Asked by Eduardo
A traffic light is suspended midway across a 60 meter wide street by a cable that is attached to vertical poles on either side of the street. When the traffic light is hung in the middle the cable sags 30 cm. If the light is 1500 kg what should be the minimum tension rating for the cable so they don't break?
wouldn't
.03 / cos(89.94) = Tension?
what am i doing wrong?
wouldn't
.03 / cos(89.94) = Tension?
what am i doing wrong?
Answers
Answered by
bobpursley
draw the figure. now divide it in two, right an left.
The left side tension is t, holding half the weight 1/2 *1500*g
let the angle from the horizontal to the wire be theta (the angle of depression at the poles).
SinTheta=750g/tension
but cosTheta=.3/30
solve for theta from second equation, then solve for tension in the first.
Justlooking at your numbers, I am not certain what .03 represents.
The left side tension is t, holding half the weight 1/2 *1500*g
let the angle from the horizontal to the wire be theta (the angle of depression at the poles).
SinTheta=750g/tension
but cosTheta=.3/30
solve for theta from second equation, then solve for tension in the first.
Justlooking at your numbers, I am not certain what .03 represents.
Answered by
Eduardo
my mistake. the weight is 150 kg not 1500, and .3 m = 30 cm
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