A block of mass 0.5 kg is suspended at the 40 cm mark of a light metre rule AB that is pivoted at point E, the 90 cm mark, and is kept at equilibrium by a string attached at point D, the 60 cm mark, as shown in the figure above. Find the tension T in the string.
W = mg = 0.5 x 10 = 5 N
⇒ T sin 30° x 30 = 5 x 50
T = 25015\frac{250}{15}15250 = 16.67N
