How Does a Climbing Man Affect a Balloon?

[The legend]

Homework Statement

A balloon of mass 100kg is stationary at a height of 100m. A man standing on a rope ladder hung from the balloon then climbs up with a velocity of 0.3m/s relative to the rope. If the man is of mass 50 kg, what happens to the balloon? (ie, it goes up/down with some velocity/ it remains stationary)

The Attempt at a Solution

I think that since the man is applying force on the rope downwards and going up with 0.3m/s, the rope goes down with 0.3m/s...so same thing happens to the balloon. Am i right?

 

[ehild]

Partly. You are right, the balloon moves with the same velocity as the rope.
The man climbs up with 3 m/s velocity relative to the rope. So what do you think the velocity of the balloon is with respect to what?

 

[The legend]

velocity of the balloon is with respect to the earth, i think..

 

[ehild]

So the man stays at the same height if he climbs upward?

Remember: The internal forces do not influence the position of the ... of a system. The system of balloon and man is stationary, the resultant of the external forces is zero. So the acceleration of the ... is zero. :)

ehild

 

[The legend]

I see now, the whole system IS stationary!
Thank you ehild!

 

[ehild]

Did you find out that I meant the centre of mass?

ehild

 

[The legend]

i didnt, i think.
Just that you hinted the balloon and man system is stationary unless an external force acts...so no external force is acting and hence they are stationary...is it right?

 

[ehild]

No, they move with respect to each other, but their centre of mass is stationary. The man moves upward with respect to the balloon. Both man and balloon move with respect their common CM so as the CM does not move with respect to the ground. You can find the velocities both the man and balloon with respect to the ground. ehild

 

[The legend]

oh, ok...
I actually don't know the concept of center of mass, so I'll look up to it in Wikipedia or the library.

But i get your idea now.

 

[ehild]

Well, you certainly know what the momentum of a mechanical system is. If m_i is the mass of the i-th point mass and v_i is its velocity, p =∑m_iv_i. And you also know that from Newton's second law, the time derivative of the momentum of this system is dp/dt = F(external) ( the resultant of the external forces) as the internal forces cancel because of Newton's third law. If this resultant of the external forces is zero, the momentum of the system is conserved, that is

p =∑m_iv_i = const.

We can define an "average" velocity of the point masses as

V=(∑m_iv_i)/∑m_i.

With that notation, p=MV, with M the total mass of the system. V is equal to the velocity of the centre of mass, a point with coordinates

X=(∑m_ix_i)/∑m_i,
Y=(∑m_iy_i)/∑m_i,
Z=(∑m_iz_i)/∑m_i.

In your problem, conservation of momentum means that

m(balloon) V(ballon)+m(man)v(man)=0

and you know the relative velocity of the man with respect to the balloon:

(man)-v(ballon)= 0.3 m/s.

Hence you can find both velocities v(man) and v(ballon) with respect to the ground.

ehild