Question

Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude of the tension in the string labeled C is 53.3 N. Calculate the magnitude of the tension in the string marked B

Answer 1

Let the Tension in the String A, B, C = Ta,Tb, Tc

Euating Horizontal Forces -
Tc * cos(theta) = Ta * cos(alpha) + Tb * cos(beta)
53.3 * 7/sqrt(7^2+1^2) = Ta * 7/sqrt(5^2 + 7^2) + Tb * 2/sqrt(5^2+2^2)
Ta * 7/sqrt(5^2 + 7^2) + Tb * 2/sqrt(5^2+2^2) = 52.76 ------------1

Equating Vertical Forces -

Tc * sin(theta) + Ta * cos(alpha) + Tb * cos(beta)
53.3 * 1/sqrt(7^2+1^2) + Ta * 5/sqrt(5^2 + 7^2) = Tb * 5/sqrt(5^2+2^2)
Tb * 5/sqrt(5^2+2^2) - Ta * 5/sqrt(5^2 + 7^2)  = 7.54 ------------2

Using eq 1 * 2
Ta = 47.5 N
Tb = 37.9 N

Tension in String B, Tb = 37.9 N