A traffic light of mass m (= 14 kg) hangs above the street by two wires, both of

Question

A traffic light of mass m (= 14 kg) hangs above the street by two wires, both of which make an angle (theta = 75*) with the vertical.

a.) Show that the tension in both ropes is the same. (Hint: use the horizontal direction.) Now, find an expression for T (the tension in each wire) in terms of the variables m, and/or g. Then use it to answer all of the following parts:

b.) How much is the tension in each wire?__N

c.) Find the tension for = (theta) = 0*.

Is this greater than or less than your answer from part b? Why? And what is the significance of your numerical answer at zero degrees? Explain.

d.) Checking another limit: Explain why it is impossible to hold up the street light with the wires being perfectly horizontal.

Explanation / Answer

(a)
Let the Tension in left wire, = Tl
Let the Tension in right wire, = Tr

Forces In Horizontal Direction will balance each other,
Tl*sin(75) = Tr*sin(75)
Tl = Tr
Hence Tension in both ropes is the same.

(b)
T* cos(75) + T* cos(75) = 14 * 9.8
T = 265.1 N

(c)
For Theta = 0

T* cos(0) + T* cos(0) = 14 * 9.8
T = 68.6 N

It is smaller than the answer in Part B,
Theta = 0 , Means That the Mass is beeen hanged by two Vertical wires.

(d)
As in this case, angle wrt Vertical will be 90. degrees.

So,
T* cos(90) + T* cos(90) = 14 * 9.8
Now, Cos(90) = 0
Therefore there won't be any Vertical component of force to balance the weight and it would be impossible to hold.

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