Trying to study for my exam but got stuck on exercise

[Voidness]

Homework Statement

Will upload an Image

Relevant Equations

Didnt understand this but I think Force equations.

 

[jbriggs444]

Homework Statement: Will upload an Image
Relevant Equations: Didnt understand this but I think Force equations.

View attachment 360563View attachment 360561

You have shown a scattering of equations. It is hard to follow your reasoning since you have not provided that part. Let me see what I can extract from the equations.

You begin by calculating the volume of the balloon and arrive at 1150 m³.

You then multiply by the density of the displaced air to arrive at the mass of the displaced air: 1472 kg.

You go on to assert that $F_\text{res}$, the net upward force on the balloon is given by the applied upward force $F$ minus the downward force of gravity $F_g$.

Algebraically, this resultant force is then also given by $F_res = ma - mg$.

You know the required acceleration $a$ is given by $\frac{2y}{t^2}$ where $y$ is the displacement that occurs (for constant acceleration starting from rest) and $t$ is the time interval over which this displacement occurs.

You evaluate $\frac{2y}{t^2}$ and get $a=\frac{2 \times 50}{15^2} = \frac{4}{9}$ m/s².

You assert that $ma = ma - mg$. This is erroneous. Can you see why?

You blindly perform algebra on $ma = ma - mg$, failing to notice the obvious algebraic manipulation that would show that $mg = 0$ from which one could immediately conclude that $m=0$.

There is an equation that looks like: $m \approx 3.1 \text{kN}$. But that would be erroneous. The units do not work out. So I may have misread that.

At this point, things are going catastrophically down hill. You arrive at a large formula that I cannot be bothered to try to read or understand.


I consider the problem itself to be poorly set. If your approach is correct, the problem setter expects us to ignore air resistance as a sphere disturbs a volume of air that outmasses the sphere by a factor of about six to one.

 

[Steve4Physics]

View attachment 360561

In addition to what @jbriggs444 said, the question seems incomplete:
- what is the initial mass (or weight) of the balloon+passengers+ballast?
- what is the initial state (e.g. stationary on the ground)?

 

[jbriggs444]

In addition to what @jbriggs444 said, the question seems incomplete:
- what is the initial mass (or weight) of the balloon+passengers+ballast?
- what is the initial state (e.g. stationary on the ground)?

If we assume/guess that the balloon is initially stationary and neutrally buoyant, both questions are answerable.

My teachers told us that when a problem appeared incomplete we were to make reasonable assumptions for the missing data, state those assumptions clearly and go on to work the exercise.

I do agree that the question should have included that background.

 

[Voidness]

Sorry for this mess. I also apologize that I am bad in physics, its my first year that I have physics.
Solution are down. I dont know how to use calculus in physics, we just started calculus in math.
The teacher said that who wants a good grade should do these exercises.

If we assume that the balloon was on the ground.

1. Solution

2. Solution

 

[Steve4Physics]

I dont know how to use calculus in physics, we just started calculus in math.

If we assume negligible air resistance, no calculus is needed for this problem.

1. Solution

View attachment 360602

2. Solution
View attachment 360604

I agree with your final answer but the working is difficult (for me anyway) to follow.

It’s worth explaining the reasoning, and defining any necessary symbols, as part of your answer. If I were answering, it might be something like this...

Assume the loaded balloon (total mass $M$) is initially at rest on the ground with neutral buoyancy. I.e. the weight ($Mg$) equals the upthrust from the displaced air ($\frac 43 \pi r^3 \rho g$) so resultant force is zero.

If a mass $m$ of ballast is removed, the new total mass is $M-m$.

The balloon’s weight has decreased by $mg$ so the resultant force is now $mg$ upwards.

Assume negligible force of air resistance while the balloon rises.

Applying Newton II (‘F = ma’):
$mg = (M- m)a$
[a couple of lines of simple algebra]
$m = M \frac a {g+a}$

[Then calculate values for $M$ and $a$ and insert in above equation.]

 

[jbriggs444]

If we assume negligible air resistance, no calculus is needed for this problem.

I am a bit curious... what happens if we try to account for air resistance?

We can assume a spherical shape. A quick trip to Google says that the coefficient of drag ($C_d$) for a sphere is 0.47. We also find the relevant formula:$$C_d = \frac{2F_d}{\rho \mu^2 A}$$Here $F_d$ is the drag force, $\rho$ is the fluid density, $\mu$ is the flow velocity and $A$ is the cross sectional area of the object in question.

From the problem statement we can harvest figures for everything but the drag force. We can solve for $F_d$ and obtain:$$F_d = \frac{C_d \rho \mu^2 A}{2}$$We can use the average ascent velocity for $\mu$, that is 50 meters in 15 seconds or about 3.3 m/s. $\rho$ is a given at 1.28 kg/m³. The coefficient of drag is 0.47. The relevant cross section is $\pi r^2 \approx 132$ m². Plugging that all in we get:$$F_d = \frac{C_d \rho \mu^2 A}{2} \approx 0.5 \times 0.47 \times 1.28 \times 3.3^2 \times 132 \approx 430 \text{N}$$Assuming that I have not blundered egregiously.

By contrast, the net buoyant force from 64 kg mass under one g is about 640 N.

So an assumption of constant vertical acceleration under negligible air resistance is not realistic. Nor is an assumption of constant speed ascent at terminal velocity. A proper solution would require solving a differential equation. Which is probably a year or more beyond the instruction that @Voidness has so far received.

A computation based on the assumption of a constant speed ascent at terminal velocity could be a useful exercise.

 

[haruspex]

If we assume that the balloon was on the ground.

If we assume the balloon is initially resting on the ground we can only calculate its minimum mass. To get the exact mass we need to assume neutral buoyancy, as @jbriggs444 stated.

The wording and format of the question in post #1 are strange. Likely a translation, but also seems computer generated in some way, with randomised constants plugged into a template. Anyone familiar with such a device?