Problem 5-57 Translational Equil.- Part 5 - B
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. \(T_1 = 49.9\; N;\) \(T_2 = 40.8 \;N\)]
Accumulated Solution
\(a = 0 \\ \sum F_x = 0, \; \sum F_y = 0\)
| \(x - \text{components}\) | \(y - \text{components}\) |
|---|---|
| \(T_{1x} = -T_1 \cos45 = -0.7071 \;T_1\) | - |
| \(T_{2x} = T_2 \cos30 = 0.8660 \;T_2\) | - |
| \(0\) | - |
| \(\text {Sum:} -0.7071 \;T_1 + 0.8660\; T_2 + 0 = 0\) | - |
Correct!
The \(y- \text{components}\) of the forces \(T_1,\) \(T_2\) and \(55.7 \;N\) are:
(A) \(T_{1y} = T_1 \sin45; \; T_{2x}= T_2 \sin30; \; 55.7\)
(B) \(T_{1y} = -T_1 \sin45; \; T_{2x} = -T_2 \sin30; \; -55.7\)
(C) \(T_{1y} = T_1 \sin45; \; T_{2x} = T_2 \sin30; \; -55.7\)
(D) \(T_{1y}= -T_1 \sin45; \; T_{2x} = -T_2 \sin30; \; 55.7\)