Questions: Un cylindre en aluminium, de hauteur h = 20 cm et de rayon r = 10 cm, repose par sa base sur une table horizontale. La masse volumique de l'aluminium est ρ = 2700 kg / m³. a- Calculez le poids du cylindre. Prendre g = 10 N / kg. b- Que vaut la force exercée par la table sur le cylindre? c- En déduire l'intensité de la force exercée par le cylindre sur la table. d- Calculez la pression exercée par le cylindre sur la table. e- On introduit entre le cylindre et la table une pièce de monnaie, de masse négligeable et de 2 cm de diamètre. Calculez la nouvelle pression exercée par le cylindre sur la table.

△ Calculate the weight of the aluminum cylinder. ○ Calculate the mass of the cylinder ☼ The volume of the cylinder is \( V = \pi r^2 h = \pi \times (0.1 \, \text{m})^2 \times 0.2 \, \text{m} = 0.002\pi \, \text{m}^3 \approx 0.00628 \, \text{m}^3 \). The mass is \( m = \rho V = 2700 \, \text{kg/m}^3 \times 0.00628 \, \text{m}^3 = 16.96 \, \text{kg} \). ○ Calculate the weight ☼ The weight is \( P = mg = 16.96 \, \text{kg} \times 10 \, \text{N/kg} = 169.6 \, \text{N} \). ✧ The weight of the cylinder is 169.6 N.

△ Determine the force exerted by the table on the cylinder. ○ Apply Newton's third law ☼ The force exerted by the table on the cylinder is equal in magnitude and opposite in direction to the weight of the cylinder, which is 169.6 N upward. ✧ The force exerted by the table on the cylinder is 169.6 N upward.

△ Determine the intensity of the force exerted by the cylinder on the table. ○ Apply Newton's third law ☼ The force exerted by the cylinder on the table is equal in magnitude and opposite in direction to the force exerted by the table on the cylinder, which is 169.6 N downward. ✧ The force exerted by the cylinder on the table is 169.6 N downward.

△ Calculate the pressure exerted by the cylinder on the table. ○ Calculate the area of the base of the cylinder ☼ The area is \( A = \pi r^2 = \pi \times (0.1 \, \text{m})^2 = 0.01\pi \, \text{m}^2 \approx 0.0314 \, \text{m}^2 \). ○ Calculate the pressure ☼ The pressure is \( P = \frac{F}{A} = \frac{169.6 \, \text{N}}{0.0314 \, \text{m}^2} = 5401.3 \, \text{Pa} \approx 5.4 \, \text{kPa} \). ✧ The pressure exerted by the cylinder on the table is 5.4 kPa.

△ Calculate the new pressure when a coin is placed between the cylinder and the table. ○ Calculate the area of the coin ☼ The radius of the coin is 0.01 m, so the area is \( A_{\text{coin}} = \pi r^2 = \pi \times (0.01 \, \text{m})^2 = 0.0001\pi \, \text{m}^2 \approx 0.000314 \, \text{m}^2 \). ○ Calculate the new pressure ☼ The new pressure is \( P_{\text{new}} = \frac{169.6 \, \text{N}}{0.000314 \, \text{m}^2} = 540127.4 \, \text{Pa} \approx 540.1 \, \text{kPa} \). ✧ The new pressure with the coin between the cylinder and the table is 540.1 kPa. ☺ a) Weight of the cylinder = 169.6 N b) Force exerted by the table on the cylinder = 169.6 N upward c) Force exerted by the cylinder on the table = 169.6 N downward d) Pressure exerted by the cylinder on the table = 5.4 kPa e) New pressure with the coin between cylinder and table = 540.1 kPa