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 s...

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 $\mathrm{<br>}$ ]

Answer Details

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 = 25015 $\frac{\mathrm{<br>}}{}$ = 16.67N