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 = \(10 ms-2\) ]
-
16.67N
- 15.67N
- 14.67N
- 18.67N
The correct answer is: A
Explanation
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
â T sin 30° x 30 = 5 x 50
â âīT=250
T = \(\frac{250}{15}\) = 16.67N
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