Chapter 5: Problem 5
A picture frame hung against a wall is suspended by two wires attached to its upper corners. If the two wires make the same angle with the vertical, what must this angle be if the tension in each wire is equal to 0.75 of the weight of the frame? (Ignore any friction between the wall and the picture frame.)
Short Answer
Expert verified
The angle is \( \theta = \cos^{-1}(\frac{2}{3}) \).
Step by step solution
01
Understand the Problem
We have a picture frame hung by two wires that make the same angle with the vertical. Each wire has a tension equal to 0.75 of the weight of the frame. Our goal is to find the angle the wires make with the vertical.
02
Set Up the Equations
Let's denote the weight of the frame as \( W \), the tension in each wire as \( T = 0.75W \), and the angle with the vertical as \( \theta \). The vertical component of the tension in each wire must add up to the weight of the frame to keep it in equilibrium.
03
Calculate Vertical Force Components
For each wire, the vertical component is \( T \cos \theta \). Therefore, the sum of the vertical components from both wires is \( 2T \cos \theta \). Since this sum must equal the weight of the frame, we have:\[ 2T \cos \theta = W \]
04
Solve the Equation for \( \cos \theta \)
Substituting \( T = 0.75W \) into the equation gives:\[ 2(0.75W) \cos \theta = W \]\[ 1.5W \cos \theta = W \]Now, divide both sides by \( W \):\[ 1.5 \cos \theta = 1 \]
05
Determine the Angle \( \theta \)
Solve for \( \cos \theta \):\[ \cos \theta = \frac{1}{1.5} = \frac{2}{3} \]Taking the inverse cosine, we find:\[ \theta = \cos^{-1}\left(\frac{2}{3}\right) \]
06
Conclusion
The wires make an angle with the vertical such that \( \theta = \cos^{-1}(\frac{2}{3}) \). This can be converted to degrees or radians for practical purposes, depending on context.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Tension in Strings
Tension is a force that is transmitted through a string, cable, or wire when it is pulled tight by forces acting from opposite ends. In this scenario, tension is crucial because it supports the picture frame, preventing it from falling.
Each wire in our picture frame problem has a tension equal to 0.75 times the weight of the picture frame. To picture it more clearly, imagine pulling on both ends of a rope. The harder you pull, the more tension is in the rope. Similarly, the picture frame exerts a force downward due to gravity, and the tension in the strings counteracts this force to hold the frame up.
Some key points about tension to keep in mind:
Each wire in our picture frame problem has a tension equal to 0.75 times the weight of the picture frame. To picture it more clearly, imagine pulling on both ends of a rope. The harder you pull, the more tension is in the rope. Similarly, the picture frame exerts a force downward due to gravity, and the tension in the strings counteracts this force to hold the frame up.
Some key points about tension to keep in mind:
- Tension acts along the length of the wire or rope.
- It is usually denoted by a capital "T" in physics equations.
- The tension is the same throughout the length of an ideal wire, provided it's light and does not stretch.
Equilibrium Forces
Equilibrium occurs when all the forces acting on an object are balanced, resulting in the object remaining still or continuing to move in a state of constant velocity. For our picture frame situation, it means the frame is neither moving up nor down, or swinging side to side.
When dealing with equilibrium, the sum of all forces in any direction must equal zero. In our photo frame exercise, the downward gravitational force is the weight, denoted as "W". The tension's vertical components from the wires act upwards, counteracting this weight. We express this balance using the equation: \[2T \cos \theta = W\]
Understanding this:
When dealing with equilibrium, the sum of all forces in any direction must equal zero. In our photo frame exercise, the downward gravitational force is the weight, denoted as "W". The tension's vertical components from the wires act upwards, counteracting this weight. We express this balance using the equation: \[2T \cos \theta = W\]
Understanding this:
- The gravitational force pulls the frame downwards, trying to create motion.
- The tension's vertical components from both wires add up to balance this force.
- As a result, the frame is held steady and doesn't move.
Trigonometric Functions
Trigonometry often comes into play in physics, especially when solving problems involving angles and forces. In this scenario, the trigonometric function we use is cosine, denoted as "cos". This function helps us explore the relationship between the angles and tensions in the strings holding our picture frame.
When a force Is applied at an angle, only a component of that force acts in a specific direction. The key trigonometric principle here is that the vertical component of tension is calculated using:\[T \cos \theta\], where \(T\) is the tension in the wire, and \(\theta\) is the angle each wire makes with the vertical.
To find the angle that holds the frame in equilibrium, we rearrange our equilibrium equation to isolate cos(θ):\[1.5 \cos \theta = 1\]
We solve for the angle using the arccosine or inverse cosine function:\[\theta = \cos^{-1}\left(\frac{2}{3}\right)\]
Some tips for using trigonometric functions:
When a force Is applied at an angle, only a component of that force acts in a specific direction. The key trigonometric principle here is that the vertical component of tension is calculated using:\[T \cos \theta\], where \(T\) is the tension in the wire, and \(\theta\) is the angle each wire makes with the vertical.
To find the angle that holds the frame in equilibrium, we rearrange our equilibrium equation to isolate cos(θ):\[1.5 \cos \theta = 1\]
We solve for the angle using the arccosine or inverse cosine function:\[\theta = \cos^{-1}\left(\frac{2}{3}\right)\]
Some tips for using trigonometric functions:
- They are crucial whenever you have forces or vectors at angles.
- Cosine is particularly useful for finding components parallel to the force's direction.
- Ensure your calculator is set to the correct mode (degrees or radians) when solving problems.