If you push any floating object down from equilibrium and release it, it bobs up and down. That looks like an oscillation, so let's analyze the restoring force driving the floating object back to equilibrium. Suppose the cube is distance y below its equilibrium position. Find an expression for the y-component of the net force exerted on it. Simplify this expression as much as possible. Does it look like a linear restoring force?

Question

contestada

Respuesta :

moya1316 moya1316 · Answer 1 · 2020-04-18T09:46:32+02:00

Answer:

[tex]F_{y}[/tex] = ( ρ_fluid g A) y

Explanation:

This exercise can be solved in two parts, the first finding the equilibrium force and the second finding the oscillating force

for the first part, let's write Newton's equilibrium equation

B₀ - W = 0

B₀ = W

ρ_fluid g V_fluid = W

the volume of the fluid is the area of the cube times the height it is submerged

V_fluid = A y

For the second part, the body introduces a quantity and below this equilibrium point, the equation is

B - W = m a

ρ_fluid g A (y₀ + y) - W = m a

ρ_fluid g A y + (ρ_fluid g A y₀ -W) = m a

ρ_fluid g A y + (B₀-W) = ma

the part in parentheses is zero since it is the force when it is in equilibrium

ρ_fluid g A y = m a

this equation the net force is

[tex]F_{y}[/tex] = ( ρ_fluid g A) y

we can see that this force varies linearly the distance and measured from the equilibrium position