Understanding the Forces in a Hot Air Balloon: Quiz #4, Question 8 Explained

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In the discussion about the forces acting on a physics student in a hot air balloon ascending at constant speed, the key focus is on understanding the relationships between the forces involved. The correct answer to the quiz question is (d), which states that the weight of the student (F2) equals the force exerted by the hot air balloon on the student (F4), although there is confusion about the nature of these forces. The normal force, which is typically considered in such scenarios, is not applicable here due to the lack of a defined surface. Additionally, the discussion clarifies that the weight of the student is the gravitational force acting downwards, while the balloon exerts an equal and opposite force upwards. Overall, the conversation emphasizes the distinction between different types of forces and their relationships in this specific context.
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A physics student in a hot air balloon ascends vertically at constant speed. Consider the following four forces that arise in this situation:
F1 = the weight of the hot air balloon F3 = the force of the student pulling on the earth
F2 = the weight of the student F4 = the force of the hot air balloon pulling on the student
8. Which one of the following relationships concerning the forces or their magnitudes is true?
(a) F4 > F2 (c) F4 > F1 (e) F3 = –F4
(b) F1 < F2 (d) F2 = –F4

Ok I know the answer is (d) because the forces form a pair someone told me but I don't see they form a pair, do they? and why do we have -F4 (I thought the normal force is positive as the balloon is pulling student up the air)
 
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We can't see the diagram, so we don't know what any of this means.
 
AlephZero said:
We can't see the diagram, so we don't know what any of this means.

thanks! I edited the question! :)
 
gcombina said:
(I thought the normal force is positive as the balloon is pulling student up the air)

Yes, the normal force is upwards, but his weight is downwards (so they have opposite directions)

gcombina said:
Ok I know the answer is (d) because the forces form a pair someone told me but I don't see they form a pair, do they?

His weight pushes down on the hot air balloon with some force, and the hot air balloon pushes up on him with an equal and opposite force (the normal force)
 
thanks! (ill get back to this question later, as i am working on another one but I am seeing your responses)
 
Nathanael said:
Yes, the normal force is upwards, but his weight is downwards (so they have opposite directions)
Since there is no defined surface, there is no "normal" force and no need to characterize any force as such.

His weight pushes down on the hot air balloon with some force, and the hot air balloon pushes up on him with an equal and opposite force (the normal force)
This skips a step.

In the physics classroom a person's "weight" is the downward force of gravity on their body. It is not the downward force that they then exert on another object.

The upward force of the balloon on the person (F4) and the downward force of gravity on the person (F2) are not a third law pair. But they do not have to be. This is not a third-law exercise. What about Newton's second law?
 

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