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A Man is standing on the ground( Mass m) next to a rope-ladder(length l) attached to a hot air balloon ( Mass 3m). If he starts climbing the ladder and covers half of the rope, find the distance that the balloon will move in the downward direction. A teacher solved this problem by using the fact that the COM's position will remain constant, but my doubt is since there's an external force acting on the system ( mg and 3mg) why will the position of com remain constant?

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The teacher is wrong.

[Edit: I added a slightly easier/better argument at the bottom]

A hot air balloon can fly due to its buoyancy (i.e. hot air is lighter than cold air), also called the Archimedes force. If the balloon is floating in mid-air, and thus not ascending or descending, this means that the gravity on the balloon is exactly matched by the the Archimedes force on it in the opposite direction: $F_A = F_g$.

When the man starts climbing the ladder of the balloon, the force due to gravity on the balloon-man system will become $mg + 3mg = \frac{4}{3}F_g$. Now the man's contribution to the Archimedes force $F_A$ is less easy to calculate (see https://en.wikipedia.org/wiki/Buoyancy#Forces_and_equilibrium). But since the man is not floating in the air, his buoyancy is considerably less, meaning that $F_{A,man} < F_{g,man} = mg$.

So we had $F_{A,balloon} = F_{g,balloon} = 3mg$. And for the man $F_{A,man} < F_{g,man} = mg$. This means that:

$F_{A,balloon+man} = F_{A,man} + F_{A,balloon} < F_{g,man} + F_{g,balloon} = 4mg = F_{g,man+balloon}$.

Or in words: the balloon-man system will accelerate slowly down to the ground, since its buoyancy is less that the gravity on it.

The reason why in reality it might work that you climb on the ladder of a balloon without the balloon descending, is because in that case the pilot of the hot-air balloon will light the fire in it, heating the air in the balloon, which will increase its buoyancy and thus compensate for your extra weight and lack of buoyancy.

Edit:

You can also see it like this: As long as the man is standing on the ground, the man-balloon system is in equilibrium, meaning no resulting external nett force, and thus indeed the COM should not move. The force of gravity on the man is balanced by the normal force (i.e. the upward force exerted by the ground supporting him) and the man's Archimedes force (although this is actually negligible for a person in air). The gravity on the balloon is counteracted by the Archimedes force on it. So

$F_{n,man} + F_{A,man} = F_{g,man}$

and

$F_{A,balloon} = F_{g,balloon}$

Thus the total net force on the man-balloon system is 0:

$F_{net} = F_{n,man} + F_{A,man} - F_{g,man} + F_{A,balloon} - F_{g,balloon} = 0$

But as soon as the man starts climbing the ladder of the balloon, the man is no longer supported by the ground, and thus the normal force $F_{n,man}$ drops out. You can see that the total external nett force on the man-balloon system will no longer be zero, and thus the COM does not have to remain at the same place.

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  • $\begingroup$ I understand all of that, but finally how do we solve the question? What's the final answer? $\endgroup$ Oct 23, 2015 at 13:09
  • $\begingroup$ Well, since we've established that there will be a resulting downward force on the man-balloon system, they will keep accelerating until there is a force balancing it again. So when they hit the ground, they'll stop. So when the man is hanging below the balloon halfway the rope ladder, they will go down until the man hits the ground again, i.e. 1/2*l. Here we're neglecting the weight of the rope, of course. Probably not the answer your teacher is looking for, but the point is the COM doesn't have to stay in rest because there is a nett force, so you can't use your teacher's method! $\endgroup$
    – Lu Kas
    Oct 23, 2015 at 14:58
  • $\begingroup$ I was thinking that if we could figure out the acceleration of COM, we could then use calculus to figure out the displacement of the COM using Accelaration= vdv/ds or d^2x/dx^2 right? $\endgroup$ Oct 23, 2015 at 16:58
  • $\begingroup$ Well not really. Your second equation should be a = d²x/dt². This shows you could integrate a twice to get x. Getting a is easy enough, you'll find a = 1/4*g. But you need to integrate over a certain time period, so over which period would you integrate? Do you understand that as long as there is a nett force, the COM will keep accelerating? As long as there is no ground to stop them, they will keep accelerating infinitely and the displacement will go to infinity. $\endgroup$
    – Lu Kas
    Oct 23, 2015 at 18:03

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