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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? (lanore any friction between the wall and the picture frame.)

Short Answer

Expert verified
The angle is approximately \(48.19^\circ\).

Step by step solution

01

Understand the Problem

We have a picture frame hanging against a wall with two wires attached to its upper corners. Each wire makes the same angle with the vertical. We need to find this angle given that the tension in each wire is 0.75 times the weight of the frame.
02

Identify Forces and Equations

The forces acting on the picture frame in the vertical direction sum to zero since the picture is in equilibrium. Each wire exerts a tension force that has both vertical and horizontal components. Given that tension in each wire is 0.75 times the weight of the frame, we let the weight of the frame be denoted by \(W\), so the tension in each wire is \(0.75W\).
03

Resolve Tension into Components

Each tension force can be resolved into a vertical and a horizontal component. The vertical component of the tension is \(0.75W \cdot \cos(\theta)\). The sum of these components from both wires must equal the weight of the picture frame, \(W\).
04

Set Up the Equation for Vertical Forces

The total vertical force is due to both wires: \(2 \times 0.75W \cdot \cos(\theta) = W\). Simplifying, we get the equation: \(1.5W \cdot \cos(\theta) = W\).
05

Solve for the Angle \(\theta\)

Divide both sides of the equation by \(W\) to get \(1.5 \cdot \cos(\theta) = 1\). Solving for \(\cos(\theta)\), we have \(\cos(\theta) = \frac{1}{1.5} = \frac{2}{3}\). Take the inverse cosine to find \(\theta\): \(\theta = \cos^{-1}\left(\frac{2}{3}\right)\).
06

Calculate Value of \(\theta\)

Use a calculator to find \(\theta\). This gives \(\theta \approx 48.19^\circ\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium
When an object is in equilibrium, it means that it is in a state of balance. In such a scenario, the sum of all forces acting on it is zero. As a result, there is no movement and the forces are in a state of balance.
For the picture frame in the original exercise, equilibrium is achieved when the net force resulting from the tension in the wires counteracts the weight of the frame perfectly.
  • The vertical forces add up to zero since they balance the weight of the frame.
  • The horizontal forces cancel each other out due to being equal and opposite.
Understanding equilibrium is crucial to analyzing any system involving forces, as it allows us to consider the forces alongside and solve for unknown quantities.
Forces
Forces are any interaction that, when unopposed, will change an object's motion. They can be push-like or pull-like and are characterized by magnitude and direction.
In the context of the hanging picture frame:
  • Tension is the key force in this scenario, acting along the wires.
  • The weight of the frame, denoted as a downward force, maintains the system under examination.
The interplay of these forces allows the picture frame to remain suspended without falling or tipping, as they are perfectly balanced through tension in the wires.
Resolution of Forces
Resolution of forces is a method used to understand the effect of a single force by breaking it down into components. In our case, each tension force has components.
  • Vertical components assist in balancing the frame鈥檚 weight.
  • Horizontal components counteract each other to maintain equilibrium.
Resolving a force involves using trigonometric functions to split the tension into vertical and horizontal components. Here, the vertical force is calculated with the equation: \[ 0.75W \cdot \cos(\theta) \] This expression helps to set up the equilibrium condition required to solve for the unknown angle.
Trigonometric Functions
Trigonometric functions such as cosine and sine are vital tools in resolving forces, especially in problems involving angles and tensions.
In the picture frame problem:
  • Cosine helps calculate the vertical component of the tension, according to the angle \( \theta \) the wires make with vertical.
  • The equation \[1.5W \cdot \cos(\theta) = W \] simplifies to allow solving for \( \theta \).
To solve for \( \theta \), we use the inverse cosine, \( \cos^{-1} \), reflecting the angles necessary for the equilibrium condition. Understanding these functions is crucial for tackling problems related to forces and angles in physics and engineering.

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Most popular questions from this chapter

A crate of 45.0 -kg tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it and observe that the crate just begins to move when your force exceeds 313 \(\mathrm{N}\) . After that you must reduce your push to 208 \(\mathrm{N}\) to keep it moving at a steady 25.0 \(\mathrm{cm} / \mathrm{s}\) , (a) What are the coefficients of static and kinetic friction between the crate and the floor? (b) What push must you exert to give it an acceleration of 1.10 \(\mathrm{m} / \mathrm{s}^{2} ?\) (c) Suppose you were performing the same experiment on this crate but were doing it on the moon instead, where the acceleration due to gravity is 1.62 \(\mathrm{m} / \mathrm{s}^{2} .\) (i) What magnitude push would cause it to move? (ii) What would its acceleration be if you maintained the push in part (b)?

Angle for Minimum Force. A box with weight w is pulled at constant speed along a level floor by a force \(\vec{F}\) that is at an angle \(\theta\) above the horizontal. The coefficient of kinetic friction between the floor and box is \(\mu_{k}\) (a) In terms of \(\theta, \mu_{k},\) and \(w,\) calculate \(F .\) (b) For \(w=400 \mathrm{N}\) and \(\mu_{k}=0.25\) , calculate \(F\) for \(\theta\) ranging from \(0^{\circ}\) to \(90^{\circ}\) in increments of \(10^{\circ} .\) Graph \(F\) versus \(\theta\) . (c) From the general expression in part (a), calculate the value of \(\theta\) for which the value of \(F,\) required to maintain constant speed, is a minimum. (Hint: At a point where a function is minimum, what are the first and second derivatives of the function? Here \(F\) is a function of \(\theta .\) ) For the special case of \(w=400 \mathrm{N}\) and \(\mu_{\mathrm{x}}=0.25\) evaluate this optimal \(\theta\) and compare your result to the graph you constructed in part \((\mathrm{b})\) .

Genesis Crash. On September \(8,2004\) , the Genesis space-craft crashed in the Utah desert because its parachute did not open. The \(210-k g\) capsule hit the ground at 311 \(\mathrm{km} / \mathrm{h}\) and penetrated the soil to a depth of 81.0 \(\mathrm{cm} .\) (a) Assuming it to be constant, what was its acceleration \(\left(\text { in } \mathrm{m} / \mathrm{s}^{2} \text { and in } g^{\prime} \mathrm{s}\right)\) during the crash? (b) What force did the ground exert on the capsule during the crash? Express the force in newtons and as a multiple of the capsule's weight. (c) For how long did this force last?

Certain streets in San Francisco make an angle of \(17.5^{\circ}\) with the horizontal. What force parallel to the street surface is required to keep a loaded 1967 Corvette of mass 1390 \(\mathrm{kg}\) from rolling down such a street?

Wheels. You find that it takes a horizontal force of 160 \(\mathrm{N}\) to slide a box along the surface of a level fioor at constant speed. The coefficient of static friction is \(0.52,\) and the coefficient of kinetic friction is \(0.47 .\) If you place the box on a dolly of mass 5.3 \(\mathrm{kg}\) and with coefficient of rolling friction \(0.018,\) what horizontal acceleration would that \(160-\mathrm{N}\) force provide?
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