Buoyancy Problem Involving Archimede's Principle

[aokidopi]

1. A wooden ball with a weight of 14.0 N hangs from a string tied to a spring scale. When the ball is at rest, exactly 50% submerged in water, the spring scale reads 5.00 N. For this problem, we will use a density of water of 1000 kg/m3, and we will use g = 10.0 m/s2.

(a) If the ball was only 20.0% submerged in water instead, what would the spring scale read?
(b) Determine the density of the ball.
(c) Determine the volume of the ball.



2. Archimede's Principle



3. I was able to find the answer to part A by applying Newton's second law on the forces acting on the wooden ball. Using the buoyant force from the given and the percentage of submergence, I set a proportion finding the buoyant force for the new percentage of submergence and resulted in 3.6 N. Since the scale measures the tension in the spring, tension would be mg-buoyant force.

However, I am having troubles with part b and c. Using Archimede's principle of buoyant force = density of fluid * volume displaced* g, I resulted in volume displaced of 3.6 E-4 cubic meters. Using a derivative of the equation, density of fluid*volume displaced=density of object*volume of object, I set the volume displaced/volume of object as 0.2, since that is the percentage of the object submerged and resulted in a object volume of 0.0018 cubic meters. Using this proportion of 0.2 as equal to density of object/density of fluid, I resulted in a object density of 200 kg/cubic meters. However, these answers were incorrect.

Any advice will be appreciated.

 

[gneill]

Answer part c first: When the ball is first submerged 50% you are essentially given the weight of the water displaced. What volume of water is that? What then is the volume of the ball?

You have the weight of the ball, thus its mass. You now have the volume too. So what's the density of the ball?

 

[BruceW]

gneill has the right idea.

P.S. aokidopi - you got part (a) correct.

 

[aokidopi]

Got it, thanks!

 

[sickboy007]

I'm a little confused still. So at 50% immersion, the force of tension is 5 N, so the F_B must be 9 N to add up to the total 14 N.
9 N / 10 m/s^2 is 0.9 kg of total displaced water.
Where do I go from here?

 

[BruceW]

That's right so far, sickboy. We also know the density of the water, so it is simple to find the volume of water displaced.

And from there, you know the volume of water displaced is half of the volume of the ball, so this gives you the volume of the ball.

 

[sickboy007]

So V=m/d, and because that is the volume at 50% submerged, it should be multiplied by 2?

 

[BruceW]

exactamundo. (yes)