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?

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.

my mistake. the weight is 150 kg not 1500, and .3 m = 30 cm

You are on the right track, but there seems to be a small error in your calculation. Let's see how we can arrive at the correct answer.

To find the minimum tension rating for the cable, we can start by considering the forces acting on the traffic light.

When the traffic light is in equilibrium, the tension force in the cable supports the weight of the light. Additionally, the weight of the light can be split into two components: one vertical component pulling downward and another horizontal component pulling towards the center of the street.

To calculate the tension in the cable, we can use trigonometry. We know that the sag in the cable is 30 cm, which is the vertical distance from the lowest point of the cable to a straight line between the two poles.

Let's define the angle between the cable and the horizontal line as θ. Since the cable sags symmetrically, the angle between the cable and the vertical line is 90 degrees minus θ.

To begin, we need to find the angle θ. We can use the given information that the cable sags 30 cm in the middle of a 60-meter wide street. Since the cable forms a right-angled triangle with the horizontal line and a vertical line, we can use trigonometry to find θ.

Using the half-width of the street as the adjacent side and half of the sag as the opposite side, we can use the tangent function:

tan(θ) = (30 cm) / (30 m)

Remember to convert the units from centimeters to meters in order to maintain consistent units throughout the calculation.

Now that we have found the value of θ, we can proceed to find the tension in the cable.

Let's call the tension T. The vertical component of the weight is T x sin(90 - θ), and since the cable is in equilibrium, this vertical weight is equal to the tension in the cable:

T x sin(90 - θ) = T

Now, let's consider the horizontal component of the weight. This component pulls towards the center of the street, so it needs to be balanced by a force pulling in the opposite direction. This force is the horizontal component of the tension, which is T x cos(90 - θ).

Since the light is in equilibrium, the horizontal component of the tension must balance the horizontal component of the weight:

T x cos(90 - θ) = (1500 kg) x g

Here, g represents the acceleration due to gravity, which is approximately 9.8 m/s².

Now we have two equations:

T = T x sin(90 - θ)
T x cos(90 - θ) = (1500 kg) x g

To find the minimum tension rating for the cable, we need to solve for T.

Combining the two equations:

T = (1500 kg) x g / cos(90 - θ)

Now, you can substitute the calculated value of θ into this equation, and solve for T:

T = (1500 kg) x g / cos(90 - θ)

By plugging in the known values and performing the calculation, you will find the minimum tension rating for the cable needed to support the traffic light without breaking.