Multiple pulleys may involve combination of both static and moving pulleys. This may involve combining characterizing aspects of two systems.
Let us consider one such system as shown in the figure.
Let us consider that accelerations of the blocks are as shown in the figure. It is important to note that we have the freedom to designate direction of acceleration without referring to any other consideration. Here, we consider all accelerations in downward direction.
We observe that for given masses, there are five unknowns
a
1
,
a
2
,
a
3
,
T
1
and
T
2
a
1
,
a
2
,
a
3
,
T
1
and
T
2
. Whereas the free body diagram corresponding to three blocks provides only three independent relations. Thus, all five unknowns can not be evaluated using three equations.
However, we can add two additional equations; one that relates three accelerations and the one that relates tensions in the two strings. Ultimately, we get five equations for five unknown quantities.
1: Accelerations
The pulley “A” is static. The accelerations of block 1 and pulley “B” are, therefore, same. The pulley “B”, however, is moving. Therefore, the accelerations of blocks 2 and 3 may not be same as discussed for the case of static pulley. The accelerations of blocks 2 and 3 with respect to moving pulley are different than their accelerations with respect to ground reference.
We need to know the relation among accelerations of block 1, 2 and 3 with respect ground. In order to obtain this relation, we first establish the relation among the positions of moving blocks and pulley with respect to some fixed reference. Then we can obtain relation among accelerations by taking second differentiation of position with respect to time. Important to understand here is that these positions are measured with respect to a fixed reference. As such, their differentiation will yield accelerations of the blocks with respect to fixed reference i.e. ground reference. Let the positions be determined from a horizontal datum drawn through the static pulley as shown in the figure.
Now, the lengths of the two strings are constant. Let they be
L
1
and
L
2
L
1
and
L
2
.
L
1
=
x
0
+
x
1
L
2
=
(
x
2
-
x
0
)
+
(
x
3
-
x
0
)
=
x
2
+
x
3
-
2
x
0
L
1
=
x
0
+
x
1
L
2
=
(
x
2
-
x
0
)
+
(
x
3
-
x
0
)
=
x
2
+
x
3
-
2
x
0
Eliminating
x
0
x
0
, we have :
⇒
L
2
=
x
2
+
x
3
-
2
L
1
+
2
x
1
⇒
x
2
+
x
3
+
2
x
1
=
2
L
1
+
L
2
=
constant
⇒
L
2
=
x
2
+
x
3
-
2
L
1
+
2
x
1
⇒
x
2
+
x
3
+
2
x
1
=
2
L
1
+
L
2
=
constant
This is the needed relation for the positions. We know that acceleration is second derivative of position with respect to time. Hence,
⇒
a
2
+
a
3
+
2
a
1
=
0
⇒
a
2
+
a
3
+
2
a
1
=
0
(3)
This gives the relation of accelerations involved in the pulley system.
2: Tensions in the strings
We can, now, find the relation between tensions in two strings by considering the free body diagram of pulley “B” as shown in the figure.
Here,
T
1
=
2
T
2
T
1
=
2
T
2
(4)This result appears to be simple and on expected line. But it is not so. Note that pulley "B" itself is accelerated. The result, on the other hand, is exactly same as for a balanced force system. In fact this equality of forces in opposite direction is possible, because we have considered that pulley has negligible mass. This aspect has been demonstrated in the force analysis of the example given earlier (you may go through the example again if you have missed the point).
3: Free body diagrams of the blocks
The free body diagrams of the blocks are as shown in the figure.
m
1
g
-
T
1
=
m
1
a
1
m
2
g
-
T
2
=
m
2
a
2
m
3
g
-
T
2
=
m
3
a
3
m
1
g
-
T
1
=
m
1
a
1
m
2
g
-
T
2
=
m
2
a
2
m
3
g
-
T
2
=
m
3
a
3
(5)
Thus, we have altogether 5 equations for 5 unknowns.
There is one important aspect of the motions of blocks of mass "
m
2
m
2
" and "
m
3
m
3
" with respect to moving pulley "B". The motion of blocks take place with respect to an accelerating pulley. Thus, interpretation of the acceleration must be specific about the reference (ground or moving pulley). We should ensure that all measurements are in the same frame. In the methods, described above we have considered accelerations with respect to ground. Thus, if acceleration is given with respect to the moving pulley in an analysis, then we must first change value with respect to the ground.
Problem :
In the arrangement shown in the figure, mass of A = 5 kg and mass of B = 10 kg. Consider string and pulley to be mass-less and friction absent everywhere in the arrangement. Find the accelerations of blocks.
Solution :
Let the acceleration of block “A” be “a” in the downward direction. Let the tensions in the string be “
T
1
T
1
” and “
T
2
T
2
”. See figure below showing forces acting on the blocks and moving pulley. The force analysis of the block “A” yields :
⇒
5
g
−
T
1
=
5
a
⇒
5
g
−
T
1
=
5
a
⇒
50
−
T
1
=
5
a
⇒
50
−
T
1
=
5
a
From constraint of string length, we see that :
x
1
−
x
0
+
x
2
−
x
0
+
x
2
=
L
x
1
−
x
0
+
x
2
−
x
0
+
x
2
=
L
⇒
x
1
+
2
x
2
=
L
+
2
x
0
=
Constant
⇒
x
1
+
2
x
2
=
L
+
2
x
0
=
Constant
Differentiating twice with respect to time, we get relation between accelerations of two blocks as :
⇒
a
1
=
-
2
a
2
⇒
a
1
=
-
2
a
2
⇒
a
2
=
-
a
1
2
=
-
a
2
⇒
a
2
=
-
a
1
2
=
-
a
2
Force analysis of the moving mass-less pulley yields :
⇒
T
2
=
2
T
1
⇒
T
2
=
2
T
1
Force analysis of the block B results (refer to the force diagram shown in the beginning) :
⇒
2
T
1
−
10
X
10
=
10
X
a
2
=
5
a
⇒
2
T
1
−
10
X
10
=
10
X
a
2
=
5
a
⇒
2
T
1
−
100
=
5
a
⇒
2
T
1
−
100
=
5
a
Simultaneous equations of forces on blocks A and B are :
⇒
50
−
T
1
=
5
a
⇒
50
−
T
1
=
5
a
⇒
2
T
1
−
100
=
5
a
⇒
2
T
1
−
100
=
5
a
Solving for “a”, we have :
⇒
a
=
0
⇒
a
=
0
Thus, accelerations of two blocks are zero. Note, however, that tensions in the strings are not zero.
