You are asked to hang a uniform beam and sign using a cable that has a breaking strength of 372 N. The store owner desires that it hang out over the sidewalk as shown. The sign has a weight of 248 N and the beam's weight is 49 N. The beam's length is 1.30 m and the sign's dimensions are 1.00 m horizontally 0.80 m vertically. What is the minimum angle θ that you can have between the beam and cable?
To find the minimum angle θ that you can have between the beam and cable, we need to consider the forces acting on the beam and sign. We want to find the angle at which the tension in the cable is equal to the breaking strength.
First, let's analyze the forces acting on the beam and sign:
1. Weight of the beam: This force acts vertically downward and has a magnitude of 49 N. We can denote it as W_beam.
2. Weight of the sign: This force acts vertically downward and has a magnitude of 248 N. We can denote it as W_sign.
3. Tension in the cable: This force acts at an angle θ and provides the upward force to support both the beam and the sign. We can denote it as T.
Now, considering the equilibrium of forces in the vertical direction, we can write the following equation:
T + W_beam + W_sign = 0
Since the tension in the cable provides an upward force, it will be positive, while the weights of the beam and sign act downward and will be negative. Rearranging the equation, we get:
T = -W_beam - W_sign
Substituting the values:
T = -49 N - 248 N
T = -297 N
Since the tension T cannot exceed the cable's breaking strength of 372 N, we have the condition:
T ≤ 372 N
Substituting the value of T:
-297 N ≤ 372 N
To find the minimum angle θ, we can use the trigonometric relationship between the tension T and the weight of the sign W_sign:
T = W_sign / sin(θ)
Rearranging the equation, we get:
sin(θ) = W_sign / T
Substituting the given values:
sin(θ) = 248 N / -297 N
Taking the inverse sine on both sides, we can find the minimum angle θ:
θ = sin^(-1)(248 N / -297 N)
Evaluating this expression, we find the minimum angle θ.