You hold a 1.0-kg wooden block under water with your hand (𝜌wood = 600 kg/m3). Calculate the forces acting on the block if the block is in static equilibrium. How do you calculate acceleration when the block is released?
let g = 9.81 m/s^2
rho water = 1000 kg/m^2 = 10^3 kg/m^3
then
for Buoyancy:
volume of block = 1 kg / 600 kg/m^3 = 1.67 * 10^-3 m^3
Archimedes says buoyant force up on block = rho water * g * 1.67*10^-3
call it B = 10^3 * 9.81 * 1.67 * 10^-3 = 16.4 Newtons up
for weight
W = m g = 1 * 9.81 = 9.81 Newtons down
that is a net force of 16.4 - 9.8 = 6.6 Newtons up
Therefore to prevent acceleration I must push down with 6.6 Newtons
If I let go a = F/m = 6.6/ 1 = 6.6 m/s^2
That is what your book says. It is not true really. The block accelerates water around it when it accelerates. The water pushes back in other words. Hydrodynamicists call the effect "added mass" for short. Therefore the block mass seems greater than its mass in a vacuum or low density gas like air.
Vb = 1m^3/600kg * 1kg =
1/600 = 1.67*10^-3 m^3. = Vol. of block = Vol. of water displaced.
a. 1.67*10^-3m^3 * 1*10^3kg/m^3 = 1.67 kg = Mass of the water.
F = M*g = 1.67 * 9.8 = 16.4 N. = upward force of the water. = downward
force of the hand.
b. F = M*a = 16.4.
1 * a = 16.4,
a = 16.4 m/s^2.
To calculate the forces acting on the wooden block when it is in static equilibrium underwater, we need to consider the buoyant force and the force due to gravity.
1. The buoyant force can be calculated using Archimedes' principle, which states that the buoyant force acting on an object submerged in a fluid is equal to the weight of the displaced fluid. The buoyant force (Fbuoy) can be calculated as:
Fbuoy = 𝜌fluid * V * g
Where:
- 𝜌fluid is the density of the fluid in which the block is immersed (in this case, water, 𝜌water ≈ 1000 kg/m3).
- V is the volume of the wooden block submerged in water.
- g is the acceleration due to gravity (approximated as 9.8 m/s2).
2. The force due to gravity is simply the weight of the block, given by:
Fgravity = m * g
Where:
- m is the mass of the block (1.0 kg in this case).
- g is the acceleration due to gravity.
Since the block is in static equilibrium, the forces must be balanced:
Fbuoy = Fgravity
Now let's calculate the volume of the block submerged in water. Since the density of wood (𝜌wood) is given as 600 kg/m3, the volume (V) can be calculated using the formula:
V = m / 𝜌wood
Where:
- m is the mass of the block.
- 𝜌wood is the density of the wood.
Plugging in the values:
V = 1.0 kg / 600 kg/m3 ≈ 0.00167 m3
Now we can calculate the forces acting on the block:
Fbuoy = 𝜌water * V * g = 1000 kg/m3 * 0.00167 m3 * 9.8 m/s2 ≈ 16.33 N
Fgravity = m * g = 1.0 kg * 9.8 m/s2 ≈ 9.8 N
Therefore, the buoyant force acting on the block is approximately 16.33 N, and the force due to gravity is approximately 9.8 N.
To calculate the acceleration when the block is released, we need to consider Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. In this case, the net force acting on the block will only be the buoyant force since there is no external force.
Therefore, the equation for acceleration (a) can be written as:
Fbuoy = m * a
Solving for acceleration:
a = Fbuoy / m
Plugging in the values:
a = 16.33 N / 1.0 kg ≈ 16.33 m/s2
Therefore, when the block is released, it will accelerate upwards with an approximate acceleration of 16.33 m/s2.