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

AI Thread Summary
The discussion focuses on a physics problem involving a man climbing a rope ladder attached to a balloon, requiring the calculation of the accelerations of the center of mass and the balloon. Participants emphasize the application of Newton's Second Law and the concept of center of mass to derive the necessary equations. The acceleration of the man relative to the ground is determined to be the sum of the balloon's acceleration and the man's acceleration relative to the ladder. Clarifications are made regarding the expressions for the accelerations, with a consensus that the center of mass acceleration is a weighted average of the individual accelerations. The conversation concludes with a confirmation of the correct approach to solving the problem.
txy
Messages
15
Reaction score
0

Homework Statement



There is a balloon of mass Mb. 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 am 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.
 
Physics news on Phys.org
txy said:
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 am.
 
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 + am .

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 ac 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.
 
Last edited:
Perfectly correct.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top