A sign with a mass of 23 kg is hanging in two wires both making an angle of 27 degrees with the vertical. Find the tension in one of the wires. Give your answer in newtons (N) and with 3 significant figures.
1. draw a free-body diagram.
2. Let T be the tension in each of the wires, which makes an angle of 27° with the vertical.
3. Sum forces on the sign in the y-direction (vertical) to get
2Tcos(27°)=23kg * 9.81
Solve for T.
To find the tension in one of the wires, we can use the concept of equilibrium. In this case, the sign is not moving vertically, so the sum of the vertical forces acting on it must be zero.
We can consider the vertical forces acting on the sign: the weight (mg) acting downwards, and the tensions (T) acting upwards in both wires. Since the sign is not moving vertically, the sum of the vertical forces is zero.
The weight (mg) can be calculated as the product of the mass (m) and the acceleration due to gravity (g). Assuming g to be approximately 9.81 m/s^2, the weight would be 23 kg * 9.81 m/s^2 = 225.63 N.
Now, let's consider the forces in each wire. Since both wires make an angle of 27 degrees with the vertical, the vertical components of the tensions in both wires will add up to balance the weight of the sign.
Using trigonometry, we can find the vertical component of the tension in one wire by multiplying the tension (T) by the sine of the angle (27 degrees). Therefore, the vertical component of the tension in one wire is T * sin(27).
Since the vertical components of the tension in both wires add up to balance the weight, we can write an equation: T * sin(27) + T * sin(27) = 225.63 N.
Simplifying this equation, we get: 2 * T * sin(27) = 225.63 N.
Now, we can solve for T by dividing both sides by 2 * sin(27): T = 225.63 N / (2 * sin(27)).
Calculating this expression, we get T ≈ 219.33 N.
Therefore, the tension in one of the wires is approximately 219.33 N, rounded to three significant figures.