Why Does a Sphere Reach Terminal Velocity if Buoyant Force Exceeds Weight?

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A sphere can reach terminal velocity even when the buoyant force exceeds its weight due to the balance of forces acting on it. The upward buoyant force and the downward gravitational force create a net force that influences the sphere's motion. As the sphere falls, it accelerates until the drag force, which opposes its motion, increases to balance the net force. This results in the sphere achieving a constant velocity, known as terminal velocity, despite the buoyant force being greater than its weight. The confusion arises from misunderstanding the role of drag force, which always acts opposite to the direction of motion.
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Homework Statement
determine the drag force
Relevant Equations
Archimedes principle formula, weight formula
in my head this is just a silly problem in which i need to determine the ↓ force (weight) and the ↑force (archimedes bouyant force) and then the difference must be the drag force ↑ (the force that involves velocity) but i can't get any sense out of this answer
how is possible for the sphere to reach terminal velocity (and thus keep falling) if the bouyant force is greater in magnitude than the weight?
how is the drag force acting downwards (in the direction of motion) ? this doesn't make any sense to me
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i think the sphere must be going up instead of down so this actually makes sense
 
It Is going up. The buoyant force Is larger than the Wright.
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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