Question

2 cubes (side 1 met) one of relative density 0.6, the other 1.15 are connected by weightless wire and placed in a water tank. In equilibrium, the height of part of lighter cube above water is?

Answer 1

Answer on Question #39465, Physics, Mechanics | Kinematics | Dynamics

Question:

2 cubes (side 1 meter) one of relative density 0.6, the other 1.15 are connected by weightless wire and placed in a water tank. In equilibrium, the height of part of lighter cube above water is?

Answer:

Newton's first law of motion:

$$m _ {1} g + m _ {2} g = F _ {b 1} + F _ {b 2},$$

where $F_{b1} + F_{b2}$ is total buoyant force

$$m = \rho V = \rho a ^ {3},$$

buoyant force equals (assuming Archimedes' principle):

$$F _ {1 b} = \rho_ {w} g V = 1 \cdot g a ^ {2} (a - x),$$

where $x$ is the height of part of lighter cube above water

$$F _ {1 b} = \rho_ {w} g V = 1 \cdot g a ^ {3},$$

therefore

$$\rho_ {1} a ^ {3} g + \rho_ {2} a ^ {3} g = g a ^ {2} (a - x) + g a ^ {3}$$ $$x = (2 - \rho_ {1} - \rho_ {2}) a = (2 - 0. 6 - 1. 1 5) \cdot 1 = 0. 2 5 m$$

Answer: 0.25 m