Question

A $5.0kg$ rock whose density is $4800kg/m^3$ is suspended by a string such that half of the rock’s volume is under water. What is the tension in the string?

Short answer

The tension within the string if a half of the rock's volume is under water, $T=44N$

Step-by-step solution

Buoyant force.

The buoyancy is caused by the pressure produced by the fluid during which the article is submerged. Furthermore, because the pressure of the fluid increases with depth, the buoyant force experienced by the item is consistently upwards.

Derive the formula of tension.

We have written Newton's first law for our scenario by projecting the force on the vertical axis, picking the positive direction upwards, and projecting the force on the vertical axis.

$$\begin{gathered} T+F_B-mg=0 \\ \Rightarrow T=mg-F_B \end{gathered}$$

The buoyant force formula is thought, as is the proven fact that the degree of water displaced by the rock is half its own volume.

Please also keep detaining mind that the degree of rock is proportional to its mass and density.

${\rho }_r=\frac{m}{V}\Rightarrow V=\frac{m}{{\rho }_r}$

We are now able to write

$$\begin{gathered} T=mg-{\rho }_{w}g\frac{V}{2} \\ =g\left(m-\frac{{\rho }_w}{2}V\right) \\ =g\left(m-\frac{{\rho }_w}{2}\frac{m}{{\rho }_r}\right) \\ =mg\left(1-\frac{{\rho }_w}{2{\rho }_r}\right) \end{gathered}$$

Find the T by substituting values.

We get by replacing our known values.
$$\begin{gathered} T=5\times 9.8\left(1-\frac{1000}{2·4800}\right) \\ =44\mathrm{N} \end{gathered}$$