MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
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 chosen
point, you can always formally calculate the cross
r
r
r
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 .
(1)
(2) The vector sum of the torques about any point S on a rigid body is zero,
 Stotal   S ,1   S ,2    0 .
(2)
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
(3)
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 .
(4)
Denote the magnitude of N pl by N pl . Equation (3) then becomes
N pl  m p g .
(5)
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 (5) for the magnitude
of the normal force of the person on the ladder) becomes
ˆj : N  m g  m g  0
gl
p
l
(6)
ˆi : f  N  0 .
s
wl
(7)
Solve Equation. (6) for the upward normal force of the ground on the ladder, which has
magnitude
N gl  mp g  ml g .
(8)
We can use Equation (7) 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 .
(9)
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 cross
r
r
r
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 .
(10)
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,
(11)
1
3
3
d
d
r
rS ,cm  cos φ
i  sin  φ
j,
1
2
2
(12)
r
rS ,w  d cos φ
i  d sin φ
j.
(13)
r
N pl  mp gφ
j,
(14)
r
N wl  N wl φ
i,
(15)
r
ml g  ml gφ
j.
(16)
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

.
 d cos φ
i  d sin  φ
j  N φ
i
(17)
1
The forces are
Then the torque about S1 is


wl
Calculating the cross products we get
r
d
d
 S   cos mp g kφ  cos ml g kφ  d sin  N wl kφ  0 .
1
3
2
(18)
We can solve this last equation for the magnitude of the normal force of the wall on the ladder
 mp m 
(19)
N wl  gcotan 
 l.
2
 3
We can now substitute Equation (19) into Equation (9) 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 
(20)
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 .
(21)
Substituting Equation (8) into Equation (21) gives
fs  s (mp g  ml g) .
(22)
We can now equate the expressions in Equations (22) and (20) 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 .
(23)
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 ,
(24)
where
d
d
r
rcm,g   cos φ
i  sin  φ
j,
2
2
(25)
d
d
r
rcm, p   cos φ
i  sin  φ
j,
6
6
(26)
d
d
r
rcm,w  cos φ
i  sin  φ
j.
2
2
(27)
r
N gl  N gl φ
j,
(28)
r
f gl  f s φ
i.
(29)
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

.(30)
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 . (31)
Recall from Eqs. (8) and (9)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


 φ d
φ r
 k   2 sin N wl  k  0 (32)
We can solve this equation for N wl
 mp m 
N wl  cotan g 
 l
3
2

(33)
identical to our previous result (Eq. (19)).
Torque about the contact point S 2 between the wall and the ladder
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 .
(34)
We now explicitly write out the vectors from our choice of point to where the forces are acting
r
rS
2,p
r
rS
2 ,cm
r
rS
2d
2d
cos φ
i
sin  φ
j,
3
3
(35)
d
d
  cos φ
i   sin  φ
j,
2
2
(36)
 d cos φ
i  d sin φ
j.
(37)

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

.
 (d cos φ
i  d sin  φ
j)  (N φ
j f φ
i)
gl
(38)
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
(39)
We need to use Eqs. (8) and (9): N gl  m p g  ml g and f s  N wl , so that Eq. (39) becomes
r 2d
d
0
cos mp gkφ  cos ml gkφ  d cos (mp g  ml g)kφ  d sin  N wl kφ .
(40)
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.
(41)
3
2
