# What is the motion of a balloon with a person climbing on a ladder?

Suppose you had a balloon of mass $M$ in the atmosphere at rest relative to the ground, with a ladder attached reaching towards the ground. A person (of mass less than $M$) begins climbing up the ladder with constant speed $v$. Describe the motion of the balloon.

Apparently the solution is as follows: take the system to be the mass and the person. Momentum is conserved so $m_{person}v+Mv_{balloon}=0$ so $v_{baloon}=-m_{person}v/M$, so the balloon moves down with a speed less than $v$.

I don't understand why momentum is conserved in this system, as the net force is not zero. Gravity acts on both the person and the balloon, and the buoyancy force acts on the balloon. Perhaps this solution is assuming the mass of the balloon is negligible?

• If the balloon is at rest relative to the ground, doesn't that mean $F_{net}=0$? Or are you thinking about when the person starts climbing? – Kyle Kanos Dec 15 '14 at 2:52
• Strictly speaking the buoyancy also acts on the person, but that's a small effect because the volume of the person is so small. Try to apply Newton's third law to the problem. What does actio=reactio do in this case? – CuriousOne Dec 15 '14 at 2:53
• @Kyle, I'm talking about when the person is climbing. The buoyancy fore changes as the balloon's height is changing, and you also have to account for gravity. – Joshua Benabou Dec 15 '14 at 3:49
• If the buoyant force of the balloon balances the gravitational force of the person, then $F_{net}=0$ still. This could be the case for someone slowly climbing the ladder. – Kyle Kanos Dec 15 '14 at 3:59

You could look at it as a center of mass problem. We assume the ladder has negligible mass. When the climber reaches the balloon, they will be together at the center of mass of the system. The climber and balloon start apart by the length of ladder $l$, while the center of mass is $\frac{l\times M}{m + M}$ above the climber, and $\frac{l\times m}{m + M}$ below the balloon. The ratio of the balloon distance to travel compared to the person distance to travel is $\frac{m}{M}$. Therefore the velocity of the balloon at any point in time is $-v_{person} \times \frac{m_{person}}{M_{balloon}}$.
Center of mass problems can be reformulated as conservation of momentum problems. At $t_0$, the balloon and person are both at rest, with the person hanging off the bottom of the ladder - net momentum 0. At $t_1$, the climber has momentum $m_{person}\times v_{person}$, so the momentum of the balloon must balance this: $M_{balloon} \times \frac{-m_{person}\times v_{person}}{M_{balloon}}$. Same result.