Strategy The system of interest is the traffic light, and its free-body diagram is shown in part (c) of the figure. The three forces involved are not parallel, and so they must be projected onto a coordinate system. The most convenient coordinate system has one axis vertical and one horizontal, and the vector projections on it are shown in part (d). There are two unknowns in this problem (T) and T2), so two equations are needed to find them. These two equations come from applying Newton's second law along the vertical and horizontal axes, noting that the net external force is zero along each axis because acceleration is zero. Solution First consider the horizontal or x-axis: Thus, as you might expect, Fhetx = Tax +Tia=0. This gives us the following relationship: Ti cos 30* = T2 cos 45". Thus, T2 = 1.225T1. Note that Th and T2 are not equal in this case because the angles on either side are not equal. It is reasonable that T2 ends up being greater than Ti because it is exerted more vertically than T . Now consider the force components along the vertical or y-axis: Fnet y = Tiy + T2y - w = 0. This implies Tiy + T2y = W. Substituting the expressions for the vertical components gives T1 sin 30" + T2sin 45" = w. There are two unknowns in this equation, but substituting the expression for T2 in terms of TI reduces this to one equation with one unknown: T1 (0.500) + (1.225?1) (0.707) = w= mg, which yields 1.366T1=(15.0kg) (9.80m/s2). Solving this last equation gives the magnitude of Th to be T1 =108 N. Finally, we find the magnitude of T2 by using the relationship between them, T2 = 1.22571, found above. Thus we obtain T2 = 132 N. Significance Both tensions would be larger if both wires were more horizontal, and they will be equal if and only if the angles on either side are the same (as they were in the earlier example of a tightrope walker).