A TRAFFIC LIGHT OF MASS M = 46.1 KG IS SUPPORTED BY TWOCABLES. ASSUME THAT THE C

Question

A TRAFFIC LIGHT OF MASS M = 46.1 KG IS SUPPORTED BY TWOCABLES. ASSUME THAT THE CABLE TO THE LEFT MAKES AN ANGLETHETA = 34.1 DEGREES ABOVE THE - X AXIS AND THAT THE CABLE TO THERIGHT LIES ALONG THE + X AXIS. CALCULATE THE TENSION IN EACH OF THE SUPPORTING CABLES T1 = T2 = A TRAFFIC LIGHT OF MASS M = 46.1 KG IS SUPPORTED BY TWOCABLES. ASSUME THAT THE CABLE TO THE LEFT MAKES AN ANGLETHETA = 34.1 DEGREES ABOVE THE - X AXIS AND THAT THE CABLE TO THERIGHT LIES ALONG THE + X AXIS. CALCULATE THE TENSION IN EACH OF THE SUPPORTING CABLES T1 = T2 =

Explanation / Answer

(a)    from the given problem we can see that thehorizontal component of the rresultant force on the light by thecables is given    by    Rx = Fx         = (T)cos34.1o - (T) cos34.1o         =0    in the y-direction we get    Ry = Fy         = (T)sin34.1o + (T) sin34.1o         = (46.1kg) (9.80 m / s2)    solve for the tension in the cables    so the resultant force is Ry invertically upward direction (b)    as the traffic light is in equilibrium theresultant of these forces must be zero so the weight is nothing butRy         =0    in the y-direction we get    Ry = Fy         = (T)sin34.1o + (T) sin34.1o         = (46.1kg) (9.80 m / s2)    solve for the tension in the cables    so the resultant force is Ry invertically upward direction (b)    as the traffic light is in equilibrium theresultant of these forces must be zero so the weight is nothing butRy         = (46.1kg) (9.80 m / s2)    solve for the tension in the cables    so the resultant force is Ry invertically upward direction (b)    as the traffic light is in equilibrium theresultant of these forces must be zero so the weight is nothing butRy

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