a uniform metre rule pivoted at R, the 70cm mark. Two forces 0.1 N and 0.4 N are applied at Q, the 60cm mark and S, the 85cm mark. If the metre rule us kept in equilibrium, by the forces, calculate the weight of the metre rule.

Answer 1

the center of mass of the rule is at 50 cm

using the 70 cm pivot ... (20 * w) + (10 * 0.1) = 15 * 0.4

solve for w

Answer 2

Please explain what led to that formular..... and how did you get 20...... What I know is moment =force * distance

Answer 3

the rule is in equilibrium , so the moments are balanced (equal and opposite)
around the pivot point

the 20 is the distance from the pivot to the center of mass of the rule

Answer 4

I don't get it well