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```				   how to solve problems related to constraint relations
```

6 years ago

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```										Dear simran,
1)Draw reference, which is fixed. This reference can be the level of fixed pulley or the ground.2) Identify all movable elements like blocks and pulleys (excluding static ones).3) Assign variables for the positions of movable elements from the chosen reference.4)Write down constraint relations for the positions of the elements. Usually, total length of the string is the “constraint” that we  need to make use for writing relation for the positions. 5)  Differentiate the relation for positions once to get relation of velocities and twice to get relation of accelerations.
No. of Equations = No. of variables

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Problem 5 :  Pulleys and string are “mass-less” and there is no friction involved in the arrangement. Find the relation for the accelerations between the hanging plank (marked “1”) and the block (marked “2”).Figure 10 Combination or Multiple pulley system   Combination or Multiple pulley system   (pq9.gif) Solution 1 :  In order to obtain the relation for the accelerations of given elements, we first need to develop constraint relations for the three string having fixed lengths. For this, we choose a horizontal reference through the center of topmost fixed pulley as shown in the figure.Figure 11 Combination or Multiple pulley system   Combination or Multiple pulley system   (pq10.gif)With reference to positions as shown in the figure, the constraint relations are :            x      1        +          x      4        =          L      1                  x      1        −          x      4        +          x      3        −          x      4        =          L      2                  x      1        −          x      3        +          x      2        −          x      3        =          L      3      Differentiating above relations twice with respect to time, we have three equations :      ⇒          a      1        +          a      4        =    0        ⇒          a      1        −          a      4        +          a      3        −          a      4        =    0        ⇒          a      1        −          a      3        +          a      2        −          a      3        =    0  Rearranging second and third equation,       ⇒          a      1        +          a      2        −    2          a      3        =    0        ⇒          a      1        +          a      3        −    2          a      4        =    0  Substituting for "            a      3      " from second equation.      ⇒          a      1        +          a      2        −    2                  2                  a          4                −                  a          1                      =    0  Substituting for "            a      4      " from first equation,       ⇒          a      1        +          a      2        −    2                  −        2                  a          1                −                  a          1                      =    0              a      1        +          a      2        +    6          a      1        =    0        ⇒          a      2        +    7          a      1        =    0        ⇒          a      2        =    −    7          a      1
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6 years ago

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