Question

A 4.1kg rock whose density is 4600kg/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?(In N)

Answer 1

In the given scenario, the weight of the block acts downwards, buoyant force acts upwards and the tension also acts upwards. Since the block is in equilibrium, the net force must be zero. This is possible when the downward forces are balanced by the upward forces. Thus, from equilibrium condition, \(F_{B}+T=w_{\text {block }}\) \(w_{w} g+T=m g\) \(\rho_{w} V_{w} g+T=m g\) Calculate the volume of the block as follows: \(V_{b}=\frac{m}{\rho_{b}}=\frac{4.1 \mathrm{~kg}}{4600 \mathrm{~kg} / \mathrm{m}^{3}}=8.913 \times 10^{-4} \mathrm{~m}^{3}\) GIven that half of the block's volume of underwater. This means the volume of water displaced is. \(V_{w}=\frac{V_{b}}{2}=\frac{8.913 \times 10^{-4} \mathrm{~m}^{3}}{2}=4.4565 \times 10^{-4} \mathrm{~m}^{3}\) Now, substitute the values in \(\rho_{w} V_{w} g+T=m g\) and solve for \(T\). \(\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(4.4565 \times 10^{-4} \mathrm{~m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)+T=(4.1 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\) \(T=35.85 \mathrm{~N}\)