Using archimedes principle find the depth submerged of a log in water. Diameter 2.3cm radius 1.15cm, mass 37g, length of log 10cm, density of water 1000kg/m^3.

Myself I can get to the submereged volume using Fb=Pvg but I don't know an equation for the volume of a segmented cylinder I know its the surface area times the length, but I don't know how to get the surface area. help me please

http://mathworld.wolfram.com/CircularSegment.html

To find the depth submerged of a log using Archimedes' principle, we need to determine the volume of the log that is submerged in the water.

To start, let's find the volume of the log. Since the log is a cylinder, we can use the formula for the volume of a cylinder, which is:

V = πr^2h

Where V is the volume, π is a constant approximately equal to 3.14, r is the radius, and h is the height (or length) of the log.

In this case, the radius of the log is 1.15 cm, and the length of the log is 10 cm. So the volume of the log is:

V = 3.14 * (1.15 cm)^2 * 10 cm
V ≈ 41.95 cm^3

Now, using Archimedes' principle, we know that the buoyant force (Fb) acting on the log is equal to the weight of the water displaced by the log. The weight of the water displaced is the density of water (ρw) times the volume of water displaced (Vw).

Fb = ρw * Vw

Given that the density of water (ρw) is 1000 kg/m^3, we need to convert the volume from cm^3 to m^3. Since 1 cm^3 = 1 × 10^-6 m^3, the volume becomes:

Vw = 41.95 cm^3 * (1 × 10^-6 m^3/cm^3)
Vw = 4.195 × 10^-5 m^3

Now substitute the values into the Archimedes' principle equation:

Fb = 1000 kg/m^3 * 4.195 × 10^-5 m^3
Fb ≈ 0.04195 kg

The buoyant force (Fb) is also equal to the weight of the log (mg), where m is the mass of the log and g is the acceleration due to gravity (approximately 9.8 m/s^2). So we have:

Fb = mg

0.04195 kg = 0.037 kg * 9.8 m/s^2

Now we can solve for the mass (m) of the log:

m ≈ 0.04195 kg / 9.8 m/s^2
m ≈ 0.00428 kg

Now we have the mass of the log, so we can calculate the density of the log using the formula:

Density of the log = mass / volume

Density of the log = 0.00428 kg / 41.95 cm^3 * (1 × 10^-6 m^3/cm^3)
Density of the log = 1.02 × 10^-4 kg/m^3

Finally, let's calculate the depth submerged (hsub) of the log. The depth submerged is equal to the volume of the log submerged (Vsub) divided by the cross-sectional area of the log (A):

hsub = Vsub / A

To find the volume submerged, we need to find the volume of the log and subtract the volume of the part that is not submerged. Since the log is fully submerged, the volume submerged is equal to the volume of the log itself:

Vsub = V

The cross-sectional area of the log is given by the formula for the area of a circle:

A = πr^2

Substituting the values:

A = 3.14 * (1.15 cm)^2
A ≈ 4.153 cm^2

Finally, we can calculate the depth submerged (hsub):

hsub = V / A

hsub ≈ 41.95 cm^3 / 4.153 cm^2
hsub ≈ 10.09 cm

Therefore, the depth submerged of the log is approximately 10.09 cm.