wolz (cmw2833) – HW #11 – Antoniewicz – (57420)
1
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001(part1of2)5.0points
A uniform horizontal beam of weight 400 N
and length 4
.
49 m has two weights hanging
from it.
One weight of 300 N is located
0
.
95637 m from the left end; the other weight
of 200 N is located 0
.
95637 m from the right
end.
What must be the magnitude of the one
additional force on the beam that will produce
equilibrium?
Correct answer: 900 N.
Explanation:
Let :
W
= 400 N
,
W
1
= 300 N
,
and
W
2
= 200 N
.
For the beam to preserve translational equi
librium, the net force acting on the beam has
to be zero, so the additional force should be
F
=
W
+
W
1
+
W
2
= 400 N + 300 N + 200 N
=
900 N
.
002(part2of2)5.0points
At what distance from the left end must this
force be applied for the beam to be in equilib
rium?
Correct answer: 2
.
10182 m.
Explanation:
The beam also has to preserve its rotational
equilibrium;
i.e.
, the net torque acting on the
beam has to be zero, so
F x
=
W
L
2
+
W
1
L
1
+
W
2
L
2
x
=
W
L
2
F
+
W
1
L
1
F
+
W
2
L
2
F
=
(400 N) (4
.
49 m)
2 (900 N)
+
(300 N) (0
.
95637 m)
900 N
+
(200 N) (3
.
53363 m)
900 N
=
2
.
10182 m
.
003(part1of4)3.0points
Two weights are suspended as shown in the
diagram.
On the left, a weight of 79 N is
suspended from its upper left by a string of
tension
T
1
angled at 44
◦
with the horizontal.
A string carrying tension
T
2
is stretched be
tween them angled at 12
◦
with the horizontal.
On the right, the unknown weight
W
2
is also
suspended from its upper right by a string of
tension
T
3
angled at 50
◦
with the horizontal.
79 N
W
2
12
◦
T
2
T
1
T
3
50
◦
44
◦
What is the tension
T
1
?
The acceleration
of gravity is 9
.
8 m
/
s
2
.
Correct answer: 145
.
822 N.
Explanation:
Let :
W
1
=
M
1
g
= 79 N
,
θ
1
= 44
◦
,
θ
2
= 12
◦
,
and
θ
3
= 50
◦
.
For the 79 N weight, horizontal equilibrium
yields
T
1
cos
θ
1
=
T
2
cos
θ
2
T
2
=
T
1
cos
θ
1
cos
θ
2
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wolz (cmw2833) – HW #11 – Antoniewicz – (57420)
2
and vertical equilibrium
T
1
sin
θ
1
=
W
1
+
T
2
sin
θ
2
T
1
sin
θ
1
=
W
1
+
T
1
cos
θ
1
cos
θ
2
sin
θ
2
=
W
1
+
T
1
cos
θ
1
tan
θ
2
.
T
1
sin
θ
1

T
1
cos
θ
1
tan
θ
2
=
W
1
T
1
=
W
1
sin
θ
1

cos
θ
1
tan
θ
2
=
79 N
sin 44
◦

cos 44
◦
tan 12
◦
=
145
.
822 N
.
004(part2of4)3.0points
What is the tension
T
2
?
Correct answer: 107
.
239 N.
Explanation:
From the horizontal equation,
T
2
=
T
1
cos
θ
1
cos
θ
2
=
(145
.
822 N) cos 44
◦
cos 12
◦
=
107
.
239 N
.
005(part3of4)2.0points
What is the tension
T
3
?
Correct answer: 163
.
188 N.
Explanation:
For the weight
W
2
horizontal equilibrium
gives
T
3
cos
θ
3
=
T
2
cos
θ
2
T
3
=
T
2
cos
θ
2
cos
θ
3
=
(107
.
239 N) cos 12
◦
cos 50
◦
=
163
.
188 N
.
006(part4of4)2.0points
What is the weight
W
2
?
Correct answer: 147
.
305 kg.
Explanation:
For the weight
W
2
vertical equilibrium gives
W
2
=
T
2
sin
θ
2
+
T
3
sin
θ
3
= (107
.
239 N) sin 12
◦
+ (163
.
188 N) sin 50
◦
=
147
.
305 N
.
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 Spring '08
 Turner
 Physics, Force, Correct Answer

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