- Example 4 5 A Traffic Light At Rest Goal Use The Second Law In An Equilibrium Problem Requiring Two Free Body Diagrams 1 (61.07 KiB) Viewed 50 times
EXAMPLE 4.5 A Traffic Light at Rest Goal Use the second law in an equilibrium problem requiring two free-body diagrams. 87.00 37.0°NA 59.0° Problem A traffic light weighing 1.00 x 102 N hangs from a vertical cable tied to two other cables that are fastened to a support, as in figure (a). The upper cables make angles of 37.0° and 53.0° with the horizontal. Find the tension in each of the three cables. Strategy There are three unknowns, so we (a) A traffic light suspended by cables. (b) A free-body need to generate three equations relating diagram for the traffic light. (c) A free-body diagram for the knot joining the cables. them, which can then be solved. One equation can be obtained by applying Newton's second law to the traffic light, which has forces in the y-direction only. Two more equations can be obtained by applying the second law to the knot joining the cables-one equation from the x-component and one equation from the y-component.
PRACTICE IT Use the worked example above to help you solve this problem. A traffic light weighing 1.01 X 102 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 @ = 35.5º and 2 = 54.5° with the horizontal. Find the tension in each of the three cables. T = Z T2 = 1 Z Tz = 101 Z EXERCISE HINTS: GETTING STARTED I I'M STUCK! Use the values from PRACTICE IT to help you work this exercise. Suppose the traffic light is hung so that the tensions T, and T, are both equal to 72 N. Find the new angles they make with respect to the x-axis. (By symmetry, these angles will be the same.) Need Help? Talk to a Tutor
SOLUTION Find T, from figure (b), using the condition of equilibrium. EF, = 0 - Tz - F, = 0 T3 = f g = 1.00 x 102 N Using figure (c), resolve all three tension forces into components and construct a table for convenience. Force X-Componenty-Component T, -Tcos 37.0° Ti sin 37.00 T, T, cos 53.00 Tz sin 53.0° 0 -1.00 x 102 N Apply the conditions for equilibrium to the knot, using the components in the table. (1) EF, = -T, cos 37.0° + T, cos 53.0º = 0 (2) EF, = T, sin 37.0° + T, sin 53.0° -1.00 x 102 N = There are two equations and two remaining unknowns. Solve Equation (1) for Tz. T2 = 11 ( COS 53.00 ) = T: (.602) = 1.33T, Substitute the result for T2 into Equation (2) T, sin 37.0° +(1.337.) (sin 53.0°) - 1.00 x 102 N = 0 T = 60.1 N T2 = 1.33T, = 1.33(60.1 N) = 79.9 N