MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.01
Problem Solving Session 7 Fixed Axis Rotational Dynamics and Static
Equilibrium
W9D3-1 Group Problem Atwood Machine Solutions
A pulley of mass m p , radius R , and moment of inertia about the center of mass I cm , is
suspended from a ceiling. An inextensible string of negligible mass is wrapped around
the pulley and attached on one end to an object of mass m1 and on the other end to an
object of mass with m1  m2 . At time t  0 , the objects are released from rest. How long
1
does it take the objects to move a distance d? If I cm  m p R 2 , what does your answer
2
reduce to?
Solution: We will use Newton’s Second Law, and the Rotational Law to find the
acceleration of all the objects in the system. We can then use kinematics to calculate the
time it takes for each object to move a certain distance. As a strategy we will answer the
following questions.
a) What are the free body force diagrams on the pulley and the two objects. Is the
tension in the string constant?
b) Write down Newton’s Second Law for the pulley and the two objects.
c) Write down the rotational equation of motion for the pulley.
d) What is the constraint condition between the magnitude of the angular
acceleration of the pulley and the magnitude of the acceleration of either object.
e) Find the direction and magnitude of the acceleration of the two objects.
f) How long does it take the objects to move a distance d?
g) If I cm 
1
m p R 2 , what does your answer reduce to?
2
a) Free body diagrams:
The two tensions in the string each produce a torque on the pulley in opposite directions.
The pulley has rotational inertial: the moment of inertia is not zero. Hence if the pulley
undergoes angular acceleration there must be a net non-zero torque. (Note the pulley is
not massless. Earlier in the semester we always assumed that the pulley is massless to
avoid the complication we are considering here.) The two tensions in the string each
produce a torque on the pulley in opposite directions. The net torque is proportional to the
difference in tensions in the string on the two sides of the pulley. Therefore the tensions
cannot be equal when the pulley undergoes angular acceleration.
b) Equation of linear motion for pulley (Pulley is at rest):
T3  T1  T2  m p g  0 .
(1)
m1g  T1  m1a .
(2)
T2  m2 g  m2 a .
(3)
Equation of linear motion for m1 :
Equation of linear motion for m2 :
c) Equation of rotational motion for pulley:
R T1  T2   I cm .
(4)
a  R .
(5)
T1  T2  m1  m2 g  (m1  m2 )a .
(6)
d) For pure rotation,
e) From (2) and (3):
From (4) and Eq (5) along with the moment of inertia:
T1  T2 
I cm I cm a
 2 .
R
R
(7)
Equating the expressions for T2  T1 in (6) and (7) we obtain:
I cm a
 m1  m2 g  (m1  m2 )a ,
R2
(8)
which we can solve for the acceleration.
a
f)
d
If I cm 
1 2
at
2
m1  m2 g
 I cm

 2  (m1  m2 )
R
.
(9)
and the results from part e) =>
I

2d  cm2  (m1  m2 )
R

.
m1  m2 g
t
2d

a
t
d m p  2(m1  m2 )
2d
.

