/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 79 A \(45-N\) lithograph is support... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 79

A \(45-N\) lithograph is supported by two wires. One wire makes a \(25^{\circ}\) angle with the vertical and the other makes a \(15^{\circ}\) angle with the vertical. Find the tension in each wire. (tutorial: hanging picture \()\)

Short Answer

Expert verified
The tension in the first wire is approximately \(40.65 \text{ N}\) and in the second wire is approximately \(24.26 \text{ N}\).

Step by step solution

01

Identify Forces and Angles

There are two wires supporting the lithograph, and each wire has a tension force that we'll denote as \( T_1 \) and \( T_2 \). The angles given are \( 25^{\circ} \) and \( 15^{\circ} \) with the vertical.
02

Set Up the Equilibrium Equations

The system is in equilibrium, so the sum of vertical forces and horizontal forces must be zero. This gives us two equations: \( T_1 \cos(15^{\circ}) + T_2 \cos(25^{\circ}) = 45 \) (vertical forces) and \( T_1 \sin(15^{\circ}) = T_2 \sin(25^{\circ}) \) (horizontal forces).
03

Solve One of the Equations for a Variable

From the horizontal forces equation \( T_1 \sin(15^{\circ}) = T_2 \sin(25^{\circ}) \), solve for \( T_1 \) in terms of \( T_2 \): \( T_1 = T_2 \frac{\sin(25^{\circ})}{\sin(15^{\circ})} \).
04

Substitute the Expression into the Vertical Equation

Substitute the expression for \( T_1 \) from Step 3 into the vertical forces equation: \( T_2 \frac{\sin(25^{\circ})}{\sin(15^{\circ})}\cos(15^{\circ}) + T_2 \cos(25^{\circ}) = 45 \).
05

Solve for T_2

Simplify and solve the equation from Step 4 to find \( T_2 \). This involves calculations like \( T_2 (\frac{\sin(25^{\circ}) \cos(15^{\circ})}{\sin(15^{\circ})} + \cos(25^{\circ})) = 45 \). Solving this equation gives \( T_2 \approx 24.26 \text{ N} \).
06

Solve for T_1

Use the expression from Step 3 to calculate \( T_1 \): \( T_1 = 24.26 \cdot \frac{\sin(25^{\circ})}{\sin(15^{\circ})} \approx 40.65 \text{ N} \).

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Key Concepts

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

Equilibrium of Forces
In physics, equilibrium of forces is a state where an object is at rest or moving with a constant velocity. This means all the forces acting on the object balance each other out, so the object does not accelerate. To determine if an object is in equilibrium, we consider the sum of all forces in every direction.

When dealing with problems involving equilibrium:
  • The sum of horizontal forces should equal zero.
  • The sum of vertical forces should also equal zero.
In the given problem with the lithograph, we use these principles to set up our equations:
  • The vertical equilibrium equation is: \( T_1 \cos(15^{\circ}) + T_2 \cos(25^{\circ}) = 45 \).
  • The horizontal equilibrium equation is: \( T_1 \sin(15^{\circ}) = T_2 \sin(25^{\circ}) \).
These equations ensure the lithograph remains stationary, balancing the forces in both the vertical and horizontal directions.
Trigonometry in Physics
Trigonometry plays a crucial role in physics, especially when dealing with forces at angles. It helps break down these forces into components鈥攁 concept vital for solving equilibrium problems.

For any force that is not acting purely vertically or horizontally, trigonometric functions are used:
  • The cosine function \( \cos \) helps us find the vertical component of a force.
  • The sine function \( \sin \) helps us find the horizontal component of a force.
In the lithograph example, both wires make angles with the vertical, requiring trigonometry to analyze forces:
  • The vertical component of tension in a wire is found using \( T \cos(\text{angle}) \).
  • The horizontal component is derived using \( T \sin(\text{angle}) \).
By understanding these fundamental trigonometric concepts, you can effectively resolve forces, allowing you to apply the equilibrium conditions accurately.
Tension in Cables
Tension is a force exerted along a rope, string, or wire when it is pulled tight by forces acting from opposite ends. Understanding how tension works is essential in physics, especially when analyzing forces supporting objects.

In the scenario described:
  • The lithograph is supported by two wires, each experiencing a tension force.
  • This tension acts along the direction of the wire, balancing the forces to keep the lithograph steady.
To solve for the tension in each cable, techniques involve:
  • Using equilibrium equations (vertical and horizontal).
  • Applying trigonometric identities to isolate and calculate the tension force in each cable (\(T_1\) and \(T_2\)).
Finally, using these properties helps us find the actual tensions that ensure the lithograph remains in static equilibrium without any movement.

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Most popular questions from this chapter

A roller coaster is towed up an incline at a steady speed of $0.50 \mathrm{m} / \mathrm{s}$ by a chain parallel to the surface of the incline. The slope is \(3.0 \%,\) which means that the elevation increases by \(3.0 \mathrm{m}\) for every \(100.0 \mathrm{m}\) of horizontal distance. The mass of the roller coaster is \(400.0 \mathrm{kg}\). Ignoring friction, what is the magnitude of the force exerted on the roller coaster by the chain?
A crow perches on a clothesline midway between two poles. Each end of the rope makes an angle of \(\theta\) below the horizontal where it connects to the pole. If the weight of the crow is \(W,\) what is the tension in the rope? Ignore the weight of the rope.
An astronaut stands at a position on the Moon such that Earth is directly over head and releases a Moon rock that was in her hand. (a) Which way will it fall? (b) What is the gravitational force exerted by the Moon on a 1.0 -kg rock resting on the Moon's surface? (c) What is the gravitational force exerted by the Earth on the same 1.0 -kg rock resting on the surface of the Moon? (d) What is the net gravitational force on the rock?
An airplane is cruising along in a horizontal level flight at a constant velocity, heading due west. (a) If the weight of the plane is $2.6 \times 10^{4} \mathrm{N},$ what is the net force on the plane? (b) With what force does the air push upward on the plane?
On her way to visit Grandmother, Red Riding Hood sat down to rest and placed her 1.2 -kg basket of goodies beside her. A wolf came along, spotted the basket, and began to pull on the handle with a force of \(6.4 \mathrm{N}\) at an angle of \(25^{\circ}\) with respect to vertical. Red was not going to let go easily, so she pulled on the handle with a force of \(12 \mathrm{N}\). If the net force on the basket is straight up, at what angle was Red Riding Hood pulling?
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