/*! 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 10 A barge is hauled along a straig... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 10

A barge is hauled along a straight-line section of canal by two horses harnessed to tow ropes and walking along the tow paths on either side of the canal. Each horse pulls with a force of \(560 \mathrm{N}\) at an angle of \(15^{\circ}\) with the centerline of the canal. Find the sum of the two forces exerted by the horses on the barge.

Short Answer

Expert verified
Answer: The total force exerted by the horses on the barge is approximately \(1080.66 \mathrm{N}\), acting along the centerline of the canal.

Step by step solution

01

Define the given forces as vectors

The given forces can be defined as vectors with magnitudes and directions. Let \(\textbf{F}_1\) and \(\textbf{F}_2\) be the forces exerted by the two horses. The magnitudes of both forces are \(560 \mathrm{N}\), and they both form an angle of \(15^\circ\) with the centerline of the canal.
02

Find the horizontal and vertical components of each force

We can use trigonometry to find the horizontal (x-axis) and vertical (y-axis) components of each force. The horizontal component of \(\textbf{F}_1\) is given by \(F_{1x} = F_1\cos (\theta)\), and the vertical component of \(\textbf{F}_1\) is given by \(F_{1y} = F_1\sin (\theta)\). Similarly, we can find the components of \(\textbf{F}_2\). Here, \(\theta = 15^\circ\). \(F_{1x} = 560\cos (15^\circ)\) \(F_{1y} = 560\sin (15^\circ)\) \(F_{2x} = 560\cos (15^\circ)\) \(F_{2y} = -560\sin (15^\circ)\) (The vertical component of \(\textbf{F}_2\) is negative since it acts in the opposite direction)
03

Add the horizontal and vertical components of the forces

Now that we have the horizontal and vertical components of each force, we can add them to find the total force exerted by the horses on the barge. \(F_{total_x} = F_{1x} + F_{2x} = 2 \times 560\cos (15^\circ)\) \(F_{total_y} = F_{1y} + F_{2y} = 560\sin (15^\circ) - 560\sin (15^\circ) = 0\) The total force has a horizontal component of \(2 \times 560\cos (15^\circ)\) and a vertical component of \(0\).
04

Find the magnitude and direction of the total force

Since the vertical component of the total force is 0, the total force is entirely horizontal and acts along the centerline of the canal. The magnitude of the total force is equal to the horizontal component: \(F_{total} = 2 \times 560\cos (15^\circ)\) \(F_{total} 鈮� 1080.66 \mathrm{N}\) The total force exerted by the horses on the barge is approximately \(1080.66 \mathrm{N}\), acting along the centerline of the canal.

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 91影视!

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

Which of the fundamental forces governs the motion of planets in the solar system? Is this the strongest or the weakest of the fundamental forces? Explain.
The coefficient of static friction between a brick and a wooden board is 0.40 and the coefficient of kinetic friction between the brick and board is $0.30 .$ You place the brick on the board and slowly lift one end of the board off the ground until the brick starts to slide down the board. (a) What angle does the board make with the ground when the brick starts to slide? (b) What is the acceleration of the brick as it slides down the board?
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.
A box full of books rests on a wooden floor. The normal force the floor exerts on the box is \(250 \mathrm{N}\). (a) You push horizontally on the box with a force of \(120 \mathrm{N},\) but it refuses to budge. What can you say about the coefficient of static friction between the box and the floor? (b) If you must push horizontally on the box with a force of at least \(150 \mathrm{N}\) to start it sliding, what is the coefficient of static friction? (c) Once the box is sliding, you only have to push with a force of \(120 \mathrm{N}\) to keep it sliding. What is the coefficient of kinetic friction?
Four separate blocks are placed side by side in a left-toright row on a table. A horizontal force, acting toward the right, is applied to the block on the far left end of the row. Draw FBDs for (a) the second block on the left and for (b) the system of four blocks.
See all solutions

Recommended explanations on Physics Textbooks

Modern Physics

Read Explanation

Measurements

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Waves Physics

Read Explanation

Rotational Dynamics

Read Explanation

Scientific Method Physics

Read Explanation
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.