9.1 Pulleys(application-i)  (Page 2/2)

The block “A”, however, has certain velocity with which it is moving up. It will keep moving up against the force of gravity. On the other hand, block “B” will resume its motion of a free fall under gravity.

The string will become taught again, when the displacements of two blocks equal after the moment the block “B” was stopped momentarily. Let “t” be that time, then :

$$v_{A}t-\frac{1}{2}g{t}^2=0+\frac{1}{2}g{t}^2$$

$$\Rightarrow v_A-\frac{1}{2}gt=\frac{1}{2}gt$$

$$\Rightarrow t=\frac{v_A}{g}$$

Clearly, we need to know the velocity of the block “A”, when block “B” was stopped. It is given in the question that the system of blocks has been in motion for 1 s. Now, the acceleration of the system (hence that of constituent blocks) is :

$$a=\frac{4g-2g}{6}=\frac{20}{6}=\frac{10}{3}m/s^2$$

Applying equation of motion for block “A”, we have :

$$v_A=u_A+at=0+\frac{10}{3}X1=\frac{10}{3}m/s$$

Now putting values in the equation for time, we have :

$$\Rightarrow t=\frac{v_A}{g}=\frac{10}{3X10}=\frac{1}{3}s$$

Moving pulley

Problem 3 : Pulleys and string are “mass-less” and there is no friction involved in the arrangement. The two blocks have equal mass “m”. Find acceleration of the blocks and tension in the string.

Solution 1 : In order to determine accelerations, we shall write constraint equation to find the relation for the accelerations of two blocks. Going by the hints given in the module, we see that pulley on the table is fixed and can be used as reference. However, one of the blocks is hanging and is not in the level of other movable block on the table.

To facilitate writing of constraint relation, we imagine here that hanging block “D” is raised on the same level as block “A”. Now, we assign variables to denote positions of two movable elements as shown in the figure.

$$x_A+x_D=L$$

Differentiating twice,

$$\Rightarrow a_A+a_D=0$$

$$\Rightarrow a_A=-a_D$$

This means that the accelerations of two blocks are same. Since string is single piece, tension in the string is same. Considering free body diagrams, we have :

For A :

$$T=ma$$

For D :

$$mg-T=ma$$

Combining, we have :

$$mg-ma=ma$$

$$\Rightarrow a=\frac{g}{2}$$

$$\Rightarrow T=ma=\frac{mg}{2}$$

Combination or multiple pulley system

Problem 4 : Pulleys and string are “mass-less” and there is no friction involved in the arrangement. If at a given time velocities of block “1” and “2” are 2 m/s upward. Find the velocity of block “3” at that instant.

Solution 1 : We see here that velocities of block “1” and “2” are given and we have to find the velocity of remaining block “3”. It is apparent that if we have the relation of velocities of movable blocks, then we can get the result. We draw the reference as a horizontal line through the center of the static pulley and denote variables for all the movable elements of the system. Let " $L_1$ " and " $L_1$ " be the lengths of two strings, then :

$$x_1+x_4=L_1$$

$$x_2-x_4+x_3-x_4=L_2$$

Differentiating once, we get the relations for velocities of the movable elements,

$$\Rightarrow v_1+v_4=0$$

$$\Rightarrow v_2-v_4+v_3-v_4=0$$

We need to eliminate " $v_4$ " to get the required relation of velocities of the blocks. Rearranging second equation, we have :

$$\Rightarrow v_2+v_3-2v_4=0$$

Substituting for " $v_4$ " from the first equation,

$$\Rightarrow v_2+v_3-2\left(-v_1\right)=0$$

$$\Rightarrow v_2+v_3+2v_1=0$$

Now, considering upward direction as positive, $v_2=v_3=2m/s$ . Putting these values in the relation for velocities,

$$\Rightarrow 2+v_3+2X2=0$$

$$\Rightarrow v_3=-6m/s$$

Negative sign means that velocity of the block “3” is vertically downward.

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”).

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.

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 :

$$\Rightarrow a_1+a_4=0$$

$$\Rightarrow a_1-a_4+a_3-a_4=0$$

$$\Rightarrow a_1-a_3+a_2-a_3=0$$

Rearranging second and third equation,

$$\Rightarrow a_1+a_2-2a_3=0$$

$$\Rightarrow a_1+a_3-2a_4=0$$

Substituting for " $a_3$ " from second equation.

$$\Rightarrow a_1+a_2-2\left(2a_4-a_1\right)=0$$

Substituting for " $a_4$ " from first equation,

$$\Rightarrow a_1+a_2-2\left(-2a_1-a_1\right)=0$$

$$a_1+a_2+6a_1=0$$

$$\Rightarrow a_2+7a_1=0$$

$$\Rightarrow a_2=-7a_1$$

Thus, we need to pull the block a lot to raise the plank a bit. This is a mechanical arrangement that allows to pull a heavy plank with smaller force.