How Much Ballast Must Be Thrown to Give a Balloon Upward Acceleration?

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To achieve upward acceleration for a descending research balloon, the amount of ballast to be thrown can be calculated using the formula m = (2aM)/(g+a), where g is gravity and M is the balloon's mass. However, the calculation may seem incorrect without considering the buoyant force, which remains constant when ballast is dropped. The discussion also touches on a separate question regarding the reading of a spring scale with weights attached; it clarifies that the scale would read the weight of the attached object, regardless of the forces acting on it. Understanding the buoyant force and the weight of the balloon is crucial for solving the initial problem accurately. Overall, both problems emphasize the importance of force dynamics in different contexts.
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A research balloon of total mass M is descending vertically with download acceleration a. How much ballast must be thrown from the car to give the balloon an upward acceleration a, assuming that the upward lift of the air on the balloon does not change?

I get m = (2aM)/(g+a), where g is Earth's gravity, M mass of balloon, and m mass of ballast. It doesn't sound right though.. Can someone please help me with this?

Ok, and as for the other... (very simple)

Two 10-lb weights are attached to a spring scale on opposite sides of each other. What is the reading of the scale? Should it be 10 lbs? Also, what if a single 10-lb weight is attached to a spring scale attached to a wall? Should the scale still be 10 lbs due to equal and opposite forces? Thanks!
 
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Why does your solution of the first problem not "sound right?"
 
you must use the archimede's equation for the balloon to find out the lift force of the air effect on the balloon. and then use the gravity equation to find the weight of the balloon.and you must know the density of the gas you ut into the balloon.

for the second one: i am actually don't know what is the "reading" means. but i think if you mention about the force when two weights attached to the spring i think there are no force effect between them.but if it is mention about the middle weight i think in real you can not have the "reading of the scale" .for the "what if" question i can say yes ! if this is a single system-no have anyforce except the force between the objcts in the system. see you!
 
Since the buoyant force does not change when dropping ballast he does not need explicitly to calculate it.
 
Could somebody explain the balloon problem?

Thanks
 
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