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A bird sitting in a wire cage which is hanging from a spring balance.if bird starts flying inside the cage,reading of spring balance will remain unchanged/be less/more than earlier?

Less than the earlier because the normal reaction applied by the bird on the cage is 0 when it flies inside the cage

I think extension remains same
[image]38645133.jpg[/image]
sorry for poor drawing of bird

In case 1 : F[ss]s[/ss] = (m[ss]b[/ss] + m[ss]l[/ss]) g

Case 2 : F[ss]s[/ss] = m[ss]l[/ss] g

i agree with eureka....

see this fr proof
http://answers.yahoo.com/question/index?qid=20090724204859AAxic4V

When the bird is not flying, the scale would register the [b]weight of the cage and the weight of the bird.[/b]

Now when the bird flies, while still inside the cage, the air "lift" produced by the wings of the bird sustains the weight of the bird.[b] The floor of the cage does not bear anymore the weight of the bird.[/b]

This argument holds through even on other cases. Now consider, if an airplane flies directly overhead, will we feel its weight?

This is from Arshad's link http://answers.yahoo.com/question/index?qid=20090724204859AAxic4V

I din't understand hoe wt is same in both ( bolded ) cases

@uttara see this to clear ur dbt:
its a very common question
http://www.askmehelpdesk.com/physics/change-weight-bird-cage-when-bird-flying-inside-cage-316976.html

Thanks Arshad

[i]"If the bird is flying, the bird is pushing down on the air. The air is pushing down on the cage.

If the bird is not rising, the weight of the cage should be approximagely the same whether

the bird is resting or flying. If the cage were very small, some of the force might be

dissipated to the earth, but we're talking first-order effects, here."[/i]

My Doubt : Shud we consider the weight due to the air pushed down by the bird's wings??

yeh all that have posted here must be correct must be correct but i have something else in my mind.....will some one verify what i post?? so i am posting now...

there are two cases for this problem....at least according to me...
CASE 1) when the cage has a solid base which can stop blowing air....
so when the bird is sitting quietly the normal reaction on the bird is balanced by its weight. as the same normal force is given to the cage, and spring balance gives an upward force kx....so
kx[ss]1[/ss]=mg+m'g

now the bird starts flying the air is pushed down so that the bird gets a reaction force m'g upward....now this air strikes the base of the cage with same force...thus net downward force is again mg+m'g
so kx[ss]2[/ss]=mg+m'g

so kx[ss]1[/ss]=kx[ss]2[/ss]
or,x[ss]1[/ss]=x[ss]2[/ss]
so same reading..

CASE 2) when the cage has a grill base through which air can pass....
for bird at rest same logic holds good..
so kx[ss]1[/ss]=mg+m'g

now when the bird flies the air gives an upward force m'g on bird but does not give any force on the cage....instead hits the ground with a force m'g
so balancing forces on the cage,
kx[ss]2[/ss]=mg
thus kx[ss]2[/ss]≤kx[ss]1[/ss]
so x[ss]2[/ss]≤x[ss]1[/ss]
so reading less.....

this q is there in fiitjee materials,,ans given is UNCHANGED.

thanks ronaldo bt was my reasoning wrong??? please verify...

HEM HEM...will sum1 tell if i am wrong???specially case 2...#12???

@ subhomoy wat u hav said is sumwat correct....and wat ronald didnt tell was dat in fiitjee ques the cage is assumed to be closed frm all sides....hope u get the point....:)

Two assumptions i made in my soln
a)Considered cage massless
b)Having solid base


@subo...yeah if the base is grilled one..then ans will be diff

practically speaking the bird must push off the ground of cage to accelerate up ...so the x must increase when it pushes off (it requires an extra normal force), then after that the x will oscillate back and forth b/w x[ss]i[/ss],finally after a long time it will same as before

no akari.....when the bird is beginning to rise up upthrust is increasing but normal reaction (which balances gravity) is decreasing....
so net force on bottom part of cage remains same always..