sanchez (ds28677) – homework 27 – Turner – (58220)
1
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001
(part 1 of 2) 10.0 points
An 11 m ladder whose weight is 337 N
is placed against a smooth vertical wall.
A
person whose weight is 342 N stands on the
ladder a distance 4
.
4 m up the ladder.
The
foot of the ladder rests on the floor 4
.
62 m
from the wall.
342 N
337 N
4
.
4 m
11 m
4
.
62 m
Note:
Figure is not to scale.
Calculate the force exerted by the wall.
Correct answer: 141
.
292 N.
Explanation:
Let :
ℓ
= 11 m
,
d
= 4
.
4 m
,
s
= 4
.
62 m
,
W
ℓ
= 337 N
,
and
W
p
= 342 N
.
Pivot
N
w
F
f
N
f
W
p
W
ℓ
θ
θ
= arccos
s
ℓ
= arccos
parenleftbigg
4
.
62 m
11 m
parenrightbigg
= 65
.
1654
◦
In equilibrium
summationdisplay
vector
F
= 0
and
summationdisplay
vector
τ
= 0
.
summationdisplay
τ
◦
:
W
p
d
cos
θ
+
W
ℓ
ℓ
2
cos
θ
−
N
w
ℓ
sin
θ
= 0
,
where
d
is the distance of the person from the
bottom of the ladder. Therefore
2
F
w
ℓ
sin
θ
= 2
W
p
d
cos
θ
+
W
ℓ
ℓ
cos
θ
F
w
=
2
W
p
d
+
W
ℓ
ℓ
2
ℓ
cos
θ
sin
θ
=
2 (342 N)(4
.
4 m) + (337 N)(11 m)
2 (11 m)
×
cos 65
.
1654
◦
sin 65
.
1654
◦
=
141
.
292 N
.
002
(part 2 of 2) 10.0 points
Calculate the normal force exerted by the
floor on the ladder.
Correct answer: 679 N.
Explanation:
Applying translational equilibrium,
summationdisplay
F
y
:
N
f
− W
p
− W
ℓ
= 0
N
f
− W
p
− W
ℓ
= 0
.
N
f
=
W
p
+
W
ℓ
= 342 N + 337 N
=
679 N
.
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sanchez (ds28677) – homework 27 – Turner – (58220)
2
003
(part 1 of 3) 10.0 points
A 17
.
8 m
,
481 N uniform ladder rests against
a frictionless wall, making an angle of 68
.
3
◦
with the horizontal.
767 N
481 N
5
.
82 m
17
.
8 m
68
.
3
◦
Note:
Figure is not drawn to scale.
Find the horizontal force exerted on the
base of the ladder by Earth when a 767 N fire
fighter is 5
.
82 m from the bottom.
Correct answer: 195
.
505 N.
Explanation:
Let :
L
= 17
.
8 m
,
δ
= 5
.
82 m
,
α
= 68
.
3
◦
,
W
ℓ
= 481 N
,
and
W
p
= 767 N
.
L
sin
α
L
cos
α
L
2
cos
α
δ
cos
α
Pivot
N
w
f
N
g
W
p
W
ℓ
Applying rotational equilibrium with the
pivot at the point of contact with the ground,
summationdisplay
τ
=
W
p
δ
cos
α
+
W
ℓ
L
2
cos
α
−N
w
L
sin
α
= 0
2
W
p
δ
cos
α
+
W
ℓ
L
cos
α
= 2
N
w
L
sin
α
N
w
=
W
p
δ
cos
α
L
sin
α
+
W
ℓ
cos
α
2 sin
α
=
W
p
δ
L
tan
α
+
W
ℓ
2 tan
α
=
(767 N) (5
.
82 m)
(17
.
8 m) tan 68
.
3
◦
+
481 N
2 tan 68
.
3
◦
= 195
.
505 N
.
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 Spring '08
 Turner
 Physics, Force, Mass, Work, Sin, Trigraph, Correct Answer, Sanchez

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