Problem 5-57 Translational Equil.- Part 4 - 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. \(T_1 = 49.9\; N;\) \(T_2 = 40.8 \;N\)]

Diagram of sign suspended by two wires.

Accumulated Solution

Diagram of vectors with all directions and angles indicated.

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

Correct!
In fact both (B) and (C) are correct

(B) \(\sum F = 0\)

(C) \(\sum F_x = 0,\)   \(\sum F_y = 0\)

It is (C) which is most useful and will be used here (and in most cases).

Let us look at the \(x\)-component first.

The \(x\)-components of the forces \(T_1,\) \(T_2\) and \(55.7\;N\) are:

(A)   \(T_{1x} = T_1 \cos45; \;   T_{2x} = T_2 \cos30;   \;0\)

(B)   \(T_{1x} = -T_1 \cos45; \;   T_{2x} = T_2 \cos30;   \; 0\)

(C)   \(T_{1x} = -T_1 \cos45;   \; T_{2x} = -T_2 \cos30;   \; 0\)

(D)   \(T_{1x} = T_1 \cos45;   \; T_{2x} = -T_2 \cos30; \; 0\)