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001
(part 1 of 2) 10.0 points
A 6 kg watermelon is placed at one end of a
4
.
7 m, 272 N scaffolding supported by two ca
bles. One supporting cable is at the opposite
end of the scaffolding, and the other is 0
.
81 m
from the watermelon.
How much tension is in the cable at the end
of the scaffolding? The acceleration of gravity
is 9
.
8 m
/
s
2
.
Correct answer: 95
.
4375 N.
Explanation:
Rotational and Translational Equilibrium
summationdisplay
τ
=
summationdisplay
τ
cw

summationdisplay
τ
ccw
= 0
Let the fulcrum be at the point of attach
ment of the cable closest to the watermelon.
mg
W
T
1
The watermelon
mg
acts down (ccw) at a
distance
x
from the fulcrum. The weight
W
acts down (cw) at a distance
ℓ
2

x
from the
fulcrum, and the rightmost tension
T
1
acts up
(ccw) at a distance
ℓ

x
from the fulcrum.
Thus
W
parenleftbigg
ℓ
2

x
parenrightbigg

m g x

T
1
(
ℓ

x
) = 0
.
Multiplying by 2,
W
(
ℓ

2
x
)

2
m g x
= 2
T
1
(
ℓ

x
)
,
T
1
=
W
(
ℓ

2
x
)

2
m g x
2 (
ℓ

x
)
=
(272 N) [4
.
7 m

2 (0
.
81 m)]
2 (4
.
7 m

0
.
81 m)

2 (6 kg) (9
.
8 m
/
s
2
) (0
.
81 m)
2 (4
.
7 m

0
.
81 m)
=
95
.
4375 N
.
002
(part 2 of 2) 10.0 points
How much tension is in the cable closest to
the watermelon?
Correct answer: 235
.
362 N.
Explanation:
Let the fulcrum be at the point of attach
ment of the rightmost cable.
mg
W
T
2
The watermelon
m g
acts down (ccw) at
a distance
ℓ
from the fulcrum.
The weight
W
acts down (ccw) at a distance
ℓ
2
from the
fulcrum, and the tension
T
2
acts up (cw) at a
distance
ℓ

x
from the fulcrum. Thus
T
2
(
ℓ

x
)

m g ℓ

W
parenleftbigg
ℓ
2
parenrightbigg
= 0
.
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 Spring '08
 Turner
 Force, Work, Correct Answer, Hinojosa, Wbear Wbear, N. Explanation

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