/*! 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 56 Interactive LearningWare 4.3 at ... [FREE SOLUTION] | 91Ó°ÊÓ

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Interactive LearningWare 4.3 at reviews the principles that play a role in this problem. During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between two fence posts. The limb exerts a downward force of \(151 \mathrm{~N}\) on the wire. The left section of the wire makes an angle of \(14.0^{\circ}\) relative to the horizontal and sustains a tension of \(447 \mathrm{~N}\). Find the magnitude and direction of the tension that the right section of the wire sustains.

Short Answer

Expert verified
Tension on the right wire: 437 N at an angle 14° below the horizontal.

Step by step solution

01

Break Down the Forces

Start by identifying the forces acting on the system. We have the downward force \( F = 151 \text{ N} \), the tension in the left part of the wire \( T_1 = 447 \text{ N} \), and the unknown tension \( T_2 \) in the right part of the wire. The angle for the left tension \( \theta_1 = 14^{\circ} \). The angle \( \theta_2 \) for \( T_2 \) is what we need to find.
02

Apply Equilibrium Conditions

Since the limb is in equilibrium, the net force in both the horizontal and vertical directions should be zero. This gives us two equations to work with:1. Horizontal component: \( T_1 \cos(\theta_1) = T_2 \cos(\theta_2) \).2. Vertical component: \( T_1 \sin(\theta_1) + T_2 \sin(\theta_2) = F \).
03

Solve for Horizontal Components

Using the horizontal balance equation, solve for \( T_2 \cos(\theta_2) \):\[ T_2 \cos(\theta_2) = T_1 \cos(14^{\circ}) = 447 \cos(14^{\circ}) \].
04

Solve for Vertical Components

Using the vertical balance equation, substitute the known forces:\[ T_1 \sin(14^{\circ}) + T_2 \sin(\theta_2) = 151 \].Substitute \( T_2 \cos(\theta_2) \) from Step 3 and solve for \( T_2 \sin(\theta_2) \).
05

Solve for the Magnitude of Tension T_2

From the equilibrium conditions, solve for both \( T_2 \sin(\theta_2) \) and \( T_2 \cos(\theta_2) \), and use these to find \( T_2 \). Use the identity:\[ T_2 = \sqrt{(T_2 \cos(\theta_2))^2 + (T_2 \sin(\theta_2))^2} \]Substitute to find \( T_2 \).
06

Solve for the Angle \(\theta_2\)

Now that you have \( T_2 \cos(\theta_2) \) and \( T_2 \sin(\theta_2) \), find \( \theta_2 \) using:\[ \tan(\theta_2) = \frac{T_2 \sin(\theta_2)}{T_2 \cos(\theta_2)} \]Solve for \( \theta_2 \).
07

Calculate and Conclude

Finally, compute the values using the trigonometric functions to find both \( T_2 \) and \( \theta_2 \). Plug in the values to get the magnitude and direction of the tension on the right wire.

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Key Concepts

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

Tension in Wires
The concept of tension in wires is a fundamental topic in physics. When a force is exerted on a wire or a cable, the wire experiences a pulling force, known as tension. In equilibrium situations like the one described here, maintaining balance means that the forces acting along the wire need to be equal but opposite to maintain a state of rest. Tension can occur in various directions, depending on how the wire or cable is positioned.
In the problem with the tree limb, the left section of the wire experiences tension due to its attached angle and the downward force exerted by the limb. To ensure balance, the tension on the right section of the wire also needs to be calculated precisely. This involves considering the angles involved and applying principles like equilibrium, where the forces horizontally and vertically must sum to zero.
Force Components
To properly understand the forces acting in equilibrium, we must break them down into their components. Particularly, forces in a system like the wire carrying the tree limb can be deconstructed into two main components: horizontal and vertical.
  • The **horizontal component** is calculated using the cosine of the angle that the tension makes with the horizontal plane. It essentially controls how much of the tension is acting left or right.
  • The **vertical component** is determined using the sine of the angle. This part of the tension affects how much force is directed upwards or downwards.
In the given exercise, it is essential to calculate these components to understand how the tensions balance each other out. By breaking down the forces on both sides of the wire into horizontal and vertical components, one can conclude how they affect the system's equilibrium. Mathematically, this process involves using trigonometric identities like \( \text{sin} \) and \( \text{cos} \) to project the forces along the respective axes.
Trigonometry in Physics
Trigonometry plays an essential role in understanding and solving physics problems involving angles and forces. In situations where objects are suspended or inclined, such as the tree limb resting on a wire, trigonometric functions help analyze the components of forces.
Here's why trigonometry is crucial:
  • **Calculating Angles**: Finding the direction of forces requires calculating angles. In this exercise, the angle tells us how the tension is distributed over the wire.
  • **Decomposing Forces**: Trigonometric functions like sine and cosine are essential for breaking down the overall force into its components. This enables us to apply equilibrium conditions since the sum of the forces in the horizontal and vertical directions must equal zero.
In the problem, trigonometry helps find the angle \( \theta_2 \) where the wire aligns in the right direction. By using \( \tan(\theta_2) = \frac{T_2 \sin(\theta_2)}{T_2 \cos(\theta_2)} \), we determine the direction effectively. Through these calculations, we clearly visualize and solve for the relationships between angles, tension, and equilibrium in physics.

