rathbun (rar2954) – HW #4 – Antoniewicz – (57380)
1
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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
.
2 kg
9 kg
6 kg
F
F
r
If
F
= 27 N and
F
r
= 6 N, what is the
magnitude of the force exerted on the block
with mass 9 kg by the block with mass 6 kg?
Correct answer: 13
.
4118 N.
Explanation:
m
1
m
2
m
3
F
F
r
Given :
F
= +27 N ˆ
ı ,
F
r
=

6 N ˆ
ı ,
m
1
= 2 kg
,
m
2
= 9 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
m
2
F
32
F
12
m
3
F
r
F
23
Let
F , 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:
F
21
=

F
12
and
F
32
=

F
23
, and
F
21
=
F
12
and
F
32
=
F
23
, where
F
≡
F
is the magnitude of
F
.
The sum of the forces acting on each block
separately are
m
1
a
= +
F

F
21
= +
F

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

F
r
= (
m
1
+
m
2
+
m
3
)
a
Solving for a, we have
a
=
F

F
r
m
1
+
m
2
+
m
3
(4)
=
27 N

6 N
2 kg + 9 kg + 6 kg
= 1
.
23529 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

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

(
m
1
+
m
2
)
F
r
m
1
+
m
2
+
m
3
(5)
=
(6 kg) (27 N)
2 kg + 9 kg + 6 kg
+
(2 kg + 9 kg) (6 N)
2 kg + 9 kg + 6 kg
= 13
.
4118 N
.
Alternative Solution:
Using Eq. 3 and
the numerical result of Eq. 4, we have
F
23
=
m
3
a
+
F
r
(3)
= (6 kg) (1
.
23529 m
/
s
2
) + 6 N
= 13
.
4118 N
.
002 (part 1 of 3) 4.0 points
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rathbun (rar2954) – HW #4 – Antoniewicz – (57380)
2
Two forces, 440 N at 8
◦
and 378 N at 25
◦
are
applied to a car in an e
ff
ort to accelerate it.
3543 kg
440 N
8
◦
378 N
25
◦
What is the magnitude of the resultant of
these two forces?
Correct answer: 784
.
512 N.
Explanation:
Given
:
m
= 3543 kg
,
F
1
= 440 N
,
θ
1
= 8
◦
,
F
2
= 378 N
,
and
θ
2
=

25
◦
.
Consider the side forces:
F
1
,y
=
F
1
sin
θ
1
F
2
,y
=
F
2
sin
θ
2
F
y,net
= (440 N) sin 8
◦
+ (378 N) sin(

25
◦
)
=

98
.
5135 N
Consider the forward forces:
F
1
,x
=
F
1
cos
θ
1
F
2
,x
=
F
2
cos
θ
2
F
x,net
= (440 N) cos 8
◦
+ (378 N) cos(

25
◦
)
= 778
.
302 N
Thus the net force is
F
net
=
(778
.
302 N)
2
+ (

98
.
5135 N)
2
=
784
.
512 N
.
003 (part 2 of 3) 3.0 points
Find the direction of the resultant force (in re
lation to forward, with counterclockwise con
sidered positive).
Answer in degrees from the positive
x
axis,
with counterclockwise positive, within the
limits of

180
◦
to 180
◦
.
Correct answer:

7
.
21385
◦
.
Explanation:
tan
θ
=
F
y,net
F
y,net
θ
= tan

1
F
y,net
F
x,net
= tan

1

98
.
5135 N
778
.
302 N
=

7
.
21385
◦
,
which is 7
.
21385
◦
to the right of forward.
004 (part 3 of 3) 3.0 points
If the car has a mass of 3543 kg, what magni
tude of acceleration does it have?
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 Spring '08
 Turner
 Physics, Force, Friction, kg, rathbun

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