Question

A metal piece of mass 160 g lies in equilibrium inside a glass of water. The piece touches the bottom of the glass at a small number of points. If the density of the metal is 8000 kg m−3, find the normal force exerted by the bottom of the glass on the metal piece.
Figure

Answer 1

The normal force exerted by the bottom of the glass on the metal piece is $1.4 \mathrm{N}$

Explanation:

Step 1:

All  Forces acting on the body are :-

(1) weight of body =mg    (acting in downward direction)

(2) normal force exerted by the bottom =R  (acting in upward direction)

(3) buoyant force = $F_{B}$ (force exerted by the liquid,acting in upward direction)

  $m g=R+F_{B}$

  $\mathrm{R}=\mathrm{mg}$$-F_{B}$

  $\mathrm{R}=\mathrm{mg}-\rho \mathrm{Vg}$

Step 2:

V= volume of liquid = $\frac{m a s s}{d e n s i t y}$

By substituting the values of given quantities in above equation we get

  $\mathrm{R}=0.16 \times 10-1000 \times \frac{0.16}{8000} \times 10$

  $R=1.4 \mathrm{N}$

Thus the normal force exerted by the bottom of glass on the metal piece is$1.4 \mathrm{N}$