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Most popular questions from this chapter

The weight of the block in the drawing is \(88.9 \mathrm{~N}\). The coefficient of static friction between the block and the vertical wall is \(0.560 .\) (a) What minimum force \(\vec{F}\) is required to prevent the block from sliding down the wall? (Hint: The static frictional force exerted on the block is directed upward, parallel to the wall.) (b) What minimum force is required to start the block moving up the wall? (Hint: The static frictional force is now directed down the wall.)

A \(325-\mathrm{kg}\) boat is sailing \(15.0^{\circ}\) north of east at a speed of \(2.00 \mathrm{~m} / \mathrm{s}\). Thirty seconds later, it is sailing \(35.0^{\circ}\) north of east at a speed of \(4.00 \mathrm{~m} / \mathrm{s}\). During this time, three forces act on the boat: a \(31.0-\mathrm{N}\) force directed \(15.0^{\circ}\) north of east (due to an auxiliary engine), a 23.0-N force directed \(15.0^{\circ}\) south of west (resistance due to the water), and \(\vec{F}_{W}\) (due to the wind). Find the magnitude and direction of the force \(\overrightarrow{\mathrm{F}}_{\mathrm{W}} .\) Express the direction as an angle with respect to due east.

A block is pressed against a vertical wall by a force \(\vec{P}\), as the drawing shows. This force can either push the block upward at a constant velocity or allow it to slide downward at a constant velocity, the magnitude of the force being different in the two cases, while the directional angle \(\theta\) is the same. Kinetic friction exists between the block and the wall. (a) Is the block in equilibrium in each case? Explain. (b) In each case what is the direction of the kinetic frictional force that acts on the block? Why? (c) In each case is the magnitude of the frictional force the same or different? Justify your answer. (d) In which case is the magnitude of the force \(\overrightarrow{\mathrm{P}}\) greater? Provide a reason for your answer. The weight of the block is \(39.0 \mathrm{~N}\), and the coefficient of kinetic friction between the block and the wall is \(0.250\). The direction of the force \(\overrightarrow{\mathrm{P}}\) is \(\theta=30.0^{\circ}\). Determine the magnitude of \(\overrightarrow{\mathrm{P}}\) when the block slides up the wall and when it slides down the wall. Check to see that your answers are consistent with your answers to the Concept Questions.

Review Interactive LearningWare 4.3 at in preparation for this problem. The helicopter in the drawing is moving horizontally to the right at a constant velocity. The weight of the helicopter is \(W=53800 \mathrm{~N}\). The lift force \(\overrightarrow{\mathrm{L}}\) generated by the rotating blade makes an angle of \(21.0^{\circ}\) with respect to the vertical. (a) What is the magnitude of the lift force? (b) Determine the magnitude of the air resistance \(\overrightarrow{\mathrm{R}}\) that opposes the motion.

A bicyclist is coasting straight down a hill at a constant speed. The mass of the rider and bicycle is \(80.0 \mathrm{~kg},\) and the hill is inclined at \(15.0^{\circ}\) with respect to the horizontal. Air resistance opposes the motion of the cyclist. Later, the bicyclist climbs the same hill at the same constant speed. How much force (directed parallel to the hill) must be applied to the bicycle in order for the bicyclist to climb the hill?

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