Question

A raft is made of plastic block with a density of 650kg/m3, and its dimensions are 2.00m x 3.00m x 5.00 m. What is the rafts apparent weight in water and the buoyant force on the raft in the water? Thanks...

Answer 1

the raft would float in water as its density is less.
only part of the volume would be immersed in water, such that the bouyant force will equal the body's weight.
hence apparant weight will be 0.

the bouyant force will be = the weight of the liquid displaced = weight of the body (because its floating) = 650 * 2 * 3 * 5 * g = 19,500 * 9.8 N

Answer 2

Thanks so much. Where I am confused is the formula ( Fb = Fg (displaced fluid). Magnitude of buoyant force=weight of fluid displaced. Relative to this formula : F(net) = Fb - Fg . Where the net force is is the objects apparent weight. Since Fb = Fg ...wouldnt the answer always be zero? Subtracting something from itself will always be zero...so confused??? Please help....

Answer 3

yes, it will be zero always, as long as the body is afloat and in equilibrium.
Imagine -
a body is under water at some depth and is stationary. why would it be stationary ? because the bouyant force due to the weight of liquid displayed has balanced the body's gravitational weight.
Now imagine, that the bouyant force isn't able to balance this gravitational weight. what happens then ? the body starts sinking. so in this case, the apparant weight of the body would also be non-zero, because Mg - Fb is now greater than 0.

Now Imagine the other boundary condition. when you forcefully immerse a lighter (than water) body inside some depth in the water. the bouyant force now is much greater than the gravitational weight of the body. So what happens now ? the apparant weight is negative. As soon as you remove the restraints, the body would pop back upwards.

so, whenever a body is floating, its apparant weight will be zero. when its sinking it would be positive. when its rising, it would be negative.

Answer 4

Great reply, great explanation...Wish the textbooks would make it that easy. Appreciate the help....

Answer 5

thanks, I am happy I could help