Question

remarks its very easy to make sign errors in this kind of problem. one way to avoid them is to always measure the angle of a vector from the positive x - direction. the trigonometric functions of the angle will then automatically give the correct signs for the components. for example, $vec{t}_1$ makes an angle of $180^{circ}-37^{circ}=143^{circ}$ with respect to the positive x - axis, and its x - component, $t_1cos143^{circ}$, is negative, as it should be. question which of these would increase if a second traffic light were attached to the first? assume the cables do not change their lengths. (select all that apply.) the angle of the cable with tension $t_2$ the tension $t_2$ the angle of the cable with tension $t_1$ the tension $t_1$ practice it use the worked example above to help you solve this problem. a traffic light weighing $1.18\times10^{2}$ n hangs from a vertical cable tied to two other cables that are fastened to a support, as shown in figure (a). the upper cables make angles of $\theta_1 = 36.0^{circ}$ and $\theta_2 = 54.0^{circ}$ with the horizontal. find the tension in each of the three cables. $t_1=$ n $t_2=$ n $t_3=$ n exercise hints: getting started | im stuck! use the values from practice it to help you work this exercise. suppose the traffic light is hung so that the tensions $t_1$ and $t_2$ are both equal to 76 n. find the new angles they make with respect to the x - axis. (by symmetry, these angles will be the same.)

Explanation:

Step1: Analyze vertical - force equilibrium

For the traffic - light system, in the vertical direction, the sum of the vertical components of the tensions in the upper cables and the tension in the vertical cable must balance the weight of the traffic light. Let the weight of the traffic light be $W$, the tension in the vertical cable be $T_3$, and the tensions in the upper cables be $T_1$ and $T_2$ with angles $\theta_1$ and $\theta_2$ with the horizontal. The vertical - force equilibrium equation is $T_1\sin\theta_1 + T_2\sin\theta_2=T_3 = W$.

Step2: Consider the effect of adding a second traffic light

When a second traffic light is added, the weight $W$ increases. Since the lengths of the cables do not change, the angles of the cables do not change (because the geometry of the attachment points is fixed). According to the vertical - force equilibrium equation, if $W$ increases and $\theta_1$ and $\theta_2$ are constant, then $T_1$ and $T_2$ must increase to balance the greater weight.

Step3: Solve for tensions in Practice It

In the Practice It problem, for vertical equilibrium $T_1\sin\theta_1+T_2\sin\theta_2 = W$ and for horizontal equilibrium $T_1\cos\theta_1=T_2\cos\theta_2$. Given $W = 1.18\times10^{2}\ N$, $\theta_1 = 36.0^{\circ}$, and $\theta_2 = 54.0^{\circ}$. From $T_1\cos\theta_1=T_2\cos\theta_2$, we have $T_2=T_1\frac{\cos\theta_1}{\cos\theta_2}=T_1\frac{\cos36.0^{\circ}}{\cos54.0^{\circ}}\approx T_1\times1.376$. Substitute into the vertical - equilibrium equation: $T_1\sin36.0^{\circ}+T_1\times1.376\times\sin54.0^{\circ}=118$. $T_1(0.588 + 1.376\times0.809)=118$. $T_1(0.588+1.113)=118$. $T_1=\frac{118}{1.701}\approx69.4\ N$. Then $T_2 = 1.376\times69.4\approx95.5\ N$. And $T_3 = 118\ N$.

Step4: Solve for angles in Exercise

In the Exercise, since $T_1 = T_2 = 76\ N$ and $T_3 = W$ (from vertical equilibrium). Let the angle with the $x$ - axis be $\theta$. The vertical - equilibrium equation is $2T_1\sin\theta=W$. First, from the Practice It problem, $W = 118\ N$. Then $\sin\theta=\frac{W}{2T_1}=\frac{118}{2\times76}=\frac{118}{152}\approx0.776$. So $\theta=\sin^{- 1}(0.776)\approx50.9^{\circ}$.

Answer:

• For the first question: the tension $T_1$, the tension $T_2$
• For the Practice It problem: $T_1\approx69.4\ N$, $T_2\approx95.5\ N$, $T_3 = 118\ N$
• For the Exercise problem: $\theta\approx50.9^{\circ}$