In the above figure, the object is lowered into the water, the following observations are made.
i) The object experiences a reduction in weight. The object of the weight in water is less than its weight in air. The apparent loss in weight of the obejct is caused by the buoyant force of the surrounding water
on the object.
Apparent loss in weight of object
=Weight of object - weight of object in water.
ii) The object displaces a vlume of water.
Volume of water displaced
= volume of the submerged part of the stone
iii) From the figure, the apparent loss in weight is due to the buoyant force.
Therefore :
Bouyant Force = Actual weight - weight in water
= (Say) 70N - 40 N
= 30 N
Courtesy of:i) The object experiences a reduction in weight. The object of the weight in water is less than its weight in air. The apparent loss in weight of the obejct is caused by the buoyant force of the surrounding water
on the object.
Apparent loss in weight of object
=Weight of object - weight of object in water.
ii) The object displaces a vlume of water.
Volume of water displaced
= volume of the submerged part of the stone
iii) From the figure, the apparent loss in weight is due to the buoyant force.
Therefore :
Bouyant Force = Actual weight - weight in water
= (Say) 70N - 40 N
= 30 N
http://physics.weber.edu/carroll/Archimedes/principle.htm
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