What is the Density Ratio of a Submerged Ball in a Fluid?

AI Thread Summary
The discussion revolves around calculating the density ratio of a submerged ball in a fluid, where a block of mass M is connected to a ball of mass 3M via a pulley. The equilibrium of forces is analyzed, leading to the equation where the buoyant force plus tension equals the gravitational force on the ball. The initial attempt incorrectly concluded the fluid density as twice that of the ball's density. Clarification reveals that the buoyant force equals the weight of the displaced fluid, leading to the conclusion that the ratio of the densities corresponds to the ratio of their weights since the volume remains constant. The final understanding corrects the initial miscalculation, confirming the importance of buoyant force in determining density ratios.
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Homework Statement



A block of mass M is attached to a light string. The string passes over a massless, frictionless pulley and is attached to a solid uniform ball of mass 3M. The ball is at rest and completely submerged in a fluid as shown. What is the ratio of the density of the fluid to the ball?

Density.jpg


Homework Equations



Fb = (rho)Vg
Fg = mg

The Attempt at a Solution



The forces acting upward must balance the forces acting downward on the ball. Thus,

Fb + Tension = Fg

Then,

(rho)Vg + T = 3Mg

We know that the tension in the string is equal to M because it is keeping the block at rest. Thus,

(rho)Vg + Mg = 3Mg
(rho)Vg = 2Mg

Divide both sides by Vg and we obtain:

(rho(fluid)) = 2(rho(ball))

This is not the correct answer. Can someone explain what I'm doing wrong?
 
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What is the buoyant force on the ball?

It has 3*M*g acting down - its weight. It is in equilibrium so the buoyant force must be 2*M*g

The weight of displaced water then is density water times Volume = 2*M*g

But in that same volume is 3*M*g

Consider then the ratio of the weights of the ball, and the displaced water. Won't that ratio be the ratio of the densities since the volume is he same?
 
LowlyPion said:
What is the buoyant force on the ball?

It has 3*M*g acting down - its weight. It is in equilibrium so the buoyant force must be 2*M*g

The weight of displaced water then is density water times Volume = 2*M*g

But in that same volume is 3*M*g

Consider then the ratio of the weights of the ball, and the displaced water. Won't that ratio be the ratio of the densities since the volume is he same?

Ahh, you're a lifesaver. Thank you!
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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