Archimedes Principle: B = δVg + T

AI Thread Summary
When a metal object is submerged in water, it experiences two upward forces: buoyancy (B = δVg) and the tension in the string (T). The downward force acting on the object is its weight, which is the gravitational force. The buoyant force arises from the displaced water, counteracting the weight of the object. Clarification is needed regarding the concept of "taking out an imaginary area," as it may confuse the understanding of forces acting on submerged objects. Understanding buoyancy is crucial for grasping the balance of forces in fluid mechanics.
Kork
Messages
33
Reaction score
0
Hi :)

When I have a container with water and put a piece of metal down in the water while it's hanging on the string, I know that there will be two forces pulling it upwords:

B = δVg
and the string force T

But the force pulling it down, is that the gravitation? Or is it a force that equals to the weight of the piece of metal?
 
Physics news on Phys.org
Kork said:
But the force pulling it down, is that the gravitation? Or is it a force that equals to the weight of the piece of metal?
Yes and yes. Gravity pulls the metal downward. (That gravitational force is the weight of the object.)
 
I don't think I quite understand the buoyancy part.

If I tak a cointainer with water and take out an imaginaray area then the forces on the area would be the buoyancy and the gravitationa pulling down? Right?
 
Thank you very much!
 
Kork said:
I don't think I quite understand the buoyancy part.

If I tak a cointainer with water and take out an imaginaray area then the forces on the area would be the buoyancy and the gravitationa pulling down? Right?
Not sure what you mean by 'take out an imaginary area'. The forces on the submerged object will include the buoyant force and the weight of the object. (Maybe you can restate the question.)
 
Thread ''splain this hydrostatic paradox in tiny words'
This is (ostensibly) not a trick shot or video*. The scale was balanced before any blue water was added. 550mL of blue water was added to the left side. only 60mL of water needed to be added to the right side to re-balance the scale. Apparently, the scale will balance when the height of the two columns is equal. The left side of the scale only feels the weight of the column above the lower "tail" of the funnel (i.e. 60mL). So where does the weight of the remaining (550-60=) 490mL go...
Consider an extremely long and perfectly calibrated scale. A car with a mass of 1000 kg is placed on it, and the scale registers this weight accurately. Now, suppose the car begins to move, reaching very high speeds. Neglecting air resistance and rolling friction, if the car attains, for example, a velocity of 500 km/h, will the scale still indicate a weight corresponding to 1000 kg, or will the measured value decrease as a result of the motion? In a second scenario, imagine a person with a...
Scalar and vector potentials in Coulomb gauge Assume Coulomb gauge so that $$\nabla \cdot \mathbf{A}=0.\tag{1}$$ The scalar potential ##\phi## is described by Poisson's equation $$\nabla^2 \phi = -\frac{\rho}{\varepsilon_0}\tag{2}$$ which has the instantaneous general solution given by $$\phi(\mathbf{r},t)=\frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}',t)}{|\mathbf{r}-\mathbf{r}'|}d^3r'.\tag{3}$$ In Coulomb gauge the vector potential ##\mathbf{A}## is given by...
Back
Top