Question

A hollow, plastic sphere is held below the surface of a freshwater lake by a cord anchored to the bottom of the lake. The sphere has a volume of 0.640 m3 and the tension in the cord is 2120 N.

a) Draw a free body diagram when sphere is at the bottom.

b) The cord breaks and the sphere rise to the surface. When the sphere comes to rest, what fraction of its volume will be submerged?

Answer 1

Part A.

Part B. Step 1: Find the Buoyancy force on the sphere:

Fb = *g*V

= density of water = 1000 kg/m^3

V = Volume of sphere = 0.640 m^3

So, Fb = 1000*9.81*0.640

Fb = 6278.4 N

So mass of sphere will be, Using Force balance in vertical direction from FBD (a):

Fy_net = Fb - W - T = 0

W = Fb - T

W = weight of sphere = 6278.4 - 2120

W = 4158.4 N

m = mass of sphere = W/g = 4158.4/9.81

m = 423.9 kg

Now when the cord breaks, then

T = 0, So Fb1 = W

*g*V1 = m*g

V1 = Submerged Sphere's Volume = m*g/(*g)

V1 = m/ = 423.9/1000

V1 = 0.4239 m^3

fraction of volume submerged = Volume submerged/Total Volume

fraction = 0.4239/0.640

fraction = 0.662

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