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March 22, 2012
Physics for Scientists&Engineers 1
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March 22, 2012
Physics for Scientists&Engineers 1
2
Physics for Scientists &
Engineers 1
Spring Semester 2012
Lecture 32
Examples Involving Static Equilibrium
Student Standing on a Ladder
Typically a ladder stands on the floor and
leans against the wall
Suppose a ladder of length
ℓ
= 3.04 m with
mass
m
l
= 13.3 kg, rests against a smooth
wall at an angle of
θ
= 24.8
°
A student with mass 62.0 kg is standing on
a rung that is
r
= 1.43 m along the ladder,
measured from where the ladder touches
the ground
PROBLEM 1
•
What friction force must act on the bottom of
the ladder to keep it from slipping?
March 22, 2012
Physics for Scientists&Engineers 1
3
Student Standing on a Ladder
SOLUTION
•
Start with a free-body diagram
•
The forces are
•
R
, the normal force exerted by the wall on the
ladder
•
N
, the normal force exerted by the floor on the
ladder
•
W
m
, the weight of the student
•
W
l
, the weight of the ladder
•
f
s
, the force of static friction between the floor
and the ladder
•
The student and ladder are in static
equilibrium so
March 22, 2012
Physics for Scientists&Engineers 1
4
F
x
,
i
i
=
0
F
y
,
i
i
=
0
τ
i
i
=
0
Student Standing on a Ladder
•
The force components in the horizontal direction are
•
The force components in the vertical direction are
•
Sum the torques assuming that the pivot point is
where the ladder touches the ground
•
Solve for
R
March 22, 2012
Physics for Scientists&Engineers 1
5
F
x
,
i
i
=
f
s
−
R
=
0
f
s
=
R
F
y
,
i
i
=
N
−
m
m
g
−
m
l
g
=
0
N
=
g m
m
+
m
l
(
)
τ
i
i
=
m
l
g
(
)
2
sin
θ
+
m
m
g
(
)
r
sin
θ
−
R
cos
θ
=
0
R
=
1
2
m
l
g
(
)
sin
θ
+
m
m
g
(
)
r
sin
θ
cos
θ
=
1
2
m
l
g
(
)
+
m
m
g
(
)
r
tan
θ
Student Standing on a Ladder
•
Putting in our numerical values
•
We found that
f
s
=
R
so our answer is
PROBLEM 2
•
Suppose that the coefficient of static friction between the ladder
and the floor is 0.31.
•
Will the ladder slip?
SOLUTION 2
•
The normal force F was
•
So the maximum static friction force is
•
Which compares with
f
s
= 162 N, so the ladder will not slip
March 22, 2012
Physics for Scientists&Engineers 1
6
R
=
1
2
130. N
(
)
+
608 N
(
)
1.43 m
3.04 m
tan24.8
° =
162 N
f
s
=
162 N
N
=
g m
m
+
m
l
(
)
=
9.81 m/s
2
(
)
62.0 kg
+
13.3 kg
(
)
=
738 N
f
s,max
=
μ
s
N
=
0.31
(
)
738 N
(
)
=
229 N

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