Problem 17: Range of Mass for Balloon Equilibrium
Analysis:
Let $B = \frac{4}{3}\pi r^3 \rho_{air} g$ be the buoyant force acting at the center of the balloon.
The forces acting on the balloon are:
1. Buoyant force $B$ (up).
2. Weight of balloon $mg$ (down).
3. Weight of stone $m_0 g$ (down, acting at the attachment point).
4. Normal force $N$ and Friction $f$ from the ceiling.
Force Balance (Along the Incline):
Since the balloon is helium-filled, the net vertical force tends to be upward. The component of the net vertical force pushing “up” the incline is balanced by static friction acting “down” the incline (preventing the balloon from sliding up).
$$f = (B – mg – m_0 g)\sin\theta$$
Torque Balance (Rotational Equilibrium):
We take torque about the center of the balloon.
The friction force $f$ provides a torque $\tau_f = f R$.
For the balloon to remain at a standstill, the torque from the stone’s weight must balance the friction torque.
Let the stone hang at an angle $\phi$ from the vertical axis of the balloon (measured from the center). The lever arm of the tension $T = m_0 g$ is $R \sin\phi$.
$$\tau_{stone} = m_0 g R \sin\phi$$
Equating torques:
$$f R = m_0 g R \sin\phi \implies f = m_0 g \sin\phi$$
Condition for Minimum Mass:
Substituting the expression for friction:
$$(B – mg – m_0 g)\sin\theta = m_0 g \sin\phi$$
To find the minimum $m_0$ required to keep the balloon stable, we need the maximum possible restoring torque from the stone. The maximum lever arm occurs when the stone hangs tangentially from the side, i.e., $\sin\phi = 1$.
$$(B – mg – m_0 g)\sin\theta = m_0 g (1)$$
$$(B – mg)\sin\theta – m_0 g \sin\theta = m_0 g$$
$$(B – mg)\sin\theta = m_0 g (1 + \sin\theta)$$
$$m_0 = \frac{(B – mg)\sin\theta}{g(1 + \sin\theta)}$$
Using $B = \frac{4}{3}\pi r^3 \rho g$ and rearranging algebraically to match the specific form (multiplying numerator and denominator terms by 3 if necessary):
$$m_0 \geq \frac{(\frac{4}{3}\pi r^3 \rho – m)\sin\theta}{1 + \sin\theta} = \frac{(4\pi r^3 \rho – 3m)\sin\theta}{3(1 + \sin\theta)}$$
Lower Limit: Derived above. If $m_0$ is lower than this, the friction torque (caused by the strong upward buoyancy pushing into the ceiling) overcomes the stone’s weight, and the balloon rolls up the ceiling. Upper Limit: If $m_0$ is too large, the net vertical force becomes downward, and the balloon loses contact with the ceiling ($N < 0$). $$N = (B - (m+m_0)g)\cos\theta \geq 0$$ $$m_0 \leq \frac{B}{g} - m$$ or expressed similarly to the key: $$m_0 \leq \frac{4\pi r^3 \rho - 3m}{3}$$ Final Answer: $$\frac{(4\pi r^3 \rho – 3m)\sin\theta}{3(1 + \sin\theta)} \leq m_0 < \frac{4\pi r^3 \rho - 3m}{3}$$