Calculating Buoyant Force: Brain Blocks Pt II

[falcon0311]

I have a box when suspended from a cable in a vacuum is 4450N and is .608m wide (it's also a cube).

W = 4450N = T1 (tension on the cable)
V = 0.2248m^3

When I suspend it into a liquid, L/2 m from the surface, of density 944 kg/m^3, I want to know the buoyant force, the new tension of the cable, the force pushing down from the top and the force pushing up from the bottom from the liquid the box is suspended in.

L/2 = 0.608m / 2 = 0.304m

I have the buoyant force as being T1 - T2 = Fb = D( V )g (of the liquid)

944kg/m^3 * 0.2248m^3 * 9.8m/s^2 = 2080N = Fb

T2 = T1 - Fb = 4450N - 2080N = 2370N = T2

But how do I find the forces from the liquid above and below the box?

Thanks for any help!

 

[needhelpperson]

I have a box when suspended from a cable in a vacuum is 4450N and is .608m wide (it's also a cube).

W = 4450N = T1 (tension on the cable)
V = 0.2248m^3

When I suspend it into a liquid, L/2 m from the surface, of density 944 kg/m^3, I want to know the buoyant force, the new tension of the cable, the force pushing down from the top and the force pushing up from the bottom from the liquid the box is suspended in.

L/2 = 0.608m / 2 = 0.304m

I have the buoyant force as being T1 - T2 = Fb = D( V )g (of the liquid)

944kg/m^3 * 0.2248m^3 * 9.8m/s^2 = 2080N = Fb

T2 = T1 - Fb = 4450N - 2080N = 2370N = T2

But how do I find the forces from the liquid above and below the box?

Thanks for any help!

D*g*h = the pressure acting on the top or bottom of the box at a given height "h"

i assume L/2 means half the box's length.

you can go D*g*(L/2) = pressure on top

you also know that P = F/A

solve for F = P * L^2

That should be the force on top. Do the same for the bottom, except use (3L/2) as the height and those should be the answers.

 

[M98Ranger]

The force from the liquid above the box can be calculated using the formula F1 = P1 * A, where P1 is the pressure at the surface of the liquid and A is the area of the box. We can find the pressure using the formula P = ρgh, where ρ is the density of the liquid, g is the acceleration due to gravity, and h is the depth of the box in the liquid. In this case, h = L/2 = 0.304m. So, P1 = 944kg/m^3 * 9.8m/s^2 * 0.304m = 2819.52 N/m^2. Therefore, F1 = 2819.52 N/m^2 * 0.608m^2 = 1715.10 N.

Similarly, the force from the liquid below the box can be calculated using the formula F2 = P2 * A, where P2 is the pressure at the bottom of the box. In this case, h = 0 since the box is resting on the bottom of the liquid. So, P2 = 944kg/m^3 * 9.8m/s^2 * 0m = 0 N/m^2. Therefore, F2 = 0 N/m^2 * 0.608m^2 = 0 N.

In summary, the force pushing down from the top is 1715.10 N and the force pushing up from the bottom is 0 N. These two forces combined with the buoyant force of 2080N will result in a net upward force of 2080N on the box, which is equal to the weight of the displaced liquid.