Question

Three 5kg masses are kept at the vertices of equilateral triangle each side of 0.25 m.Fimd the resultant gravitational force on any one mass

Answer 1

In continuation to the attachment

$\sqrt{3} \times 6.67 \times {10}^{ - 11} \times 5 \times 5 \div 0.25 \times 0.25$
= (11.55×25×10^-11)/0.0625

=11.55×400×10^-11

=4620×10^-11
=
$4.6 \times {10}^{ - 8} newton$