Find the tension in the two cords shown in the figure above. Neglect the mass of the cords, and assume that the angle is 38° and the mass m is 220 kg
[Take g = 9.8 ms-2]
Attached Figure / Diagram
A
\(T_1\) = 2864 N, \(T_2\)= 3612 N
B
\(T_1\)= 3612 N, \(T_2\) = 2864 N
C
\(T_1\)= 3502 N, \(T_2\)= 2760 N
D
\(T_1\) = 2760 N, \(T_2\) = 3502 N
Verified Correct Answer: C.
Physical Principles & Numerical Solution
W = mg = 220 x 9.8 = 2156 N
⇒Sin 38º = \(\frac{2156}{T_1}\)
⇒ \(T_1\) = \(\frac{2156}{Sin 38}\)
⇒ \(T_1\) = 3502 N
Cos 38º = \(\frac{T_2}{T_1}\)
⇒ \(T_2\) = 3502 x Cos 38º
⇒ \(T_2\) = 2760 N
; \(T_1\) = 3502 N, \(T_2\) = 2760 N.
Source: WAEC / JAMB / NECO Classroom Archive (Record #67267)