How Does a Balloon React When a Man Climbs a Ladder Attached to It?

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When a 74 kg man climbs a ladder attached to a stationary balloon with a total mass of 260 kg, the balloon moves downward due to the upward motion of the man. The discussion focuses on understanding the relationship between the velocities of the man, balloon, and their center of mass. The center of mass of the system remains at rest because there are no external forces acting on it, leading to the conclusion that its velocity is zero. This creates confusion regarding the calculation of the balloon's downward velocity while the man is moving up. Ultimately, the key point is that the balloon's movement is a reaction to the man's climbing, affecting the overall dynamics of the system.
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Homework Statement



In the figure below, a 74 kg man is on a ladder hanging from a balloon that has a total mass of 260 kg (including the basket passenger). The balloon is initially stationary relative to the ground. The man on the ladder begins to climb at 2.5 m/s relative to the ladder.

(a) In what direction does the balloon move? My answer is downwards which is correct

(b) At what speed does the balloon move?

Homework Equations



vmg = vmb - vbg

(where the subscripts mg refer to man relative to ground, mb man relative to balloon (ladder), and bg balloon relative to ground)

vcom = [mmg - Mvbg]/M + m

The Attempt at a Solution



I actually have the worked out solution to this problem but I'm finding it hard to grasp that
vcom = [mmg - Mvbg]/M + m = 0 (as per my textbook solutions). This is my understanding: the balloon is stationary relative to the ground, thus the vcom of the balloon-ground system would be 0. However, the man starts to move up the ladder causing the balloon to move downward, thus changing the vcom. How can, then, the textbook solution in solving for vbg set the aforementioned equation for vcom to 0? I mean, the vbgto be calculated is not when the balloon is stationary relative to the ground, obviously, it's changing relative to the ground (moving downwards) while the man is moving up the ladder. I would appreciate help in understanding this conceptually, especially if my thinking is wrong.

Thanks!
 
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If xp, xb and xm are the distances of passenger, balloon and the center of mass from the ground, then
xm = (m*xp + M*xb)/(M + m)
If you take the derivative with respect to time to find their velocities, we get
v(xm) = (m*vp + M*vb)/(M + m). Since vp and vb are in the opposite direction the final expression becomes
vm = ...?
 
Last edited:
rl.bhat said:
If xp, xb and xm are the distances of passenger, balloon and the center of mass from the ground, then
xm = (m*xp + M*xb)/(M + m)
If you take the derivative with respect to time to find their velocities, we get
v*xm = (m*vp + M*vb)/(M + m). Since vp and vb are in the opposite direction the final expression becomes
vm = ...?

I'm sorry, I don't understand. Why should I have to take the derivative if I already have an equation tailored to this situation? Maybe I'm not understanding something...
 
In the problem there are three center of masses.
1) Center of mass of man
2) Center of mass of balloon
3) Center of mass of ( man + balloon) system.
Since there is no external force acting on the system, center of mass of the system remains at rest. So v(cm) is zero.
 
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