A sign with a mass of (24) kg is hanging in two wires both making an angle of (52) degrees with the vertical. Find the tension in one of the wires. Give your answer in newtons (N) and with 3 significant figures.
t cos 52 = (24/2)9.81 = 12*9.81
To find the tension in one of the wires, we can use the concept of equilibrium in a system of forces. Since the sign is hanging and not moving, the net force acting on it must be zero.
First, let's consider the vertical forces acting on the sign. There are two vertical forces: the weight of the sign (mg) and the tension in one of the wires (T). The weight of the sign can be calculated using the formula:
Weight = mass × acceleration due to gravity
Weight = 24 kg × 9.8 m/s^2 (acceleration due to gravity) ≈ 235.2 N
Now, let's consider the horizontal forces acting on the sign. Since the sign is in equilibrium, there is no horizontal acceleration, and thus no horizontal force is acting on it.
Next, we can resolve the tension force into its vertical and horizontal components. Since the angle between the wire and the vertical is given as 52 degrees, the vertical component of the tension force can be calculated as:
Vertical component of tension (Tv) = T × sin(θ)
where θ is the angle between the wire and the vertical, which is 52 degrees in this case.
Finally, since the net vertical force acting on the sign is zero, the vertical component of the tension force must be equal to the weight of the sign:
Tv = Weight
T × sin(52) = 235.2 N
To find the tension in one of the wires (T), we can rearrange the equation:
T = 235.2 N / sin(52)
Now, let's calculate the value of T using a calculator:
T ≈ 297.385 N
Therefore, the tension in one of the wires is approximately 297.385 N, rounded to three significant figures.