Transcript Help Section - AdvancedPlacementPhysicsC

A ladder with length L weighing 400 N rests against
a vertical frictionless wall as shown below. The
center of gravity of the ladder is at the center of the
ladder. Find the coefficient of static friction s
between the ladder and the horizontal ground at its
base if an 80 kg man climbs one-third up before the
ladder starts to slip.
No
Friction
L
y
60
x
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Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
Question 7
Question 8
Question 9
Question 10
• Question 11
1. Problems such as the one at hand require analyzing
the statics of rigid bodies. The situation described on
the first slide is static as long as the man is not more
than one-third of the way up the ladder, because there
is no motion. Therefore, the conditions for static
equilibrium apply to the situation. What are the static
equilibrium conditions?
•
•
•
•

A:
B:
C:
D:
Fx  0
Fy  0
  0
where
 is the the torque about any chosen point.
all of the above
Choice: A
Incorrect
This is only one of the required conditions
for static equilibrium.
It is true that the sum of the forces along
the x-axis must be zero, but there are
more conditions that must be met.
Choice: B
Incorrect
This is only one of the required conditions
for static equilibrium.
It is true that the sum of the forces along
the y-axis must be zero, but there are
more conditions that must be met.
Choice: C
Incorrect
This is only one of the required conditions
for static equilibrium.
The net torque on the ladder must be
zero, but other conditions must apply.
Choice: D
Correct
All three of these conditions must
be satisfied in order for a system
to be in static equilibrium.
2. Which one of these diagrams correctly
shows all of the forces acting on the ladder?
Nw
A:
Ff = Static Friction Force
WL=Weight of ladder
WL
Wm
Wm=Weight of man
Ng=Normal force from the ground
Ff
Nw=Normal force from the wall
Nw
WL
B:
Wm
Ng
Ff
WL
Wm
C:
Ng
Ff
Choice: A
Incorrect
If the ladder slips, the base will move away from
the wall. The force of friction between the ladder
base and the ground will resist this motion.
The arrow representing friction should point in
the opposite direction.
Also, there is a normal force from the ground
pushing on the ladder.
Choice: B
Correct
This diagram correctly shows all
of the forces involved in the
situation.
NOTE: The force here called “weight of the man” is actually a normal
force acting on the ladder. It is directed downwards because the man is
standing on a horizontal step. The actual force “weight of the man” acts
on the man, of course.
Choice: C
Incorrect
There is a normal force from
the wall pushing on the ladder.
3. Applying the first condition for static equilibrium,
we find which of the following to be true for the
sum of forces in the x-direction?
• A:
F
x
 Ff Nw  0
• B:
F
 Ff Ng  0

• C:


x
F
x
 Ff  WL  Wm  0
Choice: A
Correct
This sums up all the forces
that act in the positive and
negative x-directions.
Notice that the static friction
force applied by the ground is
equal to the normal force
 from
the wall pushing on the
ladder.
F
x
 Ff Nw  0
Ff  Nw
Choice: B
Incorrect
Check the diagram from the previous question.
The normal force from the ground pushing on
the ladder acts in the positive y-direction.
Choice: C
Incorrect
Check the diagram from the previous question.
The force of gravity is directed in the negative ydirection. Therefore, the weight of the ladder
and the man are not included in this sum.
4. Once again, let’s apply the first condition for
static equilibrium, but this time for the forces in the
y-direction. Which of the following is correct?
• A:
F
y
 Nw  Wm  WL  0
• B:
F
 Ng  Wm  WL  0
• C:
F
 Ng Nw  0

y
y
Choice: A
Incorrect
Check the diagram from the question 2.
The normal force from the wall pushing on the
ladder acts in the negative x-direction, so it
should not be involved in this sum.
Choice: B
Correct
This sums up all the forces
that act in the positive and
negative y-directions.
Notice that the normal force
due to the ground pushing on
the ladder is equal to the
added weights of the man
and the ladder.

F
y
 Ng  Wm  WL  0
Ng  Wm  WL
Choice: C
Incorrect
The sum of the forces in the y-directions should
include the weight of the man and the ladder.
The normal force from the wall acting on the
ladder should not be included in the sum,
because it acts in the negative x-direction.
5. If the ladder is on the verge of slipping, then Ff
should assume its maximum value. Which of these
expressions gives this value? (where  s is
coefficient of static friction)
A: Ff,max  s Wm  WL 
B: Ff,max  sNg
C: Ff,max  sNw
Choice: A
Correct
Notice that choice B is also true, because from
the previous question we saw that:
Ng  Wm  WL
Choice B is the better
choice, because it is more
general. Choice A is correct
based on an analysis of the
details of this particular
problem.

Choice: B
Correct
The maximum static friction force is equal to the
coefficient of static friction  s times the normal force that
is perpendicular to the direction of the friction.
Notice that choice A is also true, because from the
previous question we saw that:
Ng  Wm  WL
Choice B is the better
choice, because it is more
general. Choice A is correct
based on an analysis of the
details of this particular
problem.