a
m1  m2 g
(10)
1
m p R 2 , then
2


(11)
IC-W09D3-2 Group Problem Person on a Ladder Solution
A person of mass m p is standing on a rung, one third of the way up a ladder of length d .
The mass of the ladder is ml , uniformly distributed. The ladder is initially inclined at an
angle  with respect to the horizontal. Assume that there is no friction between the
ladder and the wall but that there is friction between the base of the ladder and the floor
with a coefficient of static friction  s . In this problem you will try to find the minimum
coefficient of friction between the ladder and the floor so that the person and ladder do
not slip.
a) Draw a free-body force diagrams for the person and the ladder separately,
explicitly showing where the forces act remembering that the gravitational force
acts at the center of mass of the ladder. Include unit vectors on the diagram.
Identify any possible third law interaction pairs. Write down the Newton’s Second
Law equations for both person and ladder.
b) Determine which point to calculate torque about. Remember that if you choose a
point S where a force acts then that force has zero torque about that point. When
calculating torque about a rchosen point, you can always formally calculate the
r
r
cross product  S  rS , F  F . You may also argue geometrically if the given
information of the problem makes it easier to compute the moment arm of the
force about your chosen point or the perpendicular component of the force with
respect to a line drawn from your chosen point to the point where the force acts.
You still need to determine the direction of the torque.
c) Determine the minimum coefficient of friction between the ladder and the floor so
that the person and ladder do not slip.
Solution:
We shall apply the two conditions for static equilibrium on the ladder,
(1) The sum of the forces acting on the rigid body is zero,
Ftotal  F1  F2    0 .
(12)
(2) The vector sum of the torques about any point S on a rigid body is zero,
 Stotal   S ,1   S ,2    0 .
(13)
a) Draw a free-body force diagrams for the person and the ladder separately,
explicitly showing where the forces act remembering that the gravitational force
acts at the center of mass of the ladder. Include unit vectors on the diagram.
Identify any possible third law interaction pairs.
Consider the forces acting on the person. The gravitational force acts at the center of
mass of the person and the force on the person due to the contact between the person and
the ladder is an upward normal force, denoted Nlp in the diagrams in these solutions. The
force diagram on the person is shown below.
The equation for static equilibrium of forces on the person is
ˆj : N  m g  0
lp
p
(14)
The normal force N pl that the person exerts on the ladder is part of a third law interaction
pair. From Newton’s Third Law,
Nlp  N pl .
(15)
Denote the magnitude of N pl by N pl . Equation (14) then becomes
N pl  m p g .
(16)
There are four forces acting on the ladder. The person exerts a downward contact force
on the ladder at the point of contact on the rung of the ladder, N pl , a distance d / 3 from
the contact point with the floor. The gravitational force between the earth and the ladder,
ml g , acts at the center of mass of the ladder, which is a distance d / 2 from the contact
point with the floor since the mass of the ladder is assumed to be uniformly distributed.
At the point where the ladder is in contact with the wall, the contact force of the wall with
ladder N wl is only perpendicular because we have assumed that the contact surface is
frictionless. At the point where the ladder is contact with the floor, the contact force has
both a vertical component, the normal force N gl and a horizontal component pointing
toward the wall, the static friction, fs . The force diagram is shown in the figure below
Key point: The magnitude of the static friction force depends on the other forces and
where they act. As the person walks up the ladder, the normal force due to the person,
N pl , changes position and hence the friction force will change in magnitude possibly
causing the ladder to slip.
The equations for static equilibrium of forces on the ladder (using Equation (16) for the
magnitude of the normal force of the person on the ladder) becomes
ˆj : N  m g  m g  0
gl
p
l
(17)
ˆi : f  N  0 .
s
wl
(18)
Solve Equation. (17) for the upward normal force of the ground on the ladder, which has
magnitude
(19)
N gl  mp g  ml g .
We can use Equation (18) to find a relationship between the friction force of the ground
on the ladder and the normal force of the wall on the ladder,
f s  N wl .
(20)
b) Determine which point to calculate torque about. Remember that if you choose a
point S where a force acts then that force has zero torque about that point. When
calculating torque about a chosen point, you can always formally calculate the
r
r
r
cross product  S  rS , F  F . You may also argue geometrically if the given
information of the problem makes it easier to compute the moment arm of the
force about your chosen point or the perpendicular component of the force with
respect to a line drawn from your chosen point to the point where the force acts.
You still need to determine the direction of the torque.
Because the sum of the forces are zero, the torque about any point will be zero so we
shall calculate the torque about several points: the point of contact between the ladder and
the ground, the point of contact between the wall and the ladder, and the center of mass
of the ladder.
a) Torque about the contact point between the ladder and the ground
The torque equation about the contact point S1 between the ladder and the floor is given
by
r
r
r
r
r
r
r r
 S  rS , p  N pl  rS ,cm  ml g  rS ,w  N wl  0 .
(21)
1
1
1
We show the relevant vectors in the figures below.
1
We now explicitly write out the vectors from our choice of point to where the forces are
acting:
d
d
r
rS , p  cos φ
i  sin  φ
j,
(22)
1
3
3
d
d
r
rS ,cm  cos φ
i  sin  φ
j,
1
2
2
(23)
r
rS ,w  d cos φ
i  d sin φ
j.
(24)
r
N pl  mp gφ
j,
(25)
r
N wl  N wl φ
i,
(26)
r
ml g  ml gφ
j.
(27)
1
The forces are
Then the torque about S1 is
r r
d

d

d
d
0   S   cos φ
i  sin  φ
j  mp gφ
j   cos φ
i  sin  φ
j  ml gφ
j
1
3
2
3

2

. (28)
 d cos φ
i  d sin  φ
j  N φ
i

Calculating the cross products we get

wl
r
d
d
 S   cos mp g kφ  cos ml g kφ  d sin  N wl kφ  0 .
1
3
2
(29)
We can solve this last equation for the magnitude of the normal force of the wall on the
ladder
 mp m 
(30)
N wl  gcotan 
 l.
3
2

