Solving Newtons Law Problem for Hot Air Balloon Mass of Ballast

AI Thread Summary
A hot air balloon with a total mass of 315 kg is experiencing a downward acceleration of 1.10 m/s², and the balloonist needs to achieve zero acceleration without fuel. The upward buoyant force was calculated to be approximately 2.7 x 10^3 N. To balance the forces, the gravitational force must equal the buoyant force, leading to a required mass of 279.6 kg for equilibrium. By subtracting this from the total mass, it is determined that approximately 35 kg of ballast must be discarded. The approach taken to solve the problem is confirmed as correct.
Ballox
Messages
14
Reaction score
0

Homework Statement


A hot - air balloon experiences an acceleration of 1.10 m/s^2 [down]. The total mass of the balloon, the basket, and the contents of the basket is 315 kg.

The balloonist wishes to change the acceleration to zero. There is no fuel left to heat the air in the balloon. Determine the mass of the ballast that must be discarded overboard. [NEGLECT AIR RESISTANCE]



Homework Equations


Newtons second law equation : F = mA

Free body diagrams are also important


The Attempt at a Solution



I drew a free body diagram for the hot air balloon and I noticed there were two forces: The force of gravity and the upward (buoyant) force on the system. I calculated the upward force on the system to be approximately 2.7 * 10^3 N [up].

However I'm totally stuck on what to do next.
Please help and many thanks in advance!
 
Physics news on Phys.org
Hello Ballox,

Welcome to Physics Forums!

I assume you know the gravitational force is equal to mg. The goal is to set the gravitational force equal to the buoyant force. "g" isn't about to change any time soon, so... :wink:
 
I am not sure how buoyant forces work. Does this buoyant change when force of gravity changes? If so, what is the formula for buoyant force?
 
collinsmark said:
Hello Ballox,

Welcome to Physics Forums!

I assume you know the gravitational force is equal to mg. The goal is to set the gravitational force equal to the buoyant force. "g" isn't about to change any time soon, so... :wink:


So setting the gravitational force equal to the buoyant force would be:

mG = 2.7 * 10^3 N [up] ( I rounded this, it really should be around 2740.5 N)

Which we would solve for m as : 279.6 kg.

But this gives us the mass when when the gravitational force is equal to the buoyant force...so we have to subtract this mass from the mass of the entire system to determine the amount of mass that must be discarded overboard.

So I get :

315 KG - 279.6KG ~ 35 kg.

^
So I get the answer in the textbook! ^^;

Is this the right approach?
 
Ballox said:
So I get :

315 KG - 279.6KG ~ 35 kg.

^
So I get the answer in the textbook! ^^;

Is this the right approach?

Looks good to me! :approve:
 
collinsmark said:
Looks good to me! :approve:

Sweet. Thanks for your help.
 
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