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Chapter 4: Problem 79

A 0.50 -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

Expert verified
Answer: The angle the wires make with the horizontal is approximately 18.67°.

Step by step solution

01

Write down the given variables and determine what needs to be found

We are given: - Mass of the textbook (m) = 0.50 kg - Tension on each wire (T) = 15.4 N We need to find the angle (θ) the wires make with the horizontal.
02

Calculate the gravitational force acting on the textbook

First, let's find the gravitational force acting on the textbook. The gravitational force (Fg) can be calculated using the equation: Fg = mg where 'm' is the mass and 'g' is the gravitational acceleration (approximately \(9.81\mathrm{\,m/s^2}\)). Fg = (0.50 kg) · (9.81 m/s^2) = 4.905 N
03

Apply Newton's second law and geometry principles

Since the textbook hangs symmetrically, we can treat half of the scenario as a right triangle. Using the properties of right triangles, we can relate the horizontal and vertical components of the tension force: - Vertical component of tension (T_y): \(T_y = T\sin(\theta)\) - Horizontal component of tension (T_x): \(T_x = T\cos(\theta)\) Applying Newton's second law in the vertical direction: ΣF_y = 0 (The system is at rest) So, \(T\sin(\theta) - Fg = 0\)
04

Solve for the angle θ

Now, we will solve for the angle θ. Rearrange the equation from step 3: \(T\sin(\theta) = Fg\) Divide both sides by T: \(\sin(\theta) = \dfrac{Fg}{T}\) Now, calculate the sine of θ: \(\sin(\theta) = \dfrac{4.905\mathrm{\,N}}{15.4\mathrm{\,N}} = 0.3182\) To find the angle θ, take the inverse sine: θ = \(\sin^{-1}(0.3182)\) θ ≈ 18.67° So, the angle the wires make with the horizontal is approximately 18.67°.

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