Question

The ice cube descends down a steep plane that makes an angle of 14 ° with the horizontal direction. The density of ice is 917 kg / m³, while the length of the edge of the cube is 69 cm. Find the pressure with which the cube acts on the surface of a steep plane. When calculating, ignore the friction between the ice cube and the contact surface of the steep plane. (solution- p = 6022.7 Pa)

Answer 1

Let's first find the mass of the cube:

$$m=\rho V=917\ \dfrac{kg}{m^3}\cdot(0.69\ m)^3=301.24\ kg.$$

Let's find the force that acts on the contact surface of the steep plane:

$$F=mgcos\theta=301.24\ kg\cdot9.81\ \dfrac{m}{s^2}\cdot cos14^{\circ}=2867.4\ N.$$

Finally, we can find the pressure with which the cube acts on the surface of a steep plane:

$$P=\dfrac{F}{A}=\dfrac{2867.4\ N}{(0.69\ m)^2}=6022.7\ Pa.$$