7.2 Constrained motion

  Laws of Motion

If we want to find out the acceleration of the blocks as shown in figure, then it might be very tedious to find, because here more than one string is used.

So, for such problems we use a concept of constrained, which basically relates the motion of the blocks by taking all the string into account.

We can write constrained equation using following methods,

1. By constant string length method

2. By work done method

7.2.1 Constant string length method

• Find out the object which is moving and which is at rest.
• Assign a displacement to every movable object. Its better to assign all the displacements in one direction to avoid confusion
• Deal with one string at one time and break it into segments
• If the displacement is inward in the segment take it as negative, Similarly if the displacement is outward the segment, take it as positive
• If a position is at rest take displacement as zero (0). Example: Wall or any fixed end
• Since there is no change in original length of the string, equate the whole equation to zero
• Follow these steps for other string also and then find the relation.

Note: You can take velocities or acceleration also instead of displacements. It will not affect the constrained equation

Let us understand this concept with the help of an example.

Question 8. Find the constrained equation of the system as shown in the figure.

Solution: According to the rules,

A. Block 1, Pulley 2, 3 and Block 2 are the movable objects.

B. Let $a$, $b$, $c$ & $d$ respectively be the displacement of the movable objects.

C. For string 1, $$\begin{equation} \begin{aligned} (0 + a) + (0 + b) = 0 \\ a + b = 0\quad ...(i) \\\end{aligned} \end{equation} $$

For string 2, $$\begin{equation} \begin{aligned} ( - b + 0) + ( - b + c) = 0 \\ c = 2b\quad ...(ii) \\\end{aligned} \end{equation} $$

For string 3, $$\begin{equation} \begin{aligned} ( - c + 0) + ( - c + d) = 0 \\ d = 2c\quad ...(iii) \\\end{aligned} \end{equation} $$

From equation $(i)$, $(ii)$ & $(iii)$ we get, $$d = - 4a\quad ...(iv)$$

If we differentiate the above equation we get, $${\overrightarrow v _2} = - 4{\overrightarrow v _1}$$ Also, $${\overrightarrow a _2} = - 4{\overrightarrow a _1}$$

where,

${\overrightarrow v _1}$: Velocity of block 1

${\overrightarrow v _2}$: Velocity of block 2

${\overrightarrow a _1}$: Acceleration of block 1

${\overrightarrow a _2}$: Acceleration of block 2

Note:

• The above equation shows the relation between the motions of two blocks
• Acceleration of block 2 will be four times as block 1
• Negative (-ve) sign indicates that their motion is in opposite direction

Question 9. Find the constrained equation of the system as shown in the figure.

Solution: Pulley 2, 3, block 2 and 1 are movable objects with displacements $a$, $b$, $c$ & $d$ respectively.

For string 1, $$\begin{equation} \begin{aligned} (0 + d) + (0 + a) = 0 \\ a + d = 0\quad ...(i) \\\end{aligned} \end{equation} $$

For string 2, $$\begin{equation} \begin{aligned} ( - a + d) + ( - a + b) = 0 \\ d + b = 2a\quad ...(ii) \\\end{aligned} \end{equation} $$

For string 3, $$\begin{equation} \begin{aligned} ( - b + d) + ( - b + c) = 0 \\ d + c = 2b\quad ...(iii) \\\end{aligned} \end{equation} $$

From equation $(i)$, $(ii)$ & $(iii)$ we get, $$c = - 7d\quad ...(iv)$$

Similarly, $${\overrightarrow a _2} = - 7{\overrightarrow a _1}$$

Note:

• Acceleration of block 2 is seven times as block 1
• Negative (-ve) sign indicates that their motion is in opposite direction

Question 10. Find the constrained equation of the system as shown in the figure.

Solution: In this problem, let the displacement of block 1 and 2 be $a$ & $b$ respectively.

The pulleys attached to the block 1 will have the same displacement as the block 1.

Pulley attached to the wall is fixed.

Also, here only one string is used, so we have divided the string into 7 segments.

From the figure we can write, $$\begin{equation} \begin{aligned} \left( { - a + 0} \right) + ( - a + 0) + ( - a + 0) + ( - a + 0) + ( - a + 0) + ( - a + 0) + (0 + b) = 0 \\ b = 6a \\\end{aligned} \end{equation} $$

Also, $${\overrightarrow a _2} = 6{\overrightarrow a _1}$$

Note:

• Acceleration of block 2 is six times as block 1
• Positive (+ve) sign indicates that the actual motion is in the direction chosen

Question 11. Find the constrained equation of the system as shown in the figure.

Solution: In this problem, block 1, pulley 1, 3, 4, 5, and block 2 are movable objects with displacements $a$, $b$, $c$, $d$, $e$ & $f$ respectively.

For string 1, $$\begin{equation} \begin{aligned} (0 + b) + ( - a + b) = 0 \\ a = 2b\quad ...(i) \\\end{aligned} \end{equation} $$

For string 2, $$\begin{equation} \begin{aligned} ( - b + 0) + (0 + c) = 0 \\ b = c\quad ...(ii) \\\end{aligned} \end{equation} $$

For string 3, $$\begin{equation} \begin{aligned} ( - c + 0) + ( - c + d) = 0 \\ d = 2c\quad ...(iii) \\\end{aligned} \end{equation} $$

For string 4, $$\begin{equation} \begin{aligned} ( - d + 0) + ( - d + e) = 0 \\ e = 2d\quad ...(iv) \\\end{aligned} \end{equation} $$

For string 5, $$\begin{equation} \begin{aligned} ( - e + 0) + ( - e + f) = 0 \\ f = 2e\quad ...(v) \\\end{aligned} \end{equation} $$

From equation $(i)$, $(ii)$, $(iii)$, $(iv)$ and $(v)$ we get, $$f = 4a\quad ...(vi)$$

Also, $${\overrightarrow a _2} = 4{\overrightarrow a _1}$$

Note: Acceleration of block 2 is four times the acceleration of block 1.

7.2.2 Work done method

It is the easiest and shortest method to find the constrained equation for any system.

In this method we conserve the total work done by a system.

Rules:

• Balance the tension across the system
• Assign displacements only to blocks
• Calculate the work done of each block by taking dot product of the tension and displacement vectors
• Equate the sum of all the work done to zero because while calculating constrained equations we assume no external force acts on the systems. Therefore the total energy remains conserved

Note: We can take velocities or acceleration also instead of displacements. It will not affect the constrained equation.

Question 12. Find the constrained equation of the system as shown in the figure by work done method.

Solution: We have balanced the tension across the system.

Let the displacements of block 1 & 2 be $a$ & $b$ respectively.

Work done by block 1, $$\begin{equation} \begin{aligned} {W_1} = \overrightarrow F .\;\overrightarrow a \\ {W_1} = 4Ta\cos 0^\circ \\ {W_1} = 4Ta\quad ...(i) \\\end{aligned} \end{equation} $$

Work done by block 1, $$\begin{equation} \begin{aligned} {W_2} = \overrightarrow F .\;\overrightarrow b \\ {W_2} = Tb\cos \pi \\ {W_2} = - Tb\quad ...(ii) \\\end{aligned} \end{equation} $$

Since no external force acts on the system. So, total work done by a system is zero.

$${W_T} = 0$$ or $${W_T} = {W_1} + {W_2}$$ Therefore, $$\begin{equation} \begin{aligned} 4Ta - Tb = 0 \\ b = 4a\quad ...(iii) \\\end{aligned} \end{equation} $$ Also, $${\overrightarrow a _2} = 4{\overrightarrow a _1}$$