Question

A 0.500 -kg physics textbook is hanging from two massless wires of equal length attached to a ceiling. The tension on each wire is measured as \(15.4 \mathrm{~N}\). What is the angle of the wires with the horizontal?

Short answer

Answer: The angle of the wires with the horizontal is approximately 9.14 degrees.

Step-by-step solution

Establish a coordinate system

Choose a Cartesian coordinate system with the origin at the center of mass of the hanging textbook, the positive x-axis to the right, and the positive y-axis upward.

Draw a free body diagram

Draw a free body diagram of the textbook showing the forces acting on it: Weight (mg) acting vertically downward, and tension (T) from each wire. Label the angle between the wire and the horizontal as \(\theta\).

Calculate the weight of the textbook

Calculate the weight of the textbook as the product of its mass (0.500 kg) and the acceleration due to gravity (9.81 m/s^2). Weight (W) = mg = 0.500 kg × 9.81 m/s^2 = 4.905 N

Set up the force equilibrium equations

Since the textbook is in equilibrium, the sum of the forces in the x and y directions should be zero. Write down these two equations: Sum of forces in x-direction: \(T\cos(\theta) + T\cos(\theta) = 0\). Sum of forces in y-direction: \(T\sin(\theta) + T\sin(\theta) = W\).

Solve for the tension and angle

We are given the tension T, so just solve the equation for sin(θ) and then find the angle θ: From the x-direction equation: \(2T\cos(\theta)=0\), we get that \(\cos(\theta)=0\). So, \(\theta = 90^\circ\). However, this value of \(\theta\) doesn't make sense in our context, as it would mean both tensions are acting directly vertical which isn't the case. Now, let's use the y-direction equation: \(T\sin(\theta) = \frac{W}{2}\) \(\sin(\theta) = \frac{W}{2T}\) \(\sin(\theta) = \frac{4.905 \, \mathrm{N}}{2 \times 15.4 \, \mathrm{N}}\) Now calculate the value of \(\sin(\theta)\): \(\sin(\theta) \approx 0.1592\) Finally, find the angle θ by taking the inverse sine (sin⁻¹) of the calculated value: \(\theta = \sin^{-1}(0.1592) \approx 9.14^\circ\) The angle of the wires with the horizontal is approximately 9.14 degrees.