Suitcase question?

Question

A traffic light of mass m (= 16 kg) hangs above the street by two wires, both of which make an angle (= 40 o) with the vertical.

a.) How much is the tension in each wire?

b.) Find the tension for = 0o.

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

Answer 1

Hi Dorothy--

The vertical forces must add up to zero (because the stoplight is not moving)...things at rest stay at rest unless....well, you know.

Weight to be supported = mg = 160 N (I use 10 m's^2 for g)

This is to be supported by the vertical component (cos 40) of the wires.

F in one wire = 80/cos 40 = 104 N

b) at zero, each wire holds 1/2 the weight so F in one wire = 80N

As the angle gets larger, the vertical component is F(inwire)*cos(angle) meaning that the force in the wire is 80/cos(angle).  Eventually, the wire will snap if you try to make angle too big.

At zero degrees, the wires just split the work of supporting the weight.

-Fred

Answer 2

sum of vertical forces =0

a) weight = mg = 16*9.81 = 156.96

there are two wires here, so to calculate tension in each wire

T= (156.96/2)/cos 40 = 102.45

b) at 0 each wire holds half of the total weight

So at zero degree the weight distribution is just split equally, i.e 156.96/2 = 78.48N

Just out of curiosity, is the above guy Fred Leobinger of University of Manchester...Edit your answer Fred to let me know!