Answer = A
(One point given for answer E.)
12.
Answer = B
13.
Answer = C
14.
Answer = D
15.
Answer = C
16.
Answer = A
17.
Answer = D
18.
Answer = C
19.
Answer = A
20. Answer = E
Physics 150
Exam #3
Solution
12/4/06
1.
Answer = D
2.
Answer = D
3.
Answer = E
4.
Answer = C
5.
Answer = E
6.
Answer = A
7.
Answer = D
8.
Answer = D
9.
Answer = D
10.
Answer = B
11.
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θ
2
65.818 deg
=
θ
2
atan
T
2y
T
2x
⎛
⎜
⎝
⎞
⎠
:=
Angle:
T
2
966.845N
=
T
2
T
2x
2
T
2y
2
+
:=
Magnitude of T
2
:
T
2x
396.062N
=
T
2x
T
1
cos
θ
1
(
)
⋅
:=
Then to the horizontal forces equation:
T
2y
882 N
=
T
2y
m
scaffold
g
⋅
m
man
g
⋅
+
T
1
sin
θ
1
(
)
⋅
−
:=
Returning to the vertical forces equation:
T
1
792.125N
=
T
1
L
2
m
scaffold
⋅
g
⋅
x
man
m
man
⋅
g
⋅
+
L sin
θ
1
(
)
⋅
:=
τ
L
2
m
scaffold
⋅
g
⋅
x
man
m
man
⋅
g
⋅
+
T
1
sin
θ
1
(
)
⋅
L
⋅
−
:=
0
:=
Taking torques about the right end of the scaffold:
T
2
cos
θ
2
(
)
⋅
T
1
cos
θ
1
(
)
⋅
−
0N
:=
Total horizontal forces;
T
1
sin
θ
1
(
)
⋅
T
2
sin
θ
2
(
)
⋅
+
m
scaffold
g
⋅
−
m
man
g
⋅
−
0 N
⋅
:=
Total vertical forces:
θ
1
60 deg
⋅
:=
m
man
100 kg
⋅
:=
x
man
4 m
⋅
:=
m
scaffold
60 kg
⋅
:=
L
10 m
⋅
:=
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 Winter '08
 Naik
 Physics, Force, Fnet Fnet

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