Problem 1 : Pulley and strings are “mass-less” and there is no friction involved in the arrangement. Find acceleration of the block of mass 3 kg and tensions “ T 1 T 1 ” and “ T 2 T 2 ” as shown in the figure.

Figure 1

Static or fixed pulley

Solution 1 : We observe here that two strings are taught. Therefore, acceleration of the all constituents of the system is same. This situation can be used to treat the system as composed of three blocks only (we neglect string as it has no mass). The whole system is pulled down by a net force as given by :

F = 2 g + 3 g − g = 4 g F = 2 g + 3 g − g = 4 g

Total mass of the system is :

m = 1 + 2 + 3 = 6 k g m = 1 + 2 + 3 = 6 k g

Applying law of motion, the acceleration of the system and hence acceleration of the block of mass 3 kg is :

a = F m = 4 g 6 = 40 6 = 20 3 m / s 2 a = F m = 4 g 6 = 40 6 = 20 3 m / s 2

In order to find tension “ T 1 T 1 ”, we consider FBD of mass of 1 kg. Here,

Figure 2

Free body diagrams

∑ F y = T 1 - 1 X g = m a y = 1 X 20 3 ∑ F y = T 1 - 1 X g = m a y = 1 X 20 3

⇒ T 1 = 20 3 + 10 = 50 3 N ⇒ T 1 = 20 3 + 10 = 50 3 N

In order to find tension “ T 2 T 2 ”, we consider FBD of mass of 3 kg. Here,

∑ F y = 3 X g − T 2 = m a y = 3 X 20 3 = 20 ∑ F y = 3 X g − T 2 = m a y = 3 X 20 3 = 20

⇒ T 2 = 3 X 10 − 20 = 10 N ⇒ T 2 = 3 X 10 − 20 = 10 N

Problem 2 : Pulley and string are “mass-less” and there is no friction involved in the arrangement. The blocks are released from rest. After 1 second from the start, the block of mass 4 kg is stopped momentarily. Find the time after which string becomes tight again.

Figure 3

Static or fixed pulley

Solution 1 : We need to understand the implication of stopping the block. As soon as block of 4 kg (say block “B”) is stopped, its velocity becomes zero. Tension in the string disappears. For the other block of 2 kg (say block “A”) also, the tension disappears.

Figure 4

Static or fixed pulley

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 − 1 2 g t 2 = 0 + 1 2 g t 2 v A t − 1 2 g t 2 = 0 + 1 2 g t 2

⇒ v A − 1 2 g t = 1 2 g t ⇒ v A − 1 2 g t = 1 2 g t

⇒ t = v A g ⇒ t = 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 = 4 g − 2 g 6 = 20 6 = 10 3 m / s 2 a = 4 g − 2 g 6 = 20 6 = 10 3 m / s 2

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

v A = u A + a t = 0 + 10 3 X 1 = 10 3 m / s v A = u A + a t = 0 + 10 3 X 1 = 10 3 m / s

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

⇒ t = v A g = 10 3 X 10 = 1 3 s ⇒ t = v A g = 10 3 X 10 = 1 3 s

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.

Figure 8

Combination or Multiple pulley system

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 L 1 " and " L 1 L 1 " be the lengths of two strings, then :

Figure 9

Combination or Multiple pulley system

x 1 + x 4 = L 1 x 1 + x 4 = L 1

x 2 − x 4 + x 3 − x 4 = L 2 x 2 − x 4 + x 3 − x 4 = L 2

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

⇒ v 1 + v 4 = 0 ⇒ v 1 + v 4 = 0

⇒ v 2 − v 4 + v 3 − v 4 = 0 ⇒ v 2 − v 4 + v 3 − v 4 = 0

We need to eliminate " v 4 v 4 " to get the required relation of velocities of the blocks. Rearranging second equation, we have :

⇒ v 2 + v 3 − 2 v 4 = 0 ⇒ v 2 + v 3 − 2 v 4 = 0

Substituting for " v 4 v 4 " from the first equation,

⇒ v 2 + v 3 − 2 − v 1 = 0 ⇒ v 2 + v 3 − 2 − v 1 = 0

⇒ v 2 + v 3 + 2 v 1 = 0 ⇒ v 2 + v 3 + 2 v 1 = 0

Now, considering upward direction as positive, v 2 = v 3 = 2 m / s v 2 = v 3 = 2 m / s . Putting these values in the relation for velocities,

⇒ 2 + v 3 + 2 X 2 = 0 ⇒ 2 + v 3 + 2 X 2 = 0

⇒ v 3 = − 6 m / s ⇒ v 3 = − 6 m / 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”).

Figure 10

Combination or Multiple pulley system

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

With reference to positions as shown in the figure, the constraint relations are :

x 1 + x 4 = L 1 x 1 + x 4 = L 1

x 1 − x 4 + x 3 − x 4 = L 2 x 1 − x 4 + x 3 − x 4 = L 2

x 1 − x 3 + x 2 − x 3 = L 3 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 = 0

⇒ a 1 − a 4 + a 3 − a 4 = 0 ⇒ a 1 − a 4 + a 3 − a 4 = 0

⇒ a 1 − a 3 + a 2 − a 3 = 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 2 − 2 a 3 = 0

⇒ a 1 + a 3 − 2 a 4 = 0 ⇒ a 1 + a 3 − 2 a 4 = 0

Substituting for " a 3 a 3 " from second equation.

⇒ a 1 + a 2 − 2 2 a 4 − a 1 = 0 ⇒ a 1 + a 2 − 2 2 a 4 − a 1 = 0

Substituting for " a 4 a 4 " from first equation,

⇒ a 1 + a 2 − 2 − 2 a 1 − a 1 = 0 ⇒ a 1 + a 2 − 2 − 2 a 1 − a 1 = 0

a 1 + a 2 + 6 a 1 = 0 a 1 + a 2 + 6 a 1 = 0

⇒ a 2 + 7 a 1 = 0 ⇒ a 2 + 7 a 1 = 0

⇒ a 2 = − 7 a 1 ⇒ a 2 = − 7 a 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.