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 = \(10 ms-2\) ]
[Take g = \(10 ms-2\) ]
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option 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
⇒ T sin 30° x 30 = 5 x 50
⇒ ∴T=250
T = \(\frac{250}{15}\) = 16.67N
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 = \(\frac{250}{15}\) = 16.67N