Answer:
Explanation:
volume of barge = 1/3 A H
=1/3 x 550 x 2
= 366.67 m³
When empty the bottom of the barge is located H0 = 0.45 m below the surface of the water ,
volume above water = A /12 x ( H - H₀ )³
= 170 m³
volume below water = 366.67 -170 = 96.66 m³
Buoynant force = weight of displaced water
= 96.66x 10³x 9.8
=947.26 x 10³ N
= weight of empty barge
mass of empty barge = 96.66 x 10³kg
When fully loaded with coal the bottom of the barge is located H1 = 1.8 m below the surface
volume above water = A /12 x ( H - H₁ )³
volume below water = (A/3) H - A /12 x ( H - H₁ )³
BUOYANT FORCE = Weight of water displaced
= ρg[(A/3) H - A /12 x ( H - H₁ )³ ]
Substituting the values
Buoyant force = [366.67 - 550 / 12 x ( 2-1.8 )³]x 10³ x 9.8
= 3589.77 x 10³ N
WEIGHT OF COAL = 3589.77 x 10³ - 947.26 x 10³
= 2642.51 x 10³N
MASS OF THE COAL = 2642.51 x 10³ / 9.8
= 269.64 x 10³ kg
