Daniel B. answered • 10/31/20

A retired computer professional to teach math, physics

Let

m = 270g be the mass of the ball,

r = 10.5cm be the radius of the ball

d = 1 g/cm^{3} be density of the pool water

g = 9.8 m/s^{2} be gravitational acceleration

V = volume of the ball = (4/3)π r^{3}= (4/3)π 10.5^{3} = 4849 cm^{3}

a) By the Archimedes' principle, the force F is the difference between the

weight of the water displaced by the ball and the weight of the ball.

F = Vdg - mg =

(Vd - m)g =

(4849 cm^{3} . 1g/cm^{3} - 270g) 9.8 m/s^{2} =

44874 gm/s^{2} =

44.894N

b) Assuming no water resistance,

the acceleration comes from Newton's second law

a = F/m = 44.894 N/0.27kg = 166.2 m/s^{2}

c) By the Archimedes' principle, the volume V_{0} of the ball that will

remain submerged will be large enough to displace water equal in weight

to the weight of the ball.

V_{0}dg = mg

V_{0} = m/d = 270g/(1g/cm^{3}) = 270cm^{3}

Thus 270 cm^{3} will be submerged, and the remainder 4579 cm^{3} will stand out

from the water, which is about 94%.

It is possible that the teacher wants you to notice the relation

between a) and c).

Namely, the volume of the ball that will stand out is (Vd -m)/d.

And notice that the expression (Vd - m) appeared in the calculation of part a).

The reason for this coincidence is that the volume that will stand out

will be what "the water is trying to get rid of", as

it it causing the force pushing the ball out.

Daniel B.

It is not necessary. Notice that the two cm3 cancel each other, so there is no problem. And keeping g as m/s2 then makes the conversion to N easier. But this is a very good question, because in general we need to make sure that all units are compatible. My preferred approach is to carry the units together with all the numbers; this reduces the chances of making a mistake and makes any necessary conversion automatic.10/31/20

Gabriel L.

I didn't understand why the Thrust Force has to be subtracted by the weight of the ball to determine the Resulting Force. In the a) question.11/01/20

Daniel B.

There are two forces acting on the ball: 1) Gravity, mg, acting downward, which is the same whether the ball is in water or not. 2) Thrust force, Vdg, acting upward as long as the ball is submerged. Those two forces act in opposite direction, and therefore they are subtracted. If F = Vdg - mg is negative then gravity wins and the object sinks. If F is positive (as in our case), then the upward force, Vdg, is winning, pushing the object out of the water. As it starts emerging from water, the submerged volume V starts decreasing, decreasing the force F. When the submerged volume is reduced so much that the force F becomes 0, then the object floats.11/01/20

Anthony T.

10/31/20