/*! 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 53 When force \(F_{1}, F_{2}\), and... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 53

When force \(F_{1}, F_{2}\), and \(F_{3}\) are acting on a particle of mass \(m\) such that \(F_{2}\) and \(F_{3}\) are mutually perpendicular, then the particle remains stationary. If the force \(F_{1}\) is now removed, then the acceleration of the particle is a. \(\frac{F_{1}}{m}\) b. \(\frac{F_{2}}{m}\) c. \(\frac{F_{3}}{m}\) d. \(\frac{F_{2}+F_{1}}{m}\)

Short Answer

Expert verified
a. \(\frac{F_{1}}{m}\)

Step by step solution

01

Understanding the Given Information

The problem states that three forces, \(F_1\), \(F_2\), and \(F_3\), act on a particle such that the particle remains stationary. This implies that the net force on the particle is zero, so these forces balance each other according to Newton's first law of motion.
02

Breaking Down the Force Balance Condition

Since \(F_2\) and \(F_3\) are perpendicular, their vector sum results in a diagonal force, which we can denote as \(\mathbf{R} = \mathbf{F_2} + \mathbf{F_3}\). For equilibrium, \(F_1\) must be equal in magnitude and opposite in direction to this resultant force \(\mathbf{R}\), i.e., \(\mathbf{F_1} = -\mathbf{R}\).
03

Analyzing the Situation When \(F_1\) is Removed

When \(F_1\) is removed, the force \(\mathbf{R} = \mathