a traffic light is supported by a cable that makes ang angle of 7 degree w/ the horizontal on each side of weight. what is the maximum safe weight of the light if the maximum safe tension in the cable 1200N
Draw a free body diagram. Apply a vertical force balance. If the cable is at the breaking point tension,
2*1200*sin7 = M*g
Solve for M, in kg
M = 30 kg
To find the maximum safe weight of the traffic light, we need to calculate the tension in the cable when it is at its maximum. This can be done by resolving the weight of the light into horizontal and vertical components.
We can assume that the cable is attached to the traffic light at a single point, which implies that full weight of the light acts at that point. Let's denote the weight of the traffic light as W.
Now, we can resolve W into components:
Vertical Component: W * cos(7°)
Horizontal Component: W * sin(7°)
Since the traffic light is in equilibrium, the vertical component of the weight is balanced by the upward tension in the cable, which is equal to 1200 N.
Therefore, we can write the equation:
W * cos(7°) = 1200 N
Now, let's solve for W:
W = 1200 N / cos(7°)
Using a calculator, we find:
W ≈ 1204.393 N
So, the maximum safe weight of the traffic light is approximately 1204.393 Newtons.
To find the maximum safe weight of the traffic light, we need to determine the tension in the cable. The tension in the cable can be calculated using the formula:
Tension = Weight / Cosine(angle)
Given that the maximum safe tension in the cable is 1200N and the angles on each side of the weight are 7 degrees, we can rearrange the formula to solve for the weight:
Weight = Tension * Cosine(angle)
Plugging in the values:
Weight = 1200N * Cosine(7 degrees)
Calculating this expression, we find that the weight of the traffic light should not exceed approximately 1198.95N (to a reasonable number of significant figures).
Therefore, the maximum safe weight of the traffic light is approximately 1198.95 Newtons.