Question

Three particles of mass 95 kg form an equilateral triangle of side 1.5 m. What is the magnitude of the net gravitational force on one of the particles?A 2.7 x 10-7 N B 4.0 x 10-9 N C 4.6 x 10-7 ND 930 N E 3.8 x 10-7 N

Answer 1

C 4.6 x 10-7

Step-by-step explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

The masses are equal = 95 kg

Distance between the masses is 1.5 m

The gravitational force is given by

`F=(Gm_1m_2)/(r^2)\\\Rightarrow F=(6.67* 10^(-11)* 95^2)/(1.5^2)\\\Rightarrow F=2.67541* 10^(-7)\ N`

The force due to two masses is given by

`F=√(F_1^2+F_2^2+F_1F_2cos\theta)\\\Rightarrow F=\sqrt{(2.67541* 10^(-7))^2+(2.67541* 10^(-7))^2+2.67541* 10^(-7)* 2.67541* 10^(-7)cos60}\\\Rightarrow F=4.63395* 10^(-7)\ N`

The force is C 4.6 x 10-7