<< Chapter < Page Chapter >> Page >

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 1 2 g t = 1 2 g t

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

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

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

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.

Moving pulley

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.

Moving pulley

x A + x D = L

Differentiating twice,

a A + a D = 0

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 :

Free body diagrams

Forces on blocks "A" and "D"

For A :

T = m a

For D :

m g T = m a

Combining, we have :

m g m a = m a

a = g 2

T = m a = m g 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.

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

Combination or multiple pulley system

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,

v 1 + v 4 = 0

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 :

v 2 + v 3 2 v 4 = 0

Substituting for " v 4 " from the first equation,

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 . Putting these values in the relation for velocities,

2 + v 3 + 2 X 2 = 0

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”).

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.

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 + 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

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.

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Physics for k-12. OpenStax CNX. Sep 07, 2009 Download for free at http://cnx.org/content/col10322/1.175
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Physics for k-12' conversation and receive update notifications?

Ask