F10PHYS115UAE
Assignment 3 Solutions
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45.
INTERPRET
This problem deals with interaction between different pairs of objects. The key concepts involved
here are Newton’s second and third laws.
DEVELOP
Let the three masses be denoted, from left to right, as
12
3
,,
a
n
d,
mm
m
respectively. We take the right
direction to be
x
. Assume that the table surface is horizontal and frictionless so that the only horizontal forces are
the applied force and the contact forces between the blocks. For example,
12
F
denotes the force exerted by block 1
on block 2. Since the blocks are in contact, they all have the same acceleration
a
, to the right. Newton’s second law
can be applied to each block separately:
app
21
1
12
32
2
23
3
F
Fm
a
F
a
F
ma
EVALUATE
Adding all three equations and using Newton’s third law
12
21
(0
,
FF
etc.), one finds
app
2
123
12 N
2 m/s
(to the right)
1kg 2kg 3kg
F
a
mmm
Thus, the force block 2 exerts on block 3 is
2
23
3
(3 kg)(2.0 m/s )
6.0 N
a
(to the right).
ASSESS
If we draw a freebody diagram for
3
,
m
we will see that
23
F
is the only force acting on it (ignoring
friction). Thus, the force that accelerates
3
m
comes entirely from
23
.
F
47.
INTERPRET
The objects of interest are the airplane and the two gliders, connected by ropes. The problem
involves applying Newton’s second law to find horizontal thrust of the plane’s propeller, the tension forces in each
rope, and the net force on the first glider.
DEVELOP
Let the masses of the airplane and the two gliders be denoted, from right to left, as
3
a
n
d
m
,
respectively (see figure). We take the right direction to be
x
. Assume that the runway surface is horizontal and
frictionless so that the only horizontal forces are the thrust and the tensions in the ropes.
We may then write the horizontal component (positive in direction of the acceleration) of the equations of motion
(Newton’s second law) for the three planes (all assumed to have the same acceleration) as follows:
th
1
1
12 2
23
(airplane)
(first glider)
(second glider)
FTm
a
TT m
a
Tm
a
Note:
The tension has the same magnitude at every point in a rope of negligible mass.
EVALUATE
(a)
Add all the equations of motion (the tensions cancel in pairs due to Newton’s third law), the
thrust of the airplane is
th
1
2
3
(
)
(2200kg 310kg
260kg)(1.9 m/s )
5.26 10 N
m
m
a
(b)
The tension in the first rope is
1t
h1
(
)
(310 kg
260kg)(1.9 m/s )
1.08 10 N
TFm
amm
a
(c)
The tension in the second rope is
2
(260 kg)(1.9 m/s )
494 N.
a
(d)
The net force on the first glider is
2
net
2
(310 kg)(1.9 m/s )
589 N.
a
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 Fall '06
 ROHAN
 mechanics, Force, Mass, m/s

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