Problem 5-57 Translational Equil.- Part 6 - C

A sign outside a hair stylist's shop is suspended by two wires. The force of gravity on the sign has a magnitude of \(55.7\; N.\) If the angles between the wires and the horizontal are as shown in the figure, determine the magnitude of the tensions in the two wires.
[Ans. \(T_1 = 49.9\; N;\) \(T_2 = 40.8 \;N\)]

Diagram of sign suspended by two wires.

Accumulated Solution

Diagram of vectors with all directions and angles indicated.

\(a = 0 \\ \sum F_x = 0, \; \sum F_y = 0\)

\(x - \text{components}\) \(y - \text{components}\)
\(T_{1x} = -T_1 \cos45 = -0.7071 \;T_1\) \(T_{1y} = -T_1 \sin45 = 0.7071\; T_1\)
\(T_{2x} = T_2 \cos30 = 0.8660 \;T_2\) \(T_{2y} = T_2 \sin30 = 0.5000 \;T_2\)
\(0\) \(-55.7\)
\(\text {Sum:} -0.7071 \;T_1 + 0.8660\; T_2 + 0 = 0\) \(\text{Sum}: 0.7071\; T_1 + 0.5000\; T_2 - 55.7 = 0\)

Correct!

The final two equations are:
\(-0.7071\; T_1 + 0.8660\; T_2= 0 \\ 0.7071 \;T_1 + 0.5000 \; T_2 - 55.7 = 0\)

The solution of these two equations is:

(A)   \(T_1 = 49.9 \;N;  \; T_2 = 40.8\; N\)

(B)   \(T_1 = -49.9\; N; \;  T_2 = -40.8 \; N\)

(C)   \(T_1 = 40.8 \; N;  \; T_2 = 49.9 \;N\)

(D)   \(T_1 = -40.8 \;N;  \; T2 = -49.9 \; N\)