9.4 Working with moving pulleys (application)  (Page 2/3)

$$a_{1A}=a_1-a_A$$

$$\Rightarrow a_{1A}=0-\left(-\frac{20}{M}\right)=\frac{20}{M}\mathrm{upward}$$

This result is physically verifiable. Since pulley has downward acceleration, the block should have upward acceleration of the same magnitude such that block acceleration with respect to ground is zero. The blocks are connected by the same string that goes over the pulley. It means that relative accelerations of the blocks are equal but opposite in direction (see Working with moving pulleys ).

$$\Rightarrow a_{2A}=-a_{1A}=-\frac{20}{M}\mathrm{downward}$$

The acceleration of mass “ $m_2$ ” with respect to ground can, now, be obtained as :

$$a_{2A}=a_2-a_A$$

$$a_2=a_{2A}+a_A=-\frac{20}{M}+\left(-\frac{20}{M}\right)=-\frac{40}{M}$$

From force analysis of block of mass 2 kg, we have :

$$-m_{2}g+T_1=-m_2{a}_2$$

$$m_{2}g-T_1=m_2{a}_2$$

$$\Rightarrow 2X10-10=2X\frac{40}{M}$$

$$\Rightarrow 10M=80$$

$$\Rightarrow M=8kg$$

Problem 2 : For the arrangement as shown in the figure, find the acceleration of block on the horizontal floor. Consider all surfaces “friction-less” and strings and pulleys as “mass-less” for the force analysis.

Block and pulley system

Solution : If we look at the motions of blocks from the reference of moving pulley, then the situation is that of motion of two blocks and a static pulley. Since the blocks are connected by the same string, the accelerations of the blocks with respect to pulley are equal in magnitude but opposite in direction.

Block and pulley system

Let the acceleration of the block of mass, “ $m_1$ ”, placed on the horizon, be " $a_1$ ". Since block on the horizontal surface and the moving pulley are connected by a single piece of string, the acceleration of the moving pulley is same. As such, acceleration of the moving pulley is “ $a_1$ ”. Now, let the relative accelerations of the blocks with respect to moving pulley be “ $a_{21}$ ” and “ $a_{31}$ ” respectively.

In order to carry out force analysis of the different elements of the system, we are required to express accelerations with respect to ground. This is an important point. Note that Newton's laws of motion is applicable to inertial frame of reference - not to an accelerated frame of reference like that of moving pulley. We can convert “accelerations with respect to accelerating pulley” to “accelerations with respect to ground” by using the concept of relative acceleration. Let block of mass “ $m_2$ ” moves down and the block of mass “ $m_3$ ” moves up with respect to the moving pulley. Also, let their accelerations with respect to ground be “ $a_2$ ” and “ $a_3$ ” respectively. Then,

$$a_{21}=a_2-a_1$$

$$a_{31}=a_3-a_1$$

But magnitudes of relative accelerations with respect to moving pulley are equal.

$$\left|a_{21}\right|=\left|a_{31}\right|=a_r$$

Then, considering downward direction as positive,

$$\Rightarrow a_2=a_1+a_{21}=a_1-a_r$$

$$\Rightarrow a_3=a_1+a_{31}=a_1+a_r$$

Thus, we see that two accelerations are now expressed in terms of one relative acceleration, “ $a_r$ ”, only. The analysis of forces in the ground reference for different elements of the system are :

Free body diagrams

(i) The block of mass, “ $m_1$ ” :

$$T_1=m_1{a}_1$$

(ii) For pulley :

$$T_2=\frac{T_1}{2}$$

Combining two equations,

$$T_2=\frac{m_1{a}_1}{2}$$

(iii) For block of mass, “ $m_2$ ” :

$$m_{2}g-T_2=m_2{a}_2$$