A weight of 20 KN supported by two cords, one 3 m long and the other 4m long with points of support 5 m apart find the tensions T1 and T2 in the cords

Answers (1)
S safeer

Equating forces along x axis

\\T1sina=T1sinb\\\\T1\frac{4}{5}=T2\frac{3}{5}\\\\T2=\frac{4}{3}T1.....(1)\\\\

Equating forces along y axis

\\mg=T1cosa+T2cosb\\\\20KN=T1\frac{3}{5}+T2\frac{4}{5}......(2)

Substituting the value of T2 from equation 1 to equation 2 we get

\\20KN=T1\frac{3}{5}+T1\frac{16}{15}\\\\20KN=T1\frac{5}{3}\\\\T_1=12N\\T2=16N

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