Question

A cubical block of wood of edge $3cm$ floats in water. The lower surface of the cube just touches the free end of a vertical spring fixed at the bottom of the pot. Find the maximum weight that can be put on the block without wetting it. Density of wood$=800kg{m}^{-3}$ and spring constant of the spring$=50N{m}^{-1}$. Take $g=10m{s}^{-2}$.

Answer 1

The specific gravity of the block = 0.8.

Hence the height inside water = 3 cm $\times$ 0.8 = 2.4 cm.

The height outside after = 3 cm - 2.4 = 0.6 cm.

Suppose the maximum weight that can be put without wetting it is W.

The block in this case is completely immersed in the water.

The volume of the displaced water

= volume of the block = $27\times {10}^{-6}{m}^3$ .

Hence, the force of buoyancy

= $(27\times {10}^{-6}{m}^3)$ $\times 1(1000$ $kg/m^3)$ $\times (10$ $m/s^2)$ = 0.27 N.

The spring is compressed by 0.6 cm and hence the

upward force exerted by the spring

$=50\frac{N}{m}\times$ 0.6 cm = 0.3 N.

The force of buoyancy and the spring force taken together

balance the weight of the block plus the weight W put on the block.

The weight of the block is

$W^{\prime }=(27\times {10}^{-6}m)\times (800$ $kg/m^3)\times (10$ $m/s^2)$ = 0.22 N.

Thus, W = 0.27 N + 0.3 N - 0.22 N

= 0.35 N.