We can now substitute Equation (30) into Equation (20) and solve for the magnitude of
the friction force
fs 
(1/ 3)m2 g cos   (1/ 2)m1 g cos 
 g cot   m2 / 3  m1 / 2  .
sin 
(31)
c) Determine the minimum coefficient of friction between the ladder and the floor so
that the person and ladder do not slip.
The minimum coefficient of friction between the ladder and the floor so that the person
and ladder do not slip is given by the condition that
fs  s N gl .
(32)
Substituting Equation (19) into Equation (32) gives
fs  s (mp g  ml g) .
(33)
We can now equate the expressions in Equations (33) and (31) and solve for the
minimum coefficient of static friction such that the ladder just starts to slip,
s 
mp / 3  ml / 2
mp  ml
cotan .
(34)
Alternative Choices for Calculating the Torque:
Torque about the center of mass:
Let’s compute the torque about the center of mass cm . Then
r
r
r
r
r
r
r
r
r
 cm  rcm,g  ( N gl  fgl )  rcm, p  N pl  rcm,w  N wl  0 ,
where
(35)
d
d
r
rcm,g   cos φ
i  sin  φ
j,
2
2
(36)
d
d
r
rcm, p   cos φ
i  sin  φ
j,
6
6
(37)
d
d
r
rcm,w  cos φ
i  sin  φ
j.
2
2
(38)
r
N gl  N gl φ
j,
(39)
r
f gl  f s φ
i.
(40)
Noting that
The torque about the center of mass is then
r r
 d

 d

d
d
0   cm    cos φ
i  sin  φ
j  (N gl φ
j  fs φ
i)    cos φ
i  sin  φ
j  mp gφ
j
2
6
 2

 6

.(41)
d

d
  cos φ
i  sin  φ
j   N wl φ
i
2
2

We now calculate the cross products and get
 d
r
d
 cm    cos N gl  sin fs
2
 2
 φ d
 k   6 cos mp g
 φ d
φ r
 k   2 sin N wl  k  0 .(42)
Recall from Eqs. (19) and (20)that N gl  m p g  ml g and f s  N wl , Eq. becomes
 d

d
r
d
 cm    cos mp g  ml g  sin N wl  kφ   cos mp g
2
 2

6
(43)


 φ d
φ r
 k   2 sin N wl  k  0
We can solve this equation for N wl
 mp m 
N wl  cotan g 
 l
2
 3
identical to our previous result (Eq. (30)).
Torque about the contact point S 2 between the wall and the ladder
(44)
The torque equation about S 2 is given by
r
r
 S  rS
2
2
r
r

N
 rS
,p
pl
2
r r
 ml g  rS
,cm
2 ,g
r
r
r
 ( N gl  f gl )  0 .
(45)
We now explicitly write out the vectors from our choice of point to where the forces are
acting
2d
2d
r
(46)
rS , p   cos φ
i
sin  φ
j,
2
3
3
r
rS
2 ,cm
r
rS
d
d
  cos φ
i   sin  φ
j,
2
2
(47)
 d cos φ
i  d sin φ
j.
(48)
2 ,g
Then the torque about S 2 is
r r
 2d

 d

2d
d
0   S    cos φ
i
sin  φ
j  mp gφ
j    cos φ
i  sin  φ
j  ml gφ
j
2
3
2
 3

 2

.(49)
 (d cos φ
i  d sin  φ
j)  (N φ
j f φ
i)
gl
s
Calculating the cross products we get
r 2d
d
0
cos mp gkφ  cos ml gkφ  d cos N gl kφ  d sin  fskφ .
3
2
(50)
We need to use Eqs. (19) and (20): N gl  m p g  ml g and f s  N wl , so that Eq. (50)
becomes
r 2d
d
0
cos mp gkφ  cos ml gkφ  d cos (mp g  ml g)kφ  d sin  N wl kφ . (51)
3
2
We can solve this last equation for the magnitude of the normal force of the wall on the
ladder, finding our same result that
 mp m 
N wl  gcotan 
 l.
(52)
2
 3