ghei (mg42382) – HW 9 – opyrchal – (11130)
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001 (part 1 of 2) 10.0 points
A block of mass 2 kg and one of mass 8 kg are
connected by a massless string over a pulley
that is in the shape of a disk having a radius
of 0
.
17 m, and a mass of 9 kg. In addition, the
blocks are allowed to move on a fixed block
wedge of angle 33
◦
, as shown. The coefficient
of kinetic friction is 0
.
15 for both blocks.
2 kg
8 kg
33
◦
0
.
17 m
9 kg
What is the acceleration of the two blocks?
The acceleration of gravity is 9
.
8 m
/
s
2
.
As
sume the positive direction is to the right.
Correct answer: 2
.
06186 m
/
s
2
.
Explanation:
Let :
m
1
= 2 kg
,
m
2
= 8 kg
,
M
= 9 kg
,
and
R
= 0
.
17 m
.
m
1
m
2
θ
M
T
1
T
2
Applying Newton’s law to
m
1
,
N
1

m
1
g
=
m
1
a
y
= 0
N
1
=
m
1
g ,
where the force of friction on
m
1
is
f
1
=
μ N
1
=
μ m
1
g
= (0
.
15) (2 kg) (9
.
8 m
/
s
2
)
= 2
.
94 N
and
T
1

f
1
=
m
1
a
T
1
=
m
1
a
+
f
1
.
(1)
For the mass
m
2
, applying Newton’s law
perpendicular to the slanted surface
N
2

m
2
g
cos
θ
=
m
2
a
perp
= 0
N
2
=
m
2
g
cos
θ ,
so the force of friction is
f
2
=
μ N
2
=
μ m
2
g
cos
θ
= (0
.
15) (8 kg) (9
.
8 m
/
s
2
) cos 33
◦
= 9
.
86277 N
.
Applying Newton’s law parallel to the surface,
m
2
g
sin
θ

f
2

T
2
=
m
2
a
T
2
=

m
2
a
+
m
2
g
sin
θ

f
2
(2)
Subtraction eq. (1) from, eq. (2),
T
2

T
1
=
m
2
g
sin
θ

(
m
1
+
m
2
)
a

f
1

f
2
.
Applying Newton’s law to the pulley,
summationdisplay
τ
=
I α
+
T
1
R

T
2
R
=

M R
2
2
a
R

T
1
+
T
2
=
1
2
M a
m
2
g
sin
θ

(
m
1
+
m
2
)
a

f
1

f
2
=
1
2
M a
a
=
2
m
2
g
sin
θ

2
f
1

2
f
2
2
m
1
+ 2
m
2
+
M
.
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ghei (mg42382) – HW 9 – opyrchal – (11130)
2
2
m
2
g
sin
θ

2 (
f
1
+
f
2
)
= 2(8 kg)(9
.
8 m
/
s
2
)(sin 33
◦
)

2(2
.
94 N + 9
.
86277 N)
= 59
.
7939 kg
·
m
/
s
2
and
2
m
1
+ 2
m
2
+
M
= 2(2 kg) + 2(8 kg) + 9 kg
= 29 kg
,
then
a
=
2
m
2
g
sin
θ

2
f
1

2
f
2
2
m
1
+ 2
m
2
+
M
=
59
.
7939 kg
·
m
/
s
2
29 kg
=
2
.
06186 m
/
s
2
.
002 (part 2 of 2) 10.0 points
Find the tension in the horizontal part of the
string.
Correct answer: 7
.
06372 N.
Explanation:
From eq. (1),
T
1
=
f
1
+
m
1
a
= 2
.
94 N + (2 kg) (2
.
06186 m
/
s
2
)
=
7
.
06372 N
.
003
10.0 points
Consider a thin 42 m rod pivoted at one
end.
A uniform density spherical object
(whose mass is 3 kg and radius is 3
.
5 m)
is attached to the free end of the rod and the
moment of inertia of the rod about an end
is
I
rod
=
1
3
m L
2
and the moment of iner
tia of the sphere about its center of mass is
I
sphere
=
2
5
m r
2
.
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