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

Answer:

C 4.6 x 10-7

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

$F=\dfrac{Gm_1m_2}{r^2}\\\Rightarrow F=\dfrac{6.67\times 10^{-11}\times 95^2}{1.5^2}\\\Rightarrow F=2.67541\times 10^{-7}\ N$

The force due to two masses is given by

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

The force is C 4.6 x 10-7