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Chapter 7: Problem 47

A balloon with mass \(M\) is descending down with an acceleration \(a(a

Short Answer

Expert verified
Detach mass is \( \frac{2Ma}{g+a} \).

Step by step solution

01

Identify Forces Acting on the Balloon

The forces acting on the balloon when it descends with acceleration are the weight (gravitational force) downward, which is \( Mg \), and the net force, which equals the mass times the downward acceleration, \( Ma \).
02

Equate Forces for Descent

For the moment the balloon is descending, we have: \( Mg - T = Ma \), where \( T \) is the tension or lift force acting upward. Thus, \( T = Mg - Ma = M(g-a) \).
03

Equate Forces for Ascent

After detaching a mass \( m \), the new total mass is \( M-m \). The forces during ascent are: the lift force \( T = M(g-a) \) from Step 2 now acts upward, causing an upward acceleration, \( a \). Thus, \( T - (M-m)g = (M-m)a \).
04

Solve for the Detached Mass

Substitute \( T = M(g-a) \) from Step 2 into the equation from Step 3: \[ M(g-a) - (M-m)g = (M-m)a \]Simplify and solve for \( m \):1. \( M(g-a) - Mg + mg = Ma - ma \)2. \( mg + ma = 2Ma \)3. \( m(g+a) = 2Ma \)4. \( m = \frac{2Ma}{g+a} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's Second Law
Isaac Newton's Second Law of Motion is a fundamental concept in physics.It tells us that the force acting on an object is equal to the mass of that object multiplied by its acceleration. This can be written as the formula: \[ F = ma \] where:
  • \( F \) is the force in newtons.
  • \( m \) is the mass of the object in kilograms.
  • \( a \) is the acceleration in meters per second squared.
Newton's Second Law is a powerful tool.It helps us predict how an object will move when forces act upon it. In the context of our balloon example, this law is used to understand how much the balloon accelerates depending on the masses and forces involved.Remember, if you change the mass by removing part of it, the acceleration can also change!In the balloon problem, the change in mass affects the forces.Thus, impacting whether the balloon will ascend or descend.
Forces in Equilibrium
Forces in equilibrium occur when the total net force acting on an object is zero.This means the object is not accelerating and could be at rest or moving with constant velocity.For equilibrium, the forces in all directions must balance each other out.
Let's look at the descending balloon as an example:
  • The total force down is the gravitational force \( Mg \).
  • The net force causing the descent is \( Ma \), leaving a partial balance.
  • The lift force \( T \) acts upward.
When in equilibrium, these forces are arranged such that no change in motion occurs.While descending, equilibrium can still be approximate as forces change over time to shift balance and initiate ascent when enough mass is detached to tip the scales.
Gravitational Force
Gravitational force is the attraction between two masses, most commonly experienced as the weight of an object.This force is directed towards the center of the Earth and is calculated using the equation: \[ F_{gravity} = mg \]where:
  • \( F_{gravity} \) is the gravitational force.
  • \( m \) is the mass of the object.
  • \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \) on Earth.
In our balloon problem, gravitational force is the downward force \( Mg \).It opposes the lift force, which tries to move the balloon upward.When a mass is removed from the balloon, gravitational force decreases, helping it transition from descending to ascending.

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Most popular questions from this chapter

A horizontal force, just sufficient to move a body of mass \(4 \mathrm{~kg}\) lying on a rough horizontal surface, is applied on it. The coefficient of static and kinetic friction between the body and the surface are \(0.8\) and \(0.6\), respectively. If the force continues to act even after the block has started moving, the acceleration of the block in \(\mathrm{m} / \mathrm{s}^{2}\) is \((g=10\) \(\mathrm{m} / \mathrm{s}^{2}\) ) a. \(1 / 4\) b. \(1 / 2\) c. 2 d. 4

Two blocks \(m\) and \(M\) tied together with an inextensible string are placed at rest on a rough horizontal surface with coefficient of friction \(\mu\). The block \(m\) is pulled with a variable force \(F\) at a varying angle \(\theta\) with the horizontal. The value of \(\theta\) at which the least value of \(F\) is required to move the blocks is given by a. \(\theta=\tan ^{-1} \mu\) b. \(\theta>\tan ^{-1} \mu\) c. \(\theta<\tan ^{-1} \mu\) d. Insufficient data

A block of mass \(m\) is lying on a wedge having inclination angle \(\alpha=\tan ^{-1}\left(\frac{1}{5}\right)\). Wedge is moving with a constant acceleration \(a=2 \mathrm{~m} / \mathrm{s}^{2} .\) The minimum value of coefficient of friction \(\mu\), so that \(m\) remains stationary wr.t. to wedge is a. \(2 / 9\) b. \(5 / 12\) c. \(1 / 5\) d. \(2 / 5\)

An intersteller spacecraft far away from the influence of any star or planet is moving at high speed under the influence of fusion rockets (due to thrust exerted by fusion rockets, the spacecraft is accelerating). Suddenly the engine malfunctions and stops. The spacecraft will a. immediately stops, throwing all of the occupants to the front b. begins slowing down and eventually comes to rest c. keep moving at constant speed for a while, and then. begins to slow down d. keeps moving forever with constant speed

Two persons are holding a rope of negligible weight tightly at its ends so that it is horizontal. A \(15 \mathrm{~kg}\) weight in attached to rope at the mid point which now no more remains horizontal. The minimum tension required to completely straighten the rope is a. \(150 \mathrm{~N}\) b. \(75 \mathrm{~N}\) c. \(50 \mathrm{~N}\) d. infinitely large
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