Calculating Ball Diameter in a Beaker: Accounting for Tension and Buoyancy

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Homework Help Overview

The discussion revolves around calculating the diameter of a Ping-Pong ball submerged in water, considering the forces acting on it, including tension and buoyancy. Participants are exploring the relationship between these forces and the ball's dimensions.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to apply a specific formula for the diameter but are encountering difficulties in obtaining the correct answer. There is a suggestion to derive the formula independently and to analyze the forces acting on the ball, including tension, buoyancy, and gravity.

Discussion Status

Some participants have provided guidance on considering the force diagram and the balance of forces acting on the ball. There is an acknowledgment of the need to account for all relevant forces, but no consensus has been reached on the correct approach or solution.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available and the methods they can use. There is a focus on understanding the principles of buoyancy and tension in the context of the problem.

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Ball in a Beaker - Diameter - Please Help!

Homework Statement



A thread attaches a 2.73-g Ping-Pong ball to the bottom of a beaker. When the beaker is filled with water so that the ball is totally submerged, the tension in the thread is 6.75 mN. Determine the diameter of the ball.

Homework Equations



r = [6(T + mg)/π × 10^3g]^1/3

The Attempt at a Solution



I am using the above equation but cannot get the right answer!
 
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The equation is not correct. Try to derive the formula for the diameter yourself.

ehild
 


ehild is correct, you need to derive the equation yourself. I would recommend thinking about the force diagram first. Which forces are pulling the ping pong ball down? Which forces are pulling it up? They have to be equal and opposite to each other if the ball isn't accellerating.
 


A thread attaches a 2.74-g Ping-Pong ball to the bottom of a beaker. When the beaker is filled with water so that the ball is totally submerged, the tension in the thread is 6.87 mN. Determine the diameter of the ball.

The force pushing the ping pong ball up is the buoyant force of the water. Thus if the force pulling it down is equal, the tension is equal to the buoyant force. According to Archimedes principal, the buoyant force on a object immersed in fluid is equal to the weigh of the fluid displaced by that object. So the weight of the fluid displaced is 6.87 mn. Thus, the mass is 0.00687 N / 9.87 N = 0.000701 kg. Then, d = m*v. D of water is 1000 kg * m^3. Solving for volume, we get the volume as 7.0102 * 10^-7 m^3. Then we solve for volume with (x/2)^3 * pi * 4/3 = 7.0102 * 10^-7 m^3. I get x to = 0.011022 m. But this is not the answer; what am I doing wrong?
 


tak08810 said:
The force pushing the ping pong ball up is the buoyant force of the water. Thus if the force pulling it down is equal, the tension is equal to the buoyant force.
You are forgetting about another force that acts on the ball.
 


Redbelly98 said:
You are forgetting about another force that acts on the ball.

Ah thank you! So tension + gravity is the force going down and the buoyant force opposes it.
 

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