A 8.52 -kg block of metal is suspended from a scale and immersed in water as in Figure P9.30. The dimensions of the block are 12.0 cm x 10.0 cm x 10.0 cm. The 12.0-cm dimension is vertical, and the top of the block is 9.0 cm below the surface of the water. (a) What are the forces exerted by the water on the top and bottom of the block? (Take P0 = 1.013×105 N/m2.) (top force, bottom force)
ok so waht oyu do is you find the pressure at the level the top is at which you would use P0+gh(density)
then you use PA=F
so for your problem it would be 1.013x10^5 +1000x.09x9.8=102182
102182x(.1)^2=1021.82 N for the top
for the botom you would just add .12 to your origional height
1.013x10^5 +1000x(.09+.12)x9.8=103358
103358 x .1^2=1033.58N
To find the forces exerted by the water on the top and bottom of the block, we need to consider the buoyant force acting on the block.
The buoyant force depends on the density of the fluid (water in this case), the volume of the immersed object, and the acceleration due to gravity.
First, we need to find the volume of the block:
Volume of the block = length x width x height
= 0.12 m x 0.1 m x 0.1 m (converting cm to m)
= 0.0012 m^3
Next, we need to determine the density of the block. The density (ρ_block) can be calculated using the equation:
ρ_block = (Mass of the block) / (Volume of the block)
Given that the mass of the block is 8.52 kg, we have:
ρ_block = 8.52 kg / 0.0012 m^3
ρ_block = 7100 kg/m^3
Now, let's calculate the force exerted by the water on the top and bottom of the block.
The force exerted by a fluid on a submerged object is given by the equation:
F_buoyant = ρ_fluid x V_block x g
Where:
ρ_fluid is the density of the fluid (water in this case)
V_block is the volume of the submerged object (in this case, the volume of the block below the water's surface)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
The volume of the block submerged in the water is determined by the distance from the surface to the top of the block (9.0 cm or 0.09 m).
V_block submerged = 0.12 m x 0.1 m x 0.09 m (only considering the vertical dimension)
= 0.00108 m^3
Substituting these values into the equation for the buoyant force, we get:
F_buoyant = ρ_water x V_block submerged x g
Given that the density of water (ρ_water) is approximately 1000 kg/m^3, we have:
F_buoyant = 1000 kg/m^3 x 0.00108 m^3 x 9.8 m/s^2
Calculating this, we find that the buoyant force exerted by the water on the block is approximately 10.404 N.
Now, since the block is in equilibrium (neither sinking nor rising), the buoyant force is equal to the weight of the block.
Therefore, the force exerted by the water on the top of the block is equal to the weight of the block plus the buoyant force, and the force exerted on the bottom of the block is equal to the weight of the block minus the buoyant force.
Weight of the block = mass of the block x acceleration due to gravity
= 8.52 kg x 9.8 m/s^2
approx. 83.496 N
So, the forces exerted by the water on the top and bottom of the block are:
Top force = Weight of the block + Buoyant force
= 83.496 N + 10.404 N
= 93.9 N (approximately)
Bottom force = Weight of the block - Buoyant force
= 83.496 N - 10.404 N
= 73.092 N (approximately)
Therefore, the forces exerted by the water on the top and bottom of the block are approximately 93.9 N and 73.092 N, respectively.