Should be very simple F=ma question

[Asphyxiated]

Homework Statement

A group of folks are riding in a hot air balloon 60 m above the ground. The total mass of the balloon and its occupants is 300 kg. The balloon is stationary in the air due to the upward buoyant force of the hot air. Do a free body diagram of the balloon. If the buoyant force remains constant what mass of sand must be thrown overboard so that the balloon acquires an upward acceleration of 5m/s^2?

Homework Equations

$$F=ma$$

The Attempt at a Solution

I would like to preface this by saying that this is a calculus based physics course, so I may need to use calculus here to figure this out, but I am not sure and the instructor said we would only need F=ma to do all the problems, anyway here is my logic, which doesn't end well.

If the balloon is being keep at a constant level of 60 m about the ground then there must be a constant upward acceleration of 10m/s^2 to fight the gravitational constant of -10m/s^2 (we are rounding here for this first assignment, I know 9.8m/s^2 is closer to what it really is), so the hot air buoyant force must be:

$$F=(300)(10)=3000N=3kN$$

so, as the problem states, the 3kN is constant and the new acceleration of .5 m/s^2 should also be constant in the equation:

$$3000=m(.5) \;\; or \;\; m=\frac {F}{a}$$

which would give me a mass of 6000kg, which is obviously wrong, they are looking for the change in mass which implies derivative I would think. Would that be the correct way to think about it? I have been trying to figure it out for a couple hours today, thanks for any guidance!

Edit__________________________________________

And if so would the derivative be:

$$F'=ma'+am'$$

?

If not please show me the way to get there, or where i should start, thanks again!

 

[gneill]

You might want to consider thinking in terms of forces being applied, rather than accelerations. Accelerations are the result of forces.

Essentially, you determined the force of buoyancy by equating its magnitude with the gravitational force. If some mass, m, of sand were pitched over the side, what would be the new force due to gravity on the remaining mass of the balloon and its contents?

 

[Asphyxiated]

Well whatever mass, m, you throw overboard will decrease the force due to gravity proportionally by a constant k in which k=10 by which i mean if you were to throw 10kg of sand over the side you would lose k10N or 100N of force due to gravity.

So i could, technically say that:

$$a = \frac {F}{m}$$

and incrementally lower m till a ~ 10.5 which is 14kg, which is what they are looking for, so if i do:

$$m = \frac {10.5}{3000} = 285.71 kg$$

round up to get 286 or 14kg less than the previous mass, correct way to look at it?

 

[gneill]

Tossing out mass both lowers the mass of the rig (increasing acceleration) and lowers the force due to gravity (increasing the acceleration). The effects are compounded. How about giving the 'traditional' approach a try?

You have a constant upward force due to buoyancy, Fb = 3000N.

You have an initial mass M = 300kg.

Force due to gravity is Fg = M*g = 3000N. Initially this balances the buoyant force:

Fnet = Fb - Fg = 0

The net mass of the rig is currently Mnet = M.

The acceleration is given by a = Fnet/Mnet, or zero in this case since Fnet = 0.

Now throw out some amount of mass, m. What are the new values for Fnet and Mnet? What is the new acceleration?

 

[Asphyxiated]

Perhaps I should just wait and ask tomorrow, I am really not getting what you are telling me for some reason, sorry. If you want to try to correct me here I tried it this way, which seems to be the way you are going but definitely doesn't come out right.

So if Fb=3000N and is constant and Fg could be stated as (10)Mnet and we define Mnet as Mi-Mt where Mi is initial mass and Mt is mass thrown out and acceleration is:

$$a = \frac {F_{net}}{M_{net}}$$

$$a = \frac {F_{b}-F_{g}}{M_{i}-M_{t}}$$

$$a = \frac {F_{b}-10(M_{i}-M_{t})}{M_{i}-M_{t}}$$

$$aM_{i}-aM_{t}=F_{b}-10M_{i}+10M_{t}$$

$$-10aM_{t}=F_{b}-10aM_{i}$$

$$M_{t}= \frac {F_{b}-10aM_{i}}{-10a}$$

and when you sub in the values there I get 100 for Mt if a=5 or if a=10.5 i get around 186, so it is quite obvious to me that I really don't understand what is going on here, is that even what you meant to do, in general?

Sorry for all the confusion

 

[gneill]

Perhaps I should just wait and ask tomorrow, I am really not getting what you are telling me for some reason, sorry. If you want to try to correct me here I tried it this way, which seems to be the way you are going but definitely doesn't come out right.

So if Fb=3000N and is constant and Fg could be stated as (10)Mnet and we define Mnet as Mi-Mt where Mi is initial mass and Mt is mass thrown out and acceleration is:

$$a = \frac {F_{net}}{M_{net}}$$

$$a = \frac {F_{b}-F_{g}}{M_{i}-M_{t}}$$

$$a = \frac {F_{b}-10(M_{i}-M_{t})}{M_{i}-M_{t}}$$

$$aM_{i}-aM_{t}=F_{b}-10M_{i}+10M_{t}$$

$$-10aM_{t}=F_{b}-10aM_{i}$$

$$M_{t}= \frac {F_{b}-10aM_{i}}{-10a}$$

and when you sub in the values there I get 100 for Mt if a=5 or if a=10.5 i get around 186, so it is quite obvious to me that I really don't understand what is going on here, is that even what you meant to do, in general?

Sorry for all the confusion

That's along the lines I had in mind. You went a little awry in your final two lines of the derivation for the throw-away mass, but you're nearly there.

 

[Asphyxiated]

ok, so I redid it again, how about for the last two lines:

$$aM_{i}-aM_{t}=F_{b}-10M_{i}+10M_{t}$$

$$-aM_{t}-10M_{t}=F_{b}-10M_{i}-aM_{i}$$

$$M_{t}(-a-10)=F_{b}-10M_{i}-aM_{i}$$

$$M_{t}= \frac{F_{b}-10M_{i}-aM_{i}} {-a-10}$$

and Fb-10Mi is going to be zero so we can throw that out:

$$M_{t} = \frac {-aM_{i}}{-a-10}$$

subbing the values:

$$M_{t} = \frac {-(.5)(300)}{-.5-10} = \frac {-150}{-10.5} = 14.28 kg$$

sounds right to me, especially mathematically this time, is that the correct way to do that then?

 

[gneill]

Your algebra looks good. Why did you use .5 for the desired acceleration?

 

[Asphyxiated]

Looks like i had a typo at the start of the problem, the desired acceleration is .5 m/s^2 not 5 m/s^2. Evidently I need to watch out for that too, lol.

 

[gneill]

Too bad. The answer works out to a nice round number when the acceleration is 5m/s^2 !

 

[Asphyxiated]

Ha yeah that is true, I triple checked though and it is certainly asking for 0.5 m/s^2.

Thanks for all the help though man

 

[Kurdt]

Post #3 was the correct analysis. The force on the balloon when it is accelerating is equal to the buoyant force minus the weight of the balloon at its new mass.