A traffic light weighing 115 N hangs from a cable tied to two other cables faste

Question

A traffic light weighing 115 N hangs from a cable tied to two other cables fastened to a support as in figure (a). The upper cables make angles of thefts = 38.0degree and theta_2 = 52.0degree with the horizontal. These upper cables are not as strong as the vertical cable and will break if the tension in them exceeds 110 N. Does the traffic light remain hanging in this situation, or will one of the cables break? (a) A traffic light suspended by cables. (b) The free-body diagram for the traffic light. (c) The free-body diagram for the knot where the three cables are joined. Conceptualize Insect the drawing in figure (a). Let us assume the cables do not break and that nothing is moving. Categorize If nothing is moving, no part of the system is accelerating. We can now model the light as a particle in equilibrium on which the net force is zero. Similarly, the net force on the knot in figure (c) is zero. Analyze We construct two free-body diagrams: one for the traffic light, shown in figure (b), and one for the knot that holds the three cables together, shown in figure (c). This knot is a convenient object to choose because all the forces of interest act along lines passing through the knot. Apply the equation for the traffic light in the gamma direction: Choose the coordinate axes as shown in figure (c) and resolve the forces acting on the knot into their components: Apply the particle in equilibrium model to the knot: Sigma F_Y = 0 rightarrow T_3 = F_g = 0 T_3 = Fg = 115 N Sigma F_X = - T_1 cos 38.0degree + T_2 cos 52.0degree = 0 Sigma F_Y = T_1 sin 38.0degree + T_2 sin 52.0degree + (-115 N) = Equation (1) shows that the horizontal components of T_1 and T_2 must be equal in magnitude, and Equation (2) shows that the sum of the vertical components of T_1 and T_2 must balance the downward force T_3, which is equal in magnitude to the weight of the light. Solve Equation (1) for T_2 in terms of T_1: Substitute this value for T_2 into Equation (2): T_2 = T_1(cos 38.0degree/cos 52.0degree) = 1.28 T_1 T_1 sin 38.0degree + (1.28 T_1)(sin 52.0degree) - 115 N = 0 T_1 = 70.8 N T_2 = 1.28T_1 = 90.6 N Both values are less than 110 N, so the cables will not break. Suppose the traffic light is hung so that the tensions T_1 and T_2 are both equal to 86.1 N. Find the new angles they make with respect to the x axis. (By symmetry, these angles will be the same.) theta =

Explanation / Answer

For finding new angle:

T1 = T2 = 86.1 N

T1 sin(theta) + T2 sin(theta) = weight of the light

2 T sin(theta) = 115 N

sin(theta) = 115/2 x 86.1 = 0.6678 =>theta = sin -1 (0.6678)

theta = 41.89 deg

Hence, theta = 41.89 deg

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