ME2134 1 Report

DEPARTMENT OF MECHANICAL ENGINEERING NATIONAL UNIVERSITY OF SINGAPORE

Stability of Floating Body ME 2134 - 1

Gao Yifan A0091597N Group 2D1

Objectives   

To experimentally determine the center of gravity (C.G.) and metacentric height of a body floating on water. To investigate the effects of placing a weight vertically above the C.G. on the stability of a floating body. To investigate the effects of free surfaces on the stability of a floating body.

Records of Data & Calculation d l=

5.7+6.3+5.5+ 6.5 =6 cm 4

Without the raised mass and free surface,

θ=6.5°

T=

4.40 =0.88 s 5

GM =

mgd ≈ 0.2926 m ( ρg V ¿ ) tanθ

T th =2 π

BM =

(√

K Oy ≈ 1.0757 s g ∙ GM

)

I Oy l b3 = ≈ 0.3472 m V ¿ 12lb d l

1 KB= d l=3 cm 2 KG=KB+ BM −GM ≈ 8.4659 cm With the raised mass,

GM =

mgd ≈ 0.2230 m ( ρg V ¿ ) tanθ

T th =2 π

BM =

(

K Oy ≈ 1.6569 s √ g ∙ GM

)

I Oy l b3 = ≈ 0.3472 m V ¿ 12lb d l

1 KB= d l=3 cm 2 KG=KB+ BM −GM ≈ 15.4184 cm With free surfaces,

θ=7.0 °

T=

5.00 =1.00 s 5

G2 M =

GG2 =

mgd ≈ 0.2715 m ( ρg V ¿ ) tanθ

ρ f Σi =0.9887 cm ρ V¿

GM =G2 M +GG2 =0.2814 m

T th =2 π

(

K Oy ≈ 1.0984 s √ g ∙ G2 M

)

I Oy l b3 BM = = ≈ 0.3472 m V ¿ 12lb d l

1 KB= d l=3 cm 2 KG=KB+ BM −GM ≈ 9.5857 cm

Discussions 3. After putting on the load,

1 KB= ( d l + Δ d l ) 2 where

Δ dl =

mmax ρlb

KG=

KG ' × M +(h+ d)×mmax M +mmax

where

M =ρlb dl

BM =

I Oy l b3 = V ¿ 24 KB× lb

GM =KB+ BM −KG=0 Therefore,

m 1 dl + max + 2 ρlb

(

)

b

(

'

2

12 d l +

mmax ρlb

)

KG × ρlb d l+(h+ d)×mmax = ρlb d l +mmax

mmax ≈ 414.41 tonnes 4. (a). No free surface,

GM =1m

θ=tan −1

( Mgmgd∙GM ) ≈ 1.1458°

(b). 3 ρ l c b c3 ρ l c bc GG2 = = ≈ 0.1708 m ρ 12 V ¿ 12 M

G2 M =GM −G G 2 ≈ 0.8292 m

θ=tan −1

(

mgd ≈ 1.3817° Mg ∙G 2 M

)

(c).

bc 3 2l ( ) ρl b 3 ρ c 2 GG2 = = c c ≈ 0.1708 m ρ 12V ¿ 48 M G2 M=GM −G G2 ≈ 0 .8292 m

θ=tan −1

≈ 1.3817° ( Mgmgd ∙G M ) 2

(d).

lc 3 b 3 ρ 2 c ρ lc b c GG2 = = ≈ 0.0427 m ρ 12 V ¿ 12 M 2

G2 M =GM −G G 2 ≈ 0.9573 m

θ=tan −1

≈ 1.1989° ( Mgmgd ∙G M ) 2

(e).

ρ f l c b c 3 ρf l c b c3 GG2 = = ≈ 0.1667 m ρ 12V ¿ 12 M G2 M =GM −G G 2 ≈ 0.8333 m

θ=tan −1

(

mgd ≈ 1.3748° Mg ∙G 2 M

)