- #1
falcon0311
- 29
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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!
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!