Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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:
  • 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.
In the exercise, the problem states that each wire has a specific tension, making it essential to determine the angle to ensure that the frame remains steady in its position on the wall.
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:
  • 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.
This perfect "tug-of-war" between forces ensures the frame remains in equilibrium.
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:
  • 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.
This exercise showcases how trigonometry helps us find precise angles in complex physics problems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An 8.00-kg box sits on a ramp that is inclined at 33.0\(^\circ\) above the horizontal. The coefficient of kinetic friction between the box and the surface of the ramp is \(\mu_k =\) 0.300. A constant \(horizontal\) force \(F =\) 26.0 N is applied to the box (\(\textbf{Fig. P5.73}\)), and the box moves down the ramp. If the box is initially at rest, what is its speed 2.00 s after the force is applied?

An astronaut is inside a 2.25 \(\times\) 10\(^6\) kg rocket that is blasting off vertically from the launch pad. You want this rocket to reach the speed of sound (331 m/s) as quickly as possible, but astronauts are in danger of blacking out at an acceleration greater than 4\(g\). (a) What is the maximum initial thrust this rocket's engines can have but just barely avoid blackout? Start with a free-body diagram of the rocket. (b) What force, in terms of the astronaut's weight \(w\), does the rocket exert on her? Start with a free-body diagram of the astronaut. (c) What is the shortest time it can take the rocket to reach the speed of sound?

On the ride "Spindletop" at the amusement park Six Flags Over Texas, people stood against the inner wall of a hollow vertical cylinder with radius 2.5 m. The cylinder started to rotate, and when it reached a constant rotation rate of 0.60 rev/s, the floor dropped about 0.5 m. The people remained pinned against the wall without touching the floor. (a) Draw a force diagram for a person on this ride after the floor has dropped. (b) What minimum coefficient of static friction was required for the person not to slide downward to the new position of the floor? (c) Does your answer in part (b) depend on the person's mass? (\(Note\): When such a ride is over, the cylinder is slowly brought to rest. As it slows down, people slide down the walls to the floor.)

Jack sits in the chair of a Ferris wheel that is rotating at a constant 0.100 rev/s. As Jack passes through the highest point of his circular path, the upward force that the chair exerts on him is equal to one-fourth of his weight. What is the radius of the circle in which Jack travels? Treat him as a point mass.

A man pushes on a piano with mass 180 kg; it slides at constant velocity down a ramp that is inclined at 19.0\(^\circ\) above the horizontal floor. Neglect any friction acting on the piano. Calculate the magnitude of the force applied by the man if he pushes (a) parallel to the incline and (b) parallel to the floor.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free