Mechanical Engineering - Hydraulics and Fluid Mechanics - Discussion

Buoyancy force = weight of the water displaced by the floating body as depth of immersion is 0.6.

So,
volume of water displace by floating body is = 3*2*0.6 = 3.6 m^3.
Now,
As mass of water displaced = Density of water * Volume of water,
= 1000 kg/m^3 * 3.6 m^3,
= 3600 kg,

Weight of water displaced = Mass of water * gravitation force,
= 3600 kg * 9.81 m/s^2.
= 35316 N = 35.3 KN.

Volume = 3x2x1 = 6m^3.

Volume of water displaced = Weight of body in kN/specific weight of water in kN/m^3.

V = W/9.81 ----------- (1).

We also know that,

Depth of immersion = Volume/Cross sectional area.

0.6 = V/(3x2) -------- (2).

Comparing (1) & (2) we find.

W = 35.3 kN.

Volume = 3*2*1.

Volume of water displaced = Weight of body in kn/Specific weight of water in kn/m3.

V = w/9.81.

Depth of immersion = Volume/Cross-sectional area.

0.6 = v/3*2.

Hence from the two equations we have our term as W = 35.3 kn.

Depth of immersion =0.6.
= sp. Grvty x depth of body.
Depth of body = 1m.
Therefore sp. gravty=0.6 hence density of body, ρ.
=600kg/m^3.
Wt of body =rho V g= 600x 3x2x1x9.81 = 35316N.

At equilibrium, weight of body (W) = Force of buoyancy (Fb).

Fb = Specific gravity * Volume of immersion.

Fb = 9.81*1000*0.6*2*3.

Fb = 35.3 KN.

Hence weight of body = 35.3 KN.

Weight of body = weight of water displaced = buoyancy force.
= Density x volume x g.
= 1000 x 3 x 2 x 0.6 x 9.81.
= 35316 N = 35.316 KN.

Press= density * sp.gra * depth.

Press = 1 * 9.81 * 0.6,
Press = 5.88 KN/m2.

Wt of body = press* area.
Wt. Of body= 5.88 * 6.

Ans= 35.5 Kn.

Given,
Length 3M.
Whidth 2M.

Depth 1M and depth of immersion 0.6.

3x2x0.6 = 3.6M^3.

Weight of water 9.81x3.6 = 35.3KN.

Pressure = density * gravity * height.
9.81*1000*0.6 =5886N/M2,
Then force =pressure *area,
5886 * 3 * 2 = 35.3 KN.

Let's take the body as the submerged body then the formula is given by;

W*A*X.
9810*6*0.6.
35314N = 35.3 KN.