Task 3
Wooden cube with sides length 0.5 m has a meeting relati 0.6 float in water. Calculate the cube is submerged in water.
Settlement.
Used MKS unit system
Suppose W: weight cube
FB: buoyant force
D: depth of the cube is submerged in water.
Relative Meeting:
y object
S = ------------- => object y = S water = 0.6 x 1000 = 600.0 kg f / m3
y water
Body weight: W = y objects V = 600x (0.5) = 75.0 N
Buoyant force: FB = y Vair water is on the move objects
= 1000x (0,5 x0, 5xd) = 250.0 d
In the float, brnda weight is equal to the buoyant force.
W = FB - 75.0 = 250.0 d - d = 0.3 m
So the cube is submerged in water depth is d = 0.3 m
Problem 4
Block of wood with a length of 1.0 m, width 0.4 m and 0.3 m height in horiziontal floats in water with vertical tingginy side. Meeting S = 0.7 relative timber. calculate the volume of water in pindahkaan and location in the center of buoyancy.
Settlement.
In use the system of SI units
Volume beam: V = 1.0 x 0.4 x0, 3 = 0.12 m
Brtat beam: W = pbalok g V g V = S pair
= 0.7 x1000x9, 81x824, 04 N
The volume of water on the move objects
Weight Beams
V = -----------------------
Water Density
w w 824.04
V = ----------- = ------------- = -------------------- = 0.084 m3
y water p water x g 1000 x 9.81
The depth of the beam is immersed in water.
Volume of water on the move
d = ------------------------------------------------ -
Looks tan g beam on the surface of the water
v 0.084
d = ------ = ------------- = 0.21 m
A 1.0 x 0.4
The location of the center of buoyancy
d 0.21
OB = ------- = -------- = 0.105 m
2 2
So the location of the center of buoyancy is 0.105 m from the bottom of the beams.
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