Bar in static equilibrium + hydrostatics

[Metaleer]

Homework Statement

A thin metal bar of length L is attached to a pivot point (point A in diagram). The bar is partially submerged in water and it can move freely around an axis that forms a 90 degree angle with the plane of the diagram. The density of the bar is 0.4 g cm^(-3). What percentage of the bar is submerged when it is in static equilibrium?

Homework Equations

Vectors are in bold.

F = ma

M = r x F

The Attempt at a Solution

Since the bar is in static equilibrium, both the net force and the net moment (torque) on the bar is 0.

Force N is the normal force that the wall exerts on the bar, R is the reaction of the pivot, W is the weight of the bar and A is due to Archimedes's principle (a body immersed in a fluid experiences a buoyant force equal to the weight of the displaced fluid.)

N = 0
R + A - W = 0

(Point A is used as the torque's point of origin)

And this is where I get stuck.
Any help would be appreciated, thanks. :)

 

[anathema]

Hello!

To solve this problem, we can use the equation for buoyancy force: Fb = ρVg, where ρ is the density of the fluid (water in this case), V is the volume of the submerged portion of the bar, and g is the acceleration due to gravity. We can also use the equation for torque: τ = r x F, where r is the distance from the pivot point to the point where the force is applied, and F is the force.

Since we know that the net force and net torque are 0, we can set up the following equations:

Fb = W (since the bar is in static equilibrium, the weight of the bar must be equal to the buoyancy force)

τ = r x W (since the net torque is 0, the torque due to the weight of the bar must be balanced by the torque due to the buoyancy force)

We can rearrange the first equation to solve for V: V = W/(ρg)

And for the second equation, we can substitute in the value for V and solve for r: r = τ/W

Since the bar is partially submerged, the volume of the submerged portion will be equal to the total volume of the bar (L) multiplied by the percentage of the bar that is submerged (x): V = Lx

Now we can combine all of these equations and solve for x:

Lx = W/(ρg)

x = W/(Lρg)

x = (mg)/(Lρg)

x = (0.4gcm^(-3) x 9.8ms^(-2) x L)/(L x 1000kgm^(-3) x 9.8ms^(-2))

x = 0.004 or 0.4%

Therefore, the percentage of the bar that is submerged when it is in static equilibrium is approximately 0.4%.

I hope this helps! Let me know if you have any further questions. Good luck with your studies!

 

[m2003]

Hello, it seems like you are on the right track with your analysis of the forces and moments on the bar. To determine the percentage of the bar that is submerged, we can use the equation for buoyancy (A) to find the weight of the displaced water, and then use that weight to find the percentage of the bar's weight that is supported by the water.

First, let's consider the forces acting on the bar. As you mentioned, there is a normal force (N) from the wall and a reaction force (R) from the pivot point. In addition, there is the weight of the bar (W) and the buoyant force (A) from the water. Since the bar is in static equilibrium, we know that the sum of these forces must be equal to 0. So we can set up the following equation:

N + R + A - W = 0

Next, we can use the equation for buoyancy to determine the weight of the displaced water (A). This equation is A = ρVg, where ρ is the density of the fluid, V is the volume of the displaced fluid, and g is the acceleration due to gravity. In this case, we know the density of the bar (0.4 g cm^(-3)), and we can find the volume of the displaced water by multiplying the cross-sectional area of the bar (which is L/2 since it is submerged halfway) by the length of the bar (L). So we have:

A = (0.4 g cm^(-3)) * (L/2) * L * g

Now we can plug this into our original equation and solve for the normal force (N):

N + R + (0.4 g cm^(-3)) * (L/2) * L * g - W = 0

N = W - R - (0.4 g cm^(-3)) * (L/2) * L * g

Since we know that the bar is in static equilibrium, we can also set up an equation for the moments acting on the bar. This will help us determine the percentage of the bar that is submerged. We can choose point A as the origin for our moments, so we have:

ΣM = 0

(R * L/2) - (W * L) = 0

R = 2W

Now we can plug this value for