Pamela Meyer

## Answered question

2022-01-05

A 950-kg cylindrical can buoy floats vertically in sea-water. The diameter of the buoy is 0.900 m. Calculate the additional distance the buoy will sink when an 80.0-kg man stands on top of it.

### Answer & Explanation

Raymond Foley

Beginner2022-01-06Added 39 answers

Given data:
- The mass of the buoy is: $M=950kg$.
- The diameter of the buoy is: $D=0.900m$.
- The radius of the buoy is: $r=\frac{D}{2}=0.450m$.
- The mass of the man is: $m=80.0kg$.
- The density of seawater is: $\rho =1.03×{10}^{3}k\frac{g}{{m}^{3}}$.
Required data
We are asked to determine the additional distance the buoy will sink when the man stands on top of it.
Solution
Since the buoy is still floating after the man stands on it and according to Archimedes principle from Equation (1), the additional buoyant force exerted on the buoy, equal to weight of the additional fluid displaced by the man, is also equal to the additional weight of the man:
$B={W}_{man}$
Substitute for B from Equation (1) and for W from Equation (2):
$\rho g{V}_{disp}=mg$
$\rho {V}_{disp}=m$
where ${V}_{disp}$ is the additional volume displaced when the man stands on the buoy. It is equal to AD where $A=\pi {r}^{2}$ is the area of the buoy and d is the distance the buoy sink:
$\rho \left(\pi {r}^{2}\right)d=m$
Solve for d:
$d=\frac{m}{\pi {r}^{2}\rho }$
Substitute numerical values:
$d=\frac{80.0kg}{\pi {\left(0.450m\right)}^{2}\left(1.03×{10}^{3}k\frac{g}{{m}^{3}}\right)}$
$=0.122m$

Melissa Moore

Beginner2022-01-07Added 32 answers

Answer:
Additional depth is 0.13 m
Explanation:
mass of cylinder, $M=950kg$
Let the initial height inside water is h.
diameter $=0.9m$
radius, $r=0.45m$
mass of man, $m=80kg$
density of water $=1000k\frac{g}{{m}^{3}}$
Let the additional distance is y.
For initial stage:
Buoyant force = weight
$\text{Volume immersed}×\text{density of water}×g=Mg$
$3.14×0.45×0.45×1000×h=950$
$h=1.49m$
Now:
$3.14×0.45×0.45×1000×\left(h+y\right)=950+80$
$950+635.85y=950+80$
$y=0.13m$

karton

Expert2022-01-10Added 613 answers

Step 1
Weight of an object, W=mg equation 1
where m=80kg,

g=9.8 m/s${}^{2}$
then ${W}_{m}=\left(80kg\right)\left(9.8m/{s}^{2}\right)$
=784N
Cross sectional area of the cylindrical can is $a=\frac{\pi {d}^{2}}{4}$
where, a is cross-sectional area of the buoy
d is diameter
then, $a=\frac{\pi ×\left(0.9m{\right)}^{2}}{4}$
$=0.636{m}^{2}$
Step 2
Volume of the cylindrical can buoy which sin, V=ah
then, $V=\left(0.636{m}^{2}\right)h$
We know that, Weight of the water displaced will be equal to the buoyant force then, ${B}_{SW}={\rho }_{water}{V}_{g}$ equation 2
here ${B}_{SW}$ is Buoyant force
${\rho }_{water}$ is the density of sea water.
put all known values in equation 2
${B}_{w}=\left(1030kg/{m}^{3}\right)\left(\left(0.636{m}^{2}\right)×h\right)\left(9.8m{s}^{-2}\right)$

As of, water is replaced when man stands on top buoy,
${W}_{m}={B}_{sw}$ equation 3
put all known values in equation 3

h=0.122m
Answer: Additional distance through which the buoy sink is 0.122m.

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