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. [Take g = 10ms−2 ]

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.

[Take g = 10ms−2 ]
Correct Option: Answer is
W = mg = 0.5 x 10 = 5 N

Since it’s light, neglect the weight of the metre rule.

The effective tension T acting in the vertical direction = T sin 30°

From the second condition of equilibrium, sum of clockwise moments equal sum of anticlockwise moments

Taking moment at E
⇒ T sin 30° x 30 = 5 x 50

⇒ ∴T=250
T = 250/15 = 16.67N

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