What is the Buoyant Force on a Submerged Ping-Pong Ball?

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The discussion centers on calculating the buoyant force acting on a submerged Ping-Pong ball using Archimedes' principle. To determine the force needed to keep the ball submerged, one must first calculate the volume of the ball and then find the weight of the water it displaces. The buoyant force equals the weight of this displaced water. Additionally, the net force on the ball includes its weight, the applied force to keep it submerged, and the buoyant force, which together balance out to zero. Understanding these forces is crucial for solving the problem effectively.
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My professor briefly went over Archimede's principle and assigned some problems on it. I do not know how to attack it. Here's the first problem he gave:

A Ping-Pong ball has a diameter of 3.80 cm and average density of 0.084 g/cm^3. What force is required to hold it completely submerged in water?
 
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APool555 said:
My professor briefly went over Archimede's principle and assigned some problems on it. I do not know how to attack it. Here's the first problem he gave:

A Ping-Pong ball has a diameter of 3.80 cm and average density of 0.084 g/cm^3. What force is required to hold it completely submerged in water?

The buoyant force is equal to the weight of the water displaced by the ball. You can find that by figuring out the volume of the ball and multiplying by the density of water. The net force on the ball is its little bit of weight, plus you pushing it down, plus the buoyant force, and those add up to zero.
 
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Start by identifying all the forces acting on the ball, including the applied force. One of those forces is the buoyant force, which Archimede's principle says is equal to the weight of the displaced water. (The buoyant force is the force that the fluid--water, in this case--exerts on a submerged object.)
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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Yes, that makes sense. Thank you for pointing that out.
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