The magnitude of the tension in the string marked A is 39.5 N.
What is the tension in A?
The tension in A is determined thus:
The angle at A, θ = tan⁻¹(3/8) = 20.56
When extrapolated below negative x, the angle at B, α = tan⁻¹(5/4) = 51.34
When extrapolated below negative x, the angle at C, β = tan⁻¹(1/6) = 9.46
Taking the horizontal components of tension;
56.3cos(9.46) = A * cos(20.56) + B * cos(51.34)
0.6247B= 55.53 - 0.936A
B = (55.53 - 0.936A)/0.6247 ----(1)
Taking the vertical components of tension;
56.3 * sin(9.46) + A * sin(20.6) = B * sin(51.3)
9.25 + 0.35A = 0.78B ---- (2)
substitute the value (1) in (2)
9.25 + 0.35A = 0.78{(55.53 - 0.936A)/0.6247}
(9.25 + 0.35A) * 0.6247 = 43.31 - 0.73A
0.22A + 0.73A = 43.31 - 5.78
0.93A = 37.53
A = 39.5 N
In conclusion, the tension in A is determined by solving for the vertical and horizontal components of tension.
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