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anonymous
 one year ago
I need help with this problem. I think I know how to do it, but I am having difficulty drawing a diagram, which is required in my class: A street lamp weighs 150 N. It is supported by two wires that form an angle of 120.0° with each other. The tension in the wires are equal.
a) what is the tension in each wire supporting the street lamp?
b) If the angle between the wires supporting the street lamp is reduced to 90.0° , what is the tension in each wire?
anonymous
 one year ago
I need help with this problem. I think I know how to do it, but I am having difficulty drawing a diagram, which is required in my class: A street lamp weighs 150 N. It is supported by two wires that form an angle of 120.0° with each other. The tension in the wires are equal. a) what is the tension in each wire supporting the street lamp? b) If the angle between the wires supporting the street lamp is reduced to 90.0° , what is the tension in each wire?

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zephyr141
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444271470359:dw

zephyr141
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444271953399:dw

zephyr141
 one year ago
Best ResponseYou've already chosen the best response.0since the tension in the wires are equal we just need to find the tension T in one wire. lets do the right side. now we have to work in quadrant 1 but we have an angle that goes into two quadrants. just have to have a little know how of angles to figure that out. since the T is the same then the angles must be the same. any greater or smaller degree in theta or phi will cause the tension to increase or decrease. so we can assume that theta and phi are equal. that means there is an equal amount of the 120 degree angle in both quadrants so we can divide 120 degree by two to find how much that is. dw:1444272318399:dwdw:1444272437725:dw

zephyr141
 one year ago
Best ResponseYou've already chosen the best response.0now make a free body diagram.dw:1444272839930:dw

zephyr141
 one year ago
Best ResponseYou've already chosen the best response.0let's look at the forces in the x direction and we get:\[x:Tcos30Tcos30=0\]\[x:0=0\]and now lets look in the y direction\[y:Tsin30150=0\]\[y:Tsin30=150\]\[y:T=\frac{150}{\sin30}\]there you have it.
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