3sir even ive dine for closedbox only...pls point out my error in post no 25
Suppose that there is a bird inside a Large Box.
The mass of the bird is m.
The mass of the block without the bird is M
What is the reading on a spring balance when
1) The bird is on the floor of the block
2) The bird is flying inside the box
3) The bird is pushing the box upwards by fluttering its wings?
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29 Answers
1well, i'm givin a stupid example, but if u stand on the weighin machine, which btw, givs the mass, nd if u fly above it, wont der b any difference?
1answer to bhaiyyas prev. question :-
"@everyone.. is the mass of air measured when we put a box on a weight?
I mean an open box of mass M has a small mass of air dM
thenis the weight given by
Mg or (M+dm)g ?
"
no not in open box ,
not even in a thin closed box(thick box mein toh it will be lesser than the box's weight)
but if the closed box is weighed in vacuum , then the mass of air will be measured too........
but bhaiya plz explain that last case(in original question) a bit more elaborately.....
3
62When the box is closed, the weight will be (m+M)g. In all three situations. (The reason is well explained by gaurav and philip)
When the box is open,
Then when the bird is sitting then the weight is given by (m+M)g
when the bird is fluttering, the weight will be fluctuating between about Mg depending on the thrust of the bird's fluttering of wing. . whether it is at the top or not will not matter much. Only that the upthrust and downthrust will be more or less depending on the height from the base.
62When I was giving this question, I knew that there is going to be a lot of controversies... Lets hope that is the case :) :D
Even I will learn a lot of things..
A couple of answers that I know for sure from other sources. And a few that I gave for the sake of better understanding so that we can discusss this and reach a better understanding and good conclusion.
With this in mind, read my take on the whole situation. :)
3i think the bird will feel an upthrust because of the air in block when it is flying .............
let that be B N ................and the bird prodices flight similar to a rocket in this case[3] ...
so an equal and opposite rxn acts on the box ................ now if F is the thrust for flight created by the bird then and if the bird is flying upward (but same time not touchin the roof) then F>mg - B ....................or F+B>mg ...
LOOK AT THIS CAREFULLY ULL SEE THE FORCE ON THE BASE DUE TO THE BIRD IS F+B...
hence the readin will be greater that M+m !!!!!!!!!!!!!
if the bird is steady (ie not goin up nor comin down) then F+B =mg so readin = M+m ....
and if bird is decending then F+B<mg so reading will be less than M+m............
now if its standing on the floor then its surely M+m .......
106For (3)
For closed cage
If the cage and bird move upward with acceleration 'a' then weight recorded wud be (Mg+mg+Ma+ma)/g
If slowly lifted then M+m
For open cage....
i dont know..
11i think that if the bird sits so total mass will become M +m and the force becomes (M+m)g
for me its
(1)
106For (2)
For a closed cage..... it will be (Mg+mg)/g if the bird is flying horizontally or very slowly upwards or downwards... if it moves with some acceleration then obviously answer will be different.. if it moves with acceleration 'a',
then reading will be (Mg + mg + ma)/g if acceleration is upwards or (Mg + mg - ma)/g if acceleration is downwards.
For an open cage.... nothing can be said as it depends on movement of air ... however if air only moves horizontally. then answer will be (Mg)/g. as the air exerts no downward force.. however if the air exerts downward force (say F) then reading = (Mg+F)/g
106sry posting my answer again i hadnt read the question properly...
For 1) if we neglect air pressure, then
for both open and closed cages reading will be (Mg+mg)/g i.e. M+m
62Awesome..
a couple are wrong.. and a couple I am not fully convinced yet.
I am not spelling out the exact answers yet... will want to wait for more users to give an opinion...
Sorry guys please wait for one day for the answer to this one.....
:)
1in first case it is mg+M
second case it cant be said bcoz the bird is flying so their is change in mass at every instance
third case (M+m)g
1let nishant point some mistakes in our ans and Nishant plz see the pirates question
1ashish for open box last one..
can be anything b/w M-m to M i think why M+m ???
1there is a case which is eating at my mind rite now..
will the mass be same (oops the reading i mean ) even when the bird is accelerating in air and when she (or it if you like) is stationary..
anyways i still answer M+m each time..(for closed box)
reason matching to quite extent to gaurav bhaiya
1but will mass/force of the air pressure affect? wont d bird be heavier? nd then when it flies, that mass shud be subtracted, right?
62@everyone.. is the mass of air measured when we put a box on a weight?
I mean an open box of mass M has a small mass of air dM
thenis the weight given by
Mg or (M+dm)g ?
Now does it give more options to think for more answers ;)
62friends give reasons too :)
already there are 3 answers for each choice ;)
1I think Nishant bhaiyya wants a convincing reason as well...
which i wont be able to manage now....
boards from 2nd....
(i mean ill have to think a bit...)
1by the pressure of the air bird is doveloping by its wings to remain in air
11) M + m
2) M
3) M - m
???????????
Looks very easy but i've learnt that such questions are major traps..........