Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 121

Two canal workers pull a barge along the narrow waterway at a constant speed. One worker pulls with a force of \(105 \mathrm{N}\) at an angle of \(28^{\circ}\) with respect to the forward motion of the barge and the other worker, on the opposite tow path, pulls at an angle of \(38^{\circ}\) relative to the barge motion. Both ropes are parallel to the ground. (a) With what magnitude force should the second worker pull to make the sum of the two forces be in the forward direction? (b) What is the magnitude of the force on the barge from the two tow ropes?

Short Answer

Expert verified
Answer: (a) The second worker should pull with a force of approximately \(72.63 \mathrm{N}\), and (b) the total magnitude of the force on the barge from the two tow ropes is approximately \(148.39 \mathrm{N}\).

Step by step solution

01

Understand the problem and draw a diagram

To begin, it's always good to visualize the problem by drawing a diagram. Here, we can represent the two workers, the barge, and the forces exerted by them on the barge, along with the respective angles.
02

Analyze the forces acting on the barge

Next, we need to analyze the forces acting on the barge and separate them into their horizontal and vertical components. Let F be the force exerted by the second worker. In order for the total force to remain in the forward direction, the vertical components of the forces by the first and second workers must be equal, canceling out each other's effects.
03

Separate the forces into their components

Now, we will separate the forces into their horizontal and vertical components. For the first worker, who pulls with a force of \(105 \mathrm{N}\) at an angle of \(28^{\circ}\), the horizontal and vertical components are: Horizontal component: \(105 \cos(28^{\circ})\) Vertical component: \(105 \sin(28^{\circ})\) For the second worker, with a force of \(F\) and an angle of \(38^{\circ}\): Horizontal component: \(F \cos(38^{\circ})\) Vertical component: \(F \sin(38^{\circ})\)
04

Equate the vertical components to find the magnitude of F

In order for the forces to sum in the forward direction, the vertical components must be equal, which means: \(105 \sin(28^{\circ}) = F \sin(38^{\circ})\) To find F, we can divide both sides by \(\sin(38^{\circ})\): \(F = \frac{105 \sin(28^{\circ})}{\sin(38^{\circ})} \approx 72.63 \mathrm{N}\)
05

Find the total horizontal force

Now we need to find the total horizontal force on the barge from the two tow ropes. The horizontal components of both forces acting on the barge sum together: Total horizontal force = \(105 \cos(28^{\circ}) + F \cos(38^{\circ}) \approx 105 \cos(28^{\circ}) + 72.63 \cos(38^{\circ}) \approx 148.39 \mathrm{N}\) So, (a) the second worker should pull with a force of approximately \(72.63 \mathrm{N}\) and (b) the total magnitude of the force on the barge from the two tow ropes is approximately \(148.39 \mathrm{N}\).

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.

Start your free trial

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

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

Before hanging new William Morris wallpaper in her bedroom, Brenda sanded the walls lightly to smooth out some irregularities on the surface. The sanding block weighs \(2.0 \mathrm{N}\) and Brenda pushes on it with a force of $3.0 \mathrm{N}\( at an angle of \)30.0^{\circ}$ with respect to the vertical, and angled toward the wall. Draw an FBD for the sanding block as it moves straight up the wall at a constant speed. What is the coefficient of kinetic friction between the wall and the block?
A large wooden crate is pushed along a smooth, frictionless surface by a force of \(100 \mathrm{N}\). The acceleration of the crate is measured to be $2.5 \mathrm{m} / \mathrm{s}^{2} .$ What is the mass of the crate?
A woman who weighs \(600 \mathrm{N}\) sits on a chair with her feet on the floor and her arms resting on the chair's armrests. The chair weighs $100 \mathrm{N}\(. Each armrest exerts an upward force of \)25 \mathrm{N}$ on her arms. The seat of the chair exerts an upward force of \(500 \mathrm{N}\). (a) What force does the floor exert on her feet? (b) What force does the floor exert on the chair? (c) Consider the woman and the chair to be a single system. Draw an FBD for this system that includes all of the external forces acting on it.
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?
A fisherman is holding a fishing rod with a large fish suspended from the line of the rod. Identify the forces acting on the rod and their interaction partners.
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Force

Read Explanation

Nuclear Physics

Read Explanation

Oscillations

Read Explanation

Astrophysics

Read Explanation

Atoms and Radioactivity

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.

Sign-up for free