A block of mass M=0.525kg is floating stationary in a beaker of water. A string connected to the bottom of the beaker holds the block in place. The tension in the string is 7.10 N. The bottom of the block is floating a distance h=1.66cm above the bottom of the beaker.
What is the magnitude and direction of the buoyant force acting on the block? What is the density of the block?
To find the magnitude of the buoyant force acting on the block, we can use Archimedes' principle, which states that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.
1. Determine the volume of the water displaced by the block.
- The volume of the water displaced is equal to the volume of the submerged portion of the block.
- The height of the submerged portion is given as h = 1.66 cm = 0.0166 m.
- Since the block is floating stationary, the volume of the submerged portion is equal to the volume of the block.
- The formula for the volume of a block is V = Length x Width x Height.
2. Calculate the volume of the block.
- The dimensions of the block are not given. If you have the information about the dimensions, substitute the values into the formula.
- Let's assume the dimensions of the block are L, W, and H.
- V_block = L x W x H.
3. Determine the density of water.
- Assume the density of water is ρ_water = 1000 kg/m³.
4. Find the weight of the water displaced.
- The weight W_water = ρ_water x g x V_water, where g is the acceleration due to gravity.
- Substitute the values in the equation.
5. Determine the magnitude of the buoyant force.
- The magnitude of the buoyant force F_buoyant = W_water, as stated in Archimedes' principle.
6. Find the direction of the buoyant force.
- The direction of the buoyant force is always opposite to the gravitational force acting on the object. It acts upwards in this case.
7. Calculate the density of the block.
- The density of the block ρ_block = M_block / V_block, where M_block is the mass of the block.
Step 1:
V_water = V_block = Length x Width x Height (assuming the block is fully submerged)
Step 2:
V_block = L x W x H
Step 3:
ρ_water = 1000 kg/m³
Step 4:
W_water = ρ_water x g x V_water
Step 5:
F_buoyant = W_water
Step 6:
The buoyant force acts upward.
Step 7:
ρ_block = M_block / V_block
Let's assume the dimensions of the block are provided. Please provide the values of Length, Width, and Height of the block.
To find the magnitude and direction of the buoyant force acting on the block, we can use Archimedes' principle. According to Archimedes' principle, the buoyant force on an object immersed in a fluid is equal to the weight of the fluid displaced by the object.
The weight of the block is equal to its mass multiplied by the acceleration due to gravity (W = mg). Therefore, the weight of the block is W = (0.525 kg)(9.8 m/s^2) = 5.1375 N.
Since the block is in equilibrium and floating stationary, the buoyant force is equal in magnitude and opposite in direction to the tension in the string. Therefore, the magnitude of the buoyant force is 7.10 N and it acts in the upward direction.
To find the density of the block, we can use the concept of specific gravity. The specific gravity of a substance is defined as the ratio of its density to the density of a reference substance, usually water (density of water = 1000 kg/m^3).
The specific gravity (S.G.) is given by S.G. = density of block / density of water.
Let's assume the density of the block is represented by the symbol ρ. We know that the buoyant force acting on the block is given by the formula F_b = ρ * V * g, where V is the volume of the block and g is the acceleration due to gravity.
We can rearrange the formula as follows:
ρ = F_b / (V * g)
The volume of the block can be calculated by dividing the weight of the block by the product of the acceleration due to gravity and the density of water, wie here
V = W / (g * density of water)
Substituting the known values, we have:
V = 5.1375 N / (9.8 m/s^2 * 1000 kg/m^3) = 0.000523 m^3
Now, substituting the values of F_b, V, and g into the formula for density, we get:
ρ = 7.10 N / (0.000523 m^3 * 9.8 m/s^2) ≈ 1346 kg/m^3
Therefore, the density of the block is approximately 1346 kg/m^3.
a) Fb = mg + T
b) Can't tell. Is it submerged? How deep is the water?