A street lamp weighs 200 N. It is supported by two wires that form an angle of 130° with each other. The tensions in the wires are equal.What is the tension in each wire supporting the street lamp?If the angle between the wires supporting the street lamp is reduced to 100°, what is the tension in each wire?
Problem 1: If the tension in the two wires are equal, the angle they make with the horizontal must also be equal. Always remember that a horizontal line is a straight angle; hence, it has a measurement of 180°. Let's look for the angle the wires make with the horizontal. 130 + x + x = 180 130 + 2x = 180 2x = 180 - 130 2x = 50 (Divide both sides by 2) x = 25° Each wire makes a 25° angle with the horizontal. We can now find the tension. The mass must be divided by 2. Using trigonometry, let's find the tension force. The divided weight of the lamp shall be the opposite side, and the tension force shall be the hypotenuse. sin 25° = (200 / 2) / Ftens Ftens = 100 / sin 25° Ftens = 236.62 N (Answer) The tension is 236.62 N in each wire. Problem 2: We use the same approach. 100 + x + x = 180 100 + 2x = 180 2x = 180 - 100 2x = 80 (Divide both sides by 2) x = 40° Each wire makes a 40° angle with the horizontal. Calculate the tension force. sin 40° = (200 / 2) / Ftens Ftens = 100 / sin 40° Ftens = 155.57 N (Answer) The tension is 155.57 N in each wire. Hope this helps!
