Version 058 – FIRST EXAM – Kleinman – (58835)
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001
10.0 points
The horizontal surface on which the objects
slide is frictionless.
The acceleration of gravity is 9
.
8 m
/
s
2
.
1 kg
8 kg
6 kg
F
ℓ
F
r
If
F
ℓ
= 15 N and
F
r
= 7 N, what is the
magnitude of the force 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 :
vector
F
ℓ
= +15 N ˆ
ı ,
vector
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
of the three blocks, resulting in a common
acceleration after accounting for friction.
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 force exerted
on the system from the right, from the left,
and the force exerted on
m
2
by
m
3
, respec
tively.
Note:
vector
F
21
=
−
vector
F
12
and
vector
F
32
=
−
vector
F
23
, and
bardbl
vector
F
21
bardbl
=
bardbl
vector
F
12
bardbl
and
bardbl
vector
F
32
bardbl
=
bardbl
vector
F
23
bardbl
, where
bardbl
vector
F
bardbl ≡
F
is the magnitude of
vector
F
.
The sum of the forces 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 find the acceleration we can treat the
three masses as a single object or add the
forces acting on each component of the sys
tem, Eqs. 1, 2, and 3.
F
ℓ
−
F
r
= (
m
1
+
m
2
+
m
3
)
a
Solving for 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 for
F
23
using Eq. 3 and sub
stituting
a
from 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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Version 058 – FIRST EXAM – Kleinman – (58835)
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
 Force, Friction, Correct Answer, Grand Bahama Island, EXAM Kleinman

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