Infinite Atwood System

Classical Mechanics Level 2

Consider a system of infinite pulleys as shown in the figure. Find the acceleration (in $m/s^2$ ) of the top mass.

Details and Assumptions:
$\bullet$ Initially the masses are held stationary, and then released simultaneously.
$\bullet$ The masses of the pulleys and the strings are negligible.
$\bullet$ $\displaystyle g=9.8m/s^2$

The answer is 4.9.

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Jinay Patel
Feb 21, 2014

Let the tension in the first string be T. Then, the tension in the second pulley string is T/2 as it is massless .Let the acceleration (downward) of the second pulley be a .Then the second pulley has the acceleration (g-a) .Now, suppose the other system expect the first one . This infinite system is identical to the previous system.Therefore, $T/g = (T/2)/(g-a)$ which gives $a = g/2 = 9.8/2 = 4.9 m/s^2$

Could you explain why (g-a) please?

Bernardo Peters - 7 years, 3 months ago

Let the tension in the first string be T. Then, the tension in the second pulley string is T/2 as it is massless .Let the acceleration of the second pulley be a(downward) .Then the second mass has the acceleration (g-a) .Now, suppose the other system except the first one . This infinite system is identical to the previous system.Therefore, T/g = (T/2)/(g-a), which gives a = g/2 = 9.8/2 = 4.9 m/s^2

zeeshan khan - 6 years, 10 months ago

Please see what's wrong with this .

Megh Parikh - 6 years, 9 months ago

Vivek Sharma
Mar 8, 2014

suppose first block accelerates with acc .a= xg upward..then observing from first pulley the motion of second block is similar excpet effective g = g-a= g(1-x). now.. we have T- Mg = Ma and T/2 - Mg = Mxg(1-x).. solving these equn.. we get x=1/2.. hence acceleration a=4.9 m/s2

good explanation

Malay Pandey - 7 years, 3 months ago

That is what I did!!

Simon Mos - 7 years, 3 months ago

Rajkumar Sharma
Mar 15, 2014

Let the acceleration of top most particle is a as seen by topmost pulley. Then it will be in form as a=C g where C is any constant. If we change gravity by some factor keeping arrangement of system unchanged we will be find that acceleration will be replaced by same factor. This factor along with g is known as g(effective). Hence a=Cg(effective) Now if we observe top most particle from top most pulley(since top most pulley is in rest g(effective) will be equal to g). Then we can write..... a=Cg................(1) Now if we observe the second topmost particle from frame of the second pulley then due to infinite pulley system second particle has same acceleration a as top most pulley but effective acceleration will be g-a then
We can write .... a=C(g-a)..............(2) From equation (1) and (2)........................a=g/2=5................ANS

i got everything except the part where you say that the effective acceleration of the 2nd block is g-a. Didnt understamd why.please help

Gopesh Bajre - 7 years, 2 months ago

Alexandre Miquilino
Mar 5, 2014

Each pulley "halves" the force needed to lift/loosen the object attached to it, thus at an infinite level the forces on the right side of the pulley equal the force at the left side, and so the gravitational pull equals exactly the double of the force existent on the "main" pulley. Thus the acceleration of the object must be half of g, or approximately 4.9 m/s

Samarpit Swain
Mar 5, 2014

T/g = (T/2)/(g-a) THEREFORE, a= g/2= 4.9 m/s^2

Why can't i do like this?? assuming the other part of the string(which contains infinite masses) as a single system of mass "M" then writing the force equations as follows Mg-T=Ma------------(1) T-mg=ma-------------(2) adding (1) and (2) we get (M-m)g=(M+m)a taking M common on both the sides (1-m/M)g=(1+m/M)a as M is >>>>>m m/M=0 so,a=g=9.8m/s^2 why is this solution WRONG?? plz help me

shiva raj - 7 years, 2 months ago

Infinite Atwood System

The answer is 4.9.

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