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 56.3 N. Calculate the magnitude of the tension in the string marked A.

Answer 1

The tension in the string marked A is determined as 331.35 N.

Angle made by A with respect to vertical

tanθ = Δy/Δx

tanθ = (6 - 1)/(7 - 1) = 0.8333

θ = arc tan(0.8333) = 39.81⁰

with respect to horizontal = 90 - 39.81 = 50.19⁰

Angle made by B with respect to horizontal

tanθ = Δy/Δx

tanθ = (9 - 6)/(7 - 5) = 1.5

θ = arc tan(1.5) = 56.31 ⁰

Angle made by C with respect to horizontal

tanθ = Δy/Δx

tanθ = (6 - 4)/(14 - 7) = 0.2857

θ = arc tan(0.2857) = 15.95 ⁰

Bcosθ + Aosθ = Ccosθ

Bcos(56.31) + A[cos(50.19)] = 56.3cos(15.95)

0.55B + 0.64A = 54.13 ----- (1)

Bsinθ + Asinθ = Csinθ

Bsin(56.31) + A[sin(50.19)] = 56.3sin(15.95)

0.832B + 0.77A = 15.47---- (2)

Solve (1) and (2)

0.2A = 66.27

A = 66.27/0.2

A = 331.35 N

Thus, the tension in the string marked A is determined as 331.35 N.

Learn more about tension here: brainly.com/question/918617

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