simulations used here are the perfect illustrations of
mimicking some real life situations which tells us
that there is a certain angular range of constrained
downfall of a ladder through a rough wall and
floor, the extreme values of which indicate the
initiation and termination of motion. The analysis
of the motion after detachment from the wall also
plays a significant part of our present work. The
entire theory can be verified by performing a
simple in-house experiment with a yard-stick but in
that case one has to know the numerical values of
The African Review of Physics (2012) 7:0009
the kinetic frictional coefficients related to the wall
and the floor.
References
[1] L. D. Landau and E. M. Lifshitz, Mechanics,
3rd edition (Pergamon Press, Oxford, 1960).
[2] S. Kapranidis and R. Koo, The College
Mathematics Journal 39, 374 (2008).
[3] P. Scholten and A. Simoson, The College
Mathematics Journal 27, 49 (1996).
[4] M. Freeman and P. Palffy-Muhoray, Amer. J.
Phys. 53:3, 276 (1985).
[5] K. T. McDonald, Torque Analyses of a
Sliding Ladder (Joseph Henry Laboratories,
Princeton Univ., Princeton, 2007).
[6] D. Hughes-Hallett et.al., Calculus (Wiley,
NY, 1984).
[7] G. Strang, Calculus (Wellesley Cambridge
Press, Wellesley, MA, 1991) p.164.
[8] P. Gillett, Calculus and Analytic Geometry
(Heath, Lexington, MA, 1984) p.194.
[9] N. Deo, System Simulation with Digital
Computer (Prentice Hall of India, 1990).
Received: 15 November, 2011
Accepted: 6 April, 2012
81