Problem statement

Problem 1.59. An aerostat of mass m starts coming down with a constant acceleration w. Determine the ballast mass to be dumped for the aerostat to reach the upward acceleration of the same magnitude. The air drag is to be neglected.

We assume g = 9.8 m/s². The problem gives no number for it. This means all forces in newtons depend on g. It is a slider.

We assume the aerostat starts from rest and the ballast is dropped after t of the dump = 2 s. The problem gives no time. This means only the height and velocity curves depend on it, not the ballast. It is a slider.

We assume m = 1000 kg as a starting mass and w = 2 m/s². The problem uses letters only. Both are sliders, so any case can be tried.

Symbols: m is the starting mass, w is the size of the acceleration (down, then up), R is the upward push of the air on the aerostat, Δm is the ballast dropped.

The big idea in plain words

An aerostat floats because the air pushes it up. This push, called R, depends on the volume of the big gas envelope. Sandbags are small and dense, so dropping them leaves the volume, and so R, exactly the same. Only the mass changes.

At the start the aerostat is a little too heavy, so it sinks. After the dump it is lighter than the air lift supports, so it rises. R is the same in both cases, and that single fact solves the problem. The ideal gas law is not needed here. Only Newton's second law and the idea that R stays fixed are used.

Notice that the answer is not 2 m w / g. Dropping mass also removes inertia, so the lighter aerostat accelerates more for the same push. Less ballast is needed than the naive guess.

The solution in each frame

1. Ground frame (an inertial observer)

Take up as positive. Before the dump the acceleration is −w:

After the dump the mass is m − Δm, R is unchanged, and the acceleration is +w:

Setting the two values of R equal gives m (g − w) = (m − Δm)(g + w). Expanding, m g − m w = m g + m w − Δm (g + w), so:

2. Aerostat frame (the rider in the basket)

The rider is always at rest, so the real forces plus the pseudo-force must add to zero. The pseudo-force is minus the mass times the acceleration of the frame. Before the dump the frame accelerates down at w, so the pseudo-force m w points up: R + m w − m g = 0, which gives R = m (g − w). After the dump the frame accelerates up at w, so the pseudo-force (m − Δm) w points down: R − (m − Δm) g − (m − Δm) w = 0. The same Δm follows. The rider feels lighter than normal before the dump (g − w) and heavier after it (g + w). Choose this frame on the page to see the amber pseudo-force arrow flip.

3. Frame of the centre of mass of aerostat plus bag

Treat the aerostat (mass m − Δm) and the falling bag (mass Δm) as one system of constant mass m. The only outside forces are R up and the total weight m g down, and they do not change when the bag is released. So the centre of mass keeps accelerating down at w:

The same result again. In this frame the aerostat rises away from the centre of mass and the bag falls away from it.

4. Air frame

If the air moves at a steady speed, the aerostat only drifts sideways. The vertical forces and accelerations are unchanged, so the answer is the same.

What changes and what does not. The speed, the height and the pseudo-forces depend on the frame. The air lift R, the ballast Δm and the acceleration of the system's centre of mass do not.

The solution step by step

Step 1, before the dump. R = m (g − w) = × ( − ) = N.

Step 2, after the dump. R = (m − Δm)(g + w). Then m − Δm = R/(g + w) = m (g − w)/(g + w) = kg.

Step 3, the ballast. Δm = m − (m − Δm) = m [1 − (g − w)/(g + w)] = 2 m w / (g + w) = kg, which is % of the starting mass.

Step 4, the motion. Before the dump, v = −w t and y = −w t²/2. At the dump the speed is m/s downward. After the dump the acceleration is +w, so the aerostat needs the same time t of the dump to stop. The lowest point is y = −w t² = m at s. It then climbs and passes the start level at (2 + √2) t = s.

Step 5, the bag. The bag is released with the aerostat's velocity and then falls with acceleration g. Seen from the aerostat it moves away at a relative acceleration g + w = m/s².

Five methods

1. Force method (Newton's second law)

This is the solution above: two force equations with the same R. It takes three lines.

2. Energy method

After the dump the aerostat has speed v₀ = w t downward and brakes with acceleration w, so it stops after a distance d = v₀²/(2w). The work-energy theorem says [R − (m − Δm) g] d = ½ (m − Δm) v₀². Dividing by d gives R − (m − Δm) g = (m − Δm) w, which is Newton's law again. So this method gives no new information. It is a useful check on the algebra, not a separate route.

3. Graph method

On a velocity-time graph the line has slope −w before the dump and +w after it, so it makes a V. For the system of aerostat plus bag, the total momentum has a single straight line of slope −m w, because the outside force never changes. The momentum of the aerostat plus the momentum of the bag must follow that line. At the instant after the dump this gives (m − Δm)(+w) + Δm (−g) = −m w, the same equation as in the centre-of-mass frame. The page shows the V and the slope on the velocity graph.

4. Calculus method

Write the equation of motion with a mass that drops in a step: M(t) dv/dt = R − M(t) g, with M = m before t of the dump and m − Δm after. The ballast leaves with zero speed relative to the aerostat, so there is no rocket-thrust term. Integrating each piece with constant acceleration gives the curves on the page, and requiring dv/dt = +w after the dump gives the same Δm.

5. Check method

Units: 2 m w / (g + w) has kilograms, because w and g cancel. Limits: when w is small, Δm is close to 2 m w / g, the naive value. When w tends to 0, Δm tends to 0. When w tends to g, Δm tends to m, so the lift is tiny and almost everything must go. For w equal to or above g there is no solution. Putting Δm back gives R/(m − Δm) − g = w, as the numerical check shows.

Fastest: the force method. Most reliable: the force method followed by the check method.

What each slider changes

Control	Changes the answer?	What it changes
Starting mass m	Yes	The air lift, the ballast, the mass left, all forces and the ballast graph. Not the motion
Acceleration w	Yes	Everything: lift, ballast, motion and all graphs
Gravity g	Yes	The lift, the ballast, the forces, the fall of the bag and the graphs
Dump time	No	When the bag leaves, the lowest point, the return time and the height and velocity curves
Given-answer mode	Sets w	The page finds the acceleration that needs the chosen ballast
Frame buttons	No	The picture and the height graph, not the physics
Time bar and Play	No	The clock, with 4 s holds at the dump, the lowest point and the return

Parameters

Quantity	Symbol	Value used
Starting mass	m	kg
Acceleration	w	m/s²
Gravity	g	m/s²
Air lift	R	N
Ballast dropped	Δm	kg
Mass left	m − Δm	kg
Net force after the dump	(m − Δm) w	N up
Dump time	t of the dump	s

Exam traps

• Writing Δm = 2 m w / g. This forgets that a lighter aerostat has less inertia. The naive value on screen is kg, the true value is kg.
• Thinking the aerostat turns round at once. Dropping the ballast flips the acceleration, not the velocity. It keeps sinking for the same time t of the dump, then rises.
• Letting R change when the bag is dropped. The bag has almost no volume, so R stays the same.
• A sign error in the second equation. Before the dump mg > R. After the dump R > (m − Δm) g.
• Allowing w to reach g. Then R is zero and the aerostat would be in free fall, so no real ballast fits.