Question

5. The ruler is in equilibrium. The hanging load on the left arm of the visor, at point A1, has mass m = 500 g. What force does the force gauge hook apply to point A2 of the ruler? It is known that d1 = 10 cm, while d2 = 1/2 × d1

Answer 1

$$F_1d_1=F_2d_2,$$ $$m_1gd_1=F_2\dfrac{1}{2}d_1,$$ $$F_2=2m_1g=2\cdot0.5\ kg\cdot9.8\ \dfrac{m}{s^2}=9.8\ N.$$