Choice: C
Incorrect
The normal force relevant here is the
normal force exerted by the ground,
because it acts in a direction
perpendicular to the force of friction.
Notice the following:
Ff,max  Ff
Ff  NW
Ng  Wm  WL
Ff  sNg
Nw  s Wm  WL 
Since in this problem we are only
concerned with the ladder at the critical
point before it slips.
From combing our findings from questions
3,4, and 5.
6. We now have two unknowns, Nw and s, but we only
have one equation.
Let’s try using the second condition for static equilibrium to
find another helpful equation:   0
Which of the following concepts/values will be important to
find the net torque on the system?
• A: The axis of rotation and the angular velocity.
• B: Lever arm for each force and moment of inertia.
• C: The axis of rotation and lever arm for each force.
Choice: A
Incorrect
Angular velocity is useless in this
situation. Remember, the ladder
is not in motion.
Choice: B
Incorrect
We don’t need to know the moment of inertia.
Torque can be found in other situations using the relation:
 I
Where I is moment of inertia and  is angular acceleration.
This form of expressing torque will not be useful here, because it
does not allow us to use the variables that we are trying to find the
values of.

Choice: C
Correct
A torque can be found by multiplying a
force by it’s lever arm.
 F
We need to now choose an axis of
rotation so that we know which forces and
corresponding lever arms to use to

calculate the net torque.
7. How is the lever arm for each force
defined?
• A: It is the distance from the center of mass of
the object under consideration to each of the
forces.
• B: It is the minimum distance from the axis of
rotation to the point at which the force is
applied.
• C: It is the distance from the axis of rotation to
the center of mass of the ladder.
Choice: A
Incorrect
Torque involves a force that gives an object the tendency to rotate
about a certain point, or an axis of rotation.
Only the component of a force that is perpendicular to the distance
between the location where the force is applied and the axis of
rotation cause this rotation.
A lever arm is the minimum distance from the axis of rotation to the
point at which the force is applied.
Lever arm
Perpendicular
component of an
applied force
Axis of rotation
Choice: B
Correct
This is the length that would
contribute to a torque.
NOTE:
A torque will cause rotation only if it is not
compensated by countertorques.
Choice: C
Incorrect
The lever arm is the minimum distance from the axis of rotation to
the point at which the force is applied.
Don’t Forget:
Only the component of a force that is perpendicular to the distance
between the location where the force is applied and the axis of
rotation cause this rotation.
Lever arm
Axis of rotation
Applied force
Component of
the applied force
that causes
torque and the
rotation of the
object.
8. We can choose to place our axis of rotation anywhere on
the ladder, but it is good practice to choose an axis through
which more than one force passes. So in this case, we
choose our axis to be where the ladder contacts the
ground. Choosing this axis would make which of the forces’
lever arm zero?
A: Nw
B: Wm and WL
C: Ng and Ff
Choice: A
Incorrect
The perpendicular component of this normal
force has a non-zero lever arm and produces a
torque about the axis.
Nwsin(60)
Front side view of ladder
Nw
Axis of
rotation
Axis of
rotation
Choice: B
Incorrect
Axis of
rotation
These weights
produce torques about
the chosen axis.
L
_
3
_
2
WLcos (60)
Wmcos (60)
Choice: C
Correct
These forces act at a point on the
axis, so their corresponding lever
arms are zero.
9. Now that we have chosen the axis and noted
that the magnitude of torque is equal to the
force times the lever arm, which expression will we
get when we apply the second condition
to the system?
  0
 F

• A:
 L   L 
   NwL  WL 2  Wm 3  0
 L   L 
sin(
60))

 WL  Wm  0
w
 3   2 
•B:   
L(N
L
 L

sin(
60))

(W
cos(60))

(W
cos(60))
 
 0
  L
w
m
2
 3

• C:    L(N

Choice: A
Incorrect
None of these forces are perpendicular to the
ladder.
Only forces that act perpendicular to the ladder or
perpendicular components of the forces that act
on the ladder will contribute to a net torque on the
ladder.
Check the hints from the previous question.
Choice: B
Incorrect
The lever arms associated with
the weights are not correct. The
components of the weights which
are perpendicular to the ladder
should be used.
Choice: C
Correct
This equation contains the
correct lever arms and forces.
Also, the signs of the torques are
correct.
10. Solve the correct equation from
question 9 for the normal force
from the wall pushing on the ladder
Nw.
Answer
L
 L

   L(Nw sin( 60))  2 (WL cos(60)) 3 (Wm cos(60)) 0
L
L
L(Nw sin( 60))  (WL cos(60))  (Wm cos(60))
2
3
W
W 
L(Nw sin( 60))  L L  m cos(60)
 2
3 
W
W 
Lcos(60) L  m 
 2
3 
Nw 
Lsin( 60)
W
W 
Nw   L  m cot(60)
 2
3 
11. Calculate the value of the coefficient
of static friction between the ground and
ladder by setting the two equations for
Nw together, solving for s, inserting the
given values, and simplifying.
Nw  s Wm  WL 

WL Wm 
Nw   
cot(60)
 2
3 
Wm  mmg  80kg(9.8m/s2 )  784N
Answer:
Set equations for
Nw equal to one
another:
WL Wm 
s Wm  WL   

cot(60)
 2
3 
Solve for s

W
1
Wm 
L
s  


cot(60)
3 
Wm  WL  2
Insert given values
and simplify


400N 784N
1
s  


cot(60)
3 
784N  400N 2
s  0.225
Reflection problem
1. How would the problem change if the floor
was frictionless?
Reflection problem
2. If the man was only half as heavy, would
he be able to go farther up without the
ladder falling down?