IIT JEE , AIEEE Forum - Physics , Mechanics

madman

PLEASE READ COMPLETELY

people who have I.E.Irodov check out the question 1.71 and thos who dont check the question posted by vyppi(which is similar)

when i calculated the tension from the ground frame i got the answer given at the back of the book but when i try calculating it from the frame of the lift i get a different answer , and actually when i checked out the solution in the net they had solved it from the frame of the lift and their answer wasnt matching with the one in the book.

can some one explain how the tension differs if i calculate it from some other frame

anchitsaini

in the frame of elevator --

applying pseudo force we get the equation as --

(assuming m2 > m1)

T - m1 ( g + w) = m1a        ---1       where w=w₀ and a = a relative to the lift

m2(g+w) - T =m2a            ---2

the above 2 equations give--

a = (m2 - m1)(g+w) / (m1 + m2)       ----3

(from here on i couldn't do it)

g/ g = - w / w     ---4 (since both are opposite to each other)

so multipying eqn 3 by g/ g

we get --

a=(m2 - m1)(g+w g/ g) / (m1 + m2)
=(m2 - m1)(g- w ) / (m1 + m2)

elevator shaft is fixed to the ground, so it would see the actual acceleration --

for m1--
as m1 is moving upwards ,

a = actuall acceleration(a1) - w

so

a1=a + w
   =((m2 - m1)g + 2m2 w) / (m1+m2)    on putting value of a from 3

   =((m2 - m1)g+ 2m2 wg/ g) / (m1+m2)

   =((m2 - m1)g- 2m2 w )/ (m1+m2)

   = - [(m1 - m2)g+ 2m2 w )/ (m1+m2)

hence - a1 = component of a1 in the direction of g is probably the right answer

we can find the value of T easily now

madman

any one else?

eistien

T - (m1) ( g + p) = (m1)a

where p=w₀ and a = a relative to the lift

(m2)(g+p) - T =(m2)a

a = (m2 - m1)(g+p) / (m1 + m2)

by solving these equations you will get finally

a1=- [(m1 - m2)g+ 2m2 w )/ (m1+m2)

and then T from the abve equations

i am unable to type the full thing but hope you understand from here!!!

computer001

eistien..pl read previous explanations b4 posting...n the problem is ans is given as: m1-m2/m1+m2*(g-w_o)

madman

guys i know how to solve the problem
my question is that
shouldnt force exerted on the ceiling be the same irresperctive of the frame we are measuring it from

anchitsaini

i am not sure but i would say it should be different for the simple reason that

F = ma and a depends on the frame of reference hence force exerted on ceiling would also depend on the frame of reference

madman

but if i connect a spring balance to the top of the elevator and then connect the pulley to it what would it measure

computer001

c the force actually on the ceiling is the same but in case u r seeing from an acc frame u r assuming an extra force i.e psedo force...which does not really exist...
bcoz the lift itself is accelerating ..a person in it feels as if an extra force is acting...if u do not want to talk abt pseudo force and stufff u first find acc based on inertial frame..
then use vector addition to find out acc in wrt the acc frame..ur confusion lies in the fact abt the pseudo force which in reality does not exist but is simply introduced to make our calc easier when v talk wrt non-inertial frames

anchitsaini

now before we move further
i would like to say that force applied and force measured are two different things.

force exerted on the ceiling would be the same in all reference frames but the force measured may be different depending on the frame

computer001

@ anchit:
pl chk up ur previous pt

anchitsaini

ya computer001 for the second last thing u said

in my second last pt i wanted to mention something else but i didn't express it well enough

computer001

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