<< Chapter < Page Chapter >> Page >

But the point is made. The accelerations of two ends of a string need not be same, when in contact with a moving body with a free end.

Problem : A block of mass, “m” is connected to a string, which passes over a smooth pulley as shown in the figure. If a force “F” acts in horizontal direction, find the accelerations of the pulley and block.

Moving pulley system

Solution : Let us consider that the acceleration of pulley is “a” in the direction of applied force. Now as analyzed before, acceleration of the block is “2a” and is in the same direction as that of pulley.

Moving pulley system

As motion is confined to x-direction, we draw free body diagram considering forces in x-direction only.

Free body diagram of pulley

The external forces are (i) force, F, (ii) tension, T, in the string and (ii) tension, T, in the string.

Free body diagram

F x = F - 2 T = m p a p

As mass of the pulley is zero,

F = 2 T T = F 2

This is an unexpected result. The pulley is actually accelerated, but the forces on it form a balanced force system. This apparent contradiction of force law is due to our approximation that pulley is “mass-less”. Free body diagram of block

The external force is (i) tension, T, in the string.

Free body diagram

F x = T = m a x = 2 m a

Combining two equations, we have :

a = F 4 m

Thus, acceleration of the pulley is F/4m and that of block is F/2m.

Got questions? Get instant answers now!

Combination or multiple pulley system

Multiple pulleys may involve combination of both static and moving pulleys. This may involve combining characterizing aspects of two systems.

Let us consider one such system as shown in the figure.

Multiple pulley system

Let us consider that accelerations of the blocks are as shown in the figure. It is important to note that we have the freedom to designate direction of acceleration without referring to any other consideration. Here, we consider all accelerations in downward direction.

Multiple pulley system

We observe that for given masses, there are five unknowns a 1 , a 2 , a 3 , T 1 and T 2 . Whereas the free body diagram corresponding to three blocks provides only three independent relations. Thus, all five unknowns can not be evaluated using three equations.

However, we can add two additional equations; one that relates three accelerations and the one that relates tensions in the two strings. Ultimately, we get five equations for five unknown quantities.

1: Accelerations

The pulley “A” is static. The accelerations of block 1 and pulley “B” are, therefore, same. The pulley “B”, however, is moving. Therefore, the accelerations of blocks 2 and 3 may not be same as discussed for the case of static pulley. The accelerations of blocks 2 and 3 with respect to moving pulley are different than their accelerations with respect to ground reference.

We need to know the relation among accelerations of block 1, 2 and 3 with respect ground. In order to obtain this relation, we first establish the relation among the positions of moving blocks and pulley with respect to some fixed reference. Then we can obtain relation among accelerations by taking second differentiation of position with respect to time. Important to understand here is that these positions are measured with respect to a fixed reference. As such, their differentiation will yield accelerations of the blocks with respect to fixed reference i.e. ground reference. Let the positions be determined from a horizontal datum drawn through the static pulley as shown in the figure.

Questions & Answers

List the application of projectile
Luther Reply
How can we take advantage of our knowledge about motion?
Kenneth Reply
pls explain what is dimension of 1in length and -1 in time ,what's is there difference between them
Mercy Reply
what are scalars
Abdhool Reply
show that 1w= 10^7ergs^-1
Lawrence Reply
what's lamin's theorems and it's mathematics representative
Yusuf Reply
if the wavelength is double,what is the frequency of the wave
Ekanem Reply
What are the system of units
Jonah Reply
A stone propelled from a catapult with a speed of 50ms-1 attains a height of 100m. Calculate the time of flight, calculate the angle of projection, calculate the range attained
Samson Reply
58asagravitasnal firce
water boil at 100 and why
isaac Reply
what is upper limit of speed
Riya Reply
what temperature is 0 k
0k is the lower limit of the themordynamic scale which is equalt to -273 In celcius scale
How MKS system is the subset of SI system?
Clash Reply
which colour has the shortest wavelength in the white light spectrum
Mustapha Reply
how do we add
Jennifer Reply
if x=a-b, a=5.8cm b=3.22 cm find percentage error in x
Abhyanshu Reply
x=5.8-3.22 x=2.58

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play

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?