Mechanical Engineering - Hydraulics and Fluid Mechanics - Discussion

P = 1000*10*.6
= 6000 N/m^2

F = 6000*6
= 36000 N
= 36 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.

@Fakhre Alam check your calculations.

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.

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.

Your method is right @Rahul.

You are correct and you explained with the simple concept. Thanks @Vishwas.

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.

You are right @Vishwas.

Presur (p)=density*gravity*height.
Force/area=1000*9.81*0.6,
Force=1000*9.81*0.6*6,
F=35.316 KN.