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 = \(10 ms-2\) ]
Attached Figure / Diagram
A
16.67N
B
15.67N
C
14.67N
D
18.67N
Verified Correct Answer: A.
Physical Principles & Numerical Solution
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^\circ
From the second condition of equilibrium, sum of clockwise moments equal sum of anticlockwise moments
Taking moment at E
⇒ T sin 30^\circ x 30 = 5 x 50
⇒ ∴T=250
T = \(\frac{250}{15}\) = 16.67N
Source: WAEC / JAMB / NECO Classroom Archive (Record #68009)