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Text Elements
10kg
15kg
10kg
m
1
m
2
m
3
30
o
Find the acceleration of the system
- Determine direction of acceleration
Tug of war
m
1
10kg
F
g
= 9.8(10) = 98
m
2
15kg
F
g∥
= 9.8(15)(sin30) = 73.5
30
o
wins the tug of war
m
1
- Free body diagrams
10kg
m
1
F
g
T
1
System moves this way:
m
2
10kg
T
1
F
g
F
n
F
f
T
2
m
3
15kg
F
f
30
o
T
2
F
g∥
F
Net
=
F
g
T
1
F
Net
=
T
1
m
1
=
a
F
f
T
2
m
2
=
a
F
Net
=
T
2
F
f
F
g∥
m
3
=
a
10
=
a
15
=
a
Accelerations are the same. 2 Strings, 2 different tensions
- Vector equations Add them up
m :
1
m :
2
m :
3
10a + 10a + 15a =
= 98 -
10
a
T
1
T
1
19.6
T
2
T
2
- 102.9
= 98 -
10
a
T
1
10
=
a
T
1
19.6
T
2
15
=
a
T
2
- 102.9
98 -
T
1
T
1
19.6
T
2
T
2
- 102.9
35a
= -24.5
a = -0.7
Acceleration is negative. This is because, the system is actually not accelerating. We find it is not accelerating because in the first step, where we did the tug of war, we did not account for friction. The system never overcomes friction, thus it does not move. So acceleration should be 0