Homework 7
1)
(2m)(10N) + (1m)(30N) = (1m)(5N) + (d4)(15N), d4 = [(20+305)/15]m = (45/15)m =
3m
2)
Draw pulley systems with mechanical advantages of 1, 2, 3, 4, 5, 6, 7, 8, and 9.
3)
Repeat the upper tension problem on slide 4 of lecture series 2, with blocks 1, 2, &3 having a
mass of 5, 6, & 9kg respectively.
T
1
= (6kg +9kg)(1ms
2
) = 15 N, for
T
2
= (9kg)(1ms
2
) = 9N
4)
Repeat the lower tension problem on slide 4 of lecture series 2, with a 5kg mass and a horizontal
tension of 7N. Calculate the original angle A as well as the new angle with the above mass and
force.
T = F
MWt
/CosA = F/SinA =
√
[(50N)
2
+(7N)
2
] = 50.5N, A = Cos
1
(50/50.5) = Sin
1
(7/50.5) = 8.0 deg.
2
1
4
5
6
7
9
3
8
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View Full Document5)
Repeat the upper tension problem on slide 6, of lecture series 2, with m
1
= 3kg and m
2
= 6kg.
This is the dynamic case in which the system is moving up the incline. Therefore, this is similar to
the lower example on the same slide where: T = [2(M
1
X
M
2
)g/(M
1
+M
2
)], however the global
equation must be changed to account for the mass 1 being on an incline:
Global Eqn.: m
1
a + m
2
a = m
2
g – (Sin30
o
)m
1
g
⟹
a = [m
2
g 
½
m
1
g]/[m
1
+ m
2
]
Tension Eqn.: T  (Sin30
o
)m
1
g = m
1
a
⟹
T = ½
m
1
g + m
1
a
T =
½
m
1
g + m
1
[m
2
g 
½
m
1
g]/[m
1
+ m
2
]
=
m
1
g
(½
+ [m
2

½
m
1
]/[m
1
+ m
2
])
=
m
1
g
(½
[m
1
+ m
2
]/[m
1
+ m
2
] + [m
2

½
m
1
]/[m
1
+ m
2
])
=
m
1
g
(½
m
1
+
½
m
2
+ m
2

½
m
1
)/[m
1
+ m
2
])
=
(1.5)
m
1
m
2
g/[m
1
+ m
2
]) = (1.5)(3kg)(6kg)(10m/s
2
)/(3kg+6kg)] = (270/9)N = 30N
6)
Repeat the lower tension problem on slide 6 of lecture series 2, with m
1
= 3kg and m
2
= 6kg.
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 Summer '11
 Ramos
 Force, Work

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