Mechanical Engineering - Hydraulics and Fluid Mechanics - Discussion
Discussion Forum : Hydraulics and Fluid Mechanics - Section 4 (Q.No. 37)
37.
A uniform body 3 m long, 2 m wide and 1 m deep floats in water. If the depth of immersion is 0.6 m, then the weight of the body is
Discussion:
23 comments Page 3 of 3.
Rahul Saikia said:
10 years ago
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.
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.
Ramdeen said:
1 decade ago
@Fakhre Alam check your calculations.
Fakhre Alam said:
1 decade ago
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 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.
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