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Version 058 – FIRST EXAM – Kleinman – (58835)
1
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001
10.0 points
The horizontal sur±ace on which the objects
slide is ±rictionless.
The acceleration o± gravity is 9
.
8 m
/
s
2
.
1 kg
8 kg
6 kg
F
ℓ
F
r
I±
F
ℓ
= 15 N and
F
r
= 7 N, what is the
magnitude o± the ±orce exerted on the block
with mass 8 kg by the block with mass 6 kg?
1. 6.94118
2. 7.8
3. 6.35294
4. 10.2
5. 7.33333
6. 13.0
7. 9.33333
8. 11.6471
9. 10.0
10. 11.4
Correct answer: 10
.
2 N.
Explanation:
m
1
m
2
m
3
F
ℓ
F
r
Given :
v
F
ℓ
= +15 N ˆ
ı,
v
F
r
=
−
7 N ˆ
ı ,
m
1
= 1 kg
,
m
2
= 8 kg
,
m
3
= 6 kg
,
and
g
= 9
.
8 m
/
s
2
.
Note:
F
is acting on the combined mass
o± the three blocks, resulting in a common
acceleration a±ter accounting ±or ±riction.
m
1
F
21
F
ℓ
m
2
F
32
F
12
m
3
F
r
F
23
Let
F
ℓ
, F
r
, F
32
represent the ±orce exerted
on the system ±rom the right, ±rom the le±t,
and the ±orce exerted on
m
2
by
m
3
, respec
tively.
Note:
v
F
21
=
−
v
F
12
and
v
F
32
=
−
v
F
23
, and
b
v
F
21
b
=
b
v
F
12
b
and
b
v
F
32
b
=
b
v
F
23
b
, where
b
v
F
b ≡
F
is the magnitude o±
v
F
.
The sum o± the ±orces acting on each block
separately are
m
1
a
= +
F
ℓ
−
F
21
= +
F
ℓ
−
F
12
(1)
m
2
a
= +
F
12
−
F
32
= +
F
12
−
F
23
(2)
m
3
a
= +
F
23
−
F
r
(3)
To fnd the acceleration we can treat the
three masses as a single object or add the
±orces acting on each component o± the sys
tem, Eqs. 1, 2, and 3.
F
ℓ
−
F
r
= (
m
1
+
m
2
+
m
3
)
a
Solving ±or a, we have
a
=
F
ℓ
−
F
r
m
1
+
m
2
+
m
3
(4)
=
15 N
−
7 N
1 kg + 8 kg + 6 kg
= 0
.
533333 m
/
s
2
.
We can solve ±or
F
23
using Eq. 3 and sub
stituting
a
±rom Eq. 4. The result is
F
23
=
m
3
a
+
F
r
(3)
=
m
3
(
F
ℓ
−
F
r
)
m
1
+
m
2
+
m
3
+
F
r
(
m
1
+
m
2
+
m
3
)
m
1
+
m
2
+
m
3
=
m
3
F
ℓ
−
(
m
1
+
m
2
)
F
r
m
1
+
m
2
+
m
3
(5)
=
(6 kg) (15 N)
1 kg + 8 kg + 6 kg
+
(1 kg + 8 kg) (7 N)
1 kg + 8 kg + 6 kg
= 10
.
2 N
.
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2
Alternative Solution:
Using Eq. 3 and
the numerical result of Eq. 4, we have
F
23
=
m
3
a
+
F
r
(3)
= (6 kg) (0
.
533333 m
/
s
2
) + 7 N
= 10
.
2 N
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 Fall '09
 KLEINMAN

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