Man climbs rope ladder attached to balloon with acceleration relative to ladder

[txy]

Homework Statement

There is a balloon of mass M_b. A rope ladder of negligible mass is hung from it. A man of mass m stands on the rope ladder. A buoyant force F acts on the balloon, causing the man-balloon-ladder system to accelerate upwards. Now, the man climbs up the rope ladder towards the balloon with an acceleration of a_m relative to the rope ladder. Find the acceleration relative to the ground of

1. the center of mass of the man-balloon-ladder system;
2. the balloon.

The acceleration due to gravity is g.

Homework Equations

I think it's just intelligent application of Newton's Second Law of motion and concepts of the center of mass.

The Attempt at a Solution

My book provided the answers, but did not state clearly which expression belongs to which acceleration.
There is a $$\frac{F - m a_{m}}{M_{b} + m} - g$$
and a $$\frac{F}{M_{b} + m} - g$$ .

I think the second expression is for the acceleration of the center of mass. If I consider the whole system, I get
$$F - (M_{b} + m) g = (M_{b} + m) a_{c}$$, where $$a_{c}$$ is the acceleration of the center of mass.

I'm not sure how to obtain the second acceleration expression.

 

[Doc Al]

I think the second expression is for the acceleration of the center of mass. If I consider the whole system, I get
$$F - (M_{b} + m) g = (M_{b} + m) a_{c}$$, where $$a_{c}$$ is the acceleration of the center of mass.

Right.

I'm not sure how to obtain the second acceleration expression.

Hint: If the acceleration of the balloon with respect to the ground is "a", what's the acceleration of the man with respect to the ground? Use those results to express the acceleration of the center of mass in terms of "a" and a_m.

 

[txy]

Oops I didn't phrase my question properly. I should have written "I'm not sure how to obtain the other acceleration expression." to avoid ambiguity. Thankfully you understood what I meant.

Why haven't I thought of finding the acceleration of man with respect to the ground before?

If a = acceleration of balloon with respect to ground,
then
acceleration of man with respect to ground = a + a_m .

So net force on center of mass = net force on whole system = vector sum of net forces on individual objects in the system.
So
$$(M_{b} + m) a_{c} = M_{b} a + m(a + a_{m}) = (M_{b} + m) a + m a_{m}$$
And so
$$a = a_{c} - (\frac{m}{M_{b} + m}) a_{m}$$
Then I sub in the expression for a_c that I've obtained earlier and I'll get the expression given in the answer.

At first I wasn't sure about how the acceleration of center of mass is like the "weighted average" of the accelerations of the individual objects in the system. Now I understand. I hope my steps above are correct. Thanks a lot for pointing me in the right direction, with regards to the relative acceleration thing.

 

[Doc Al]

Perfectly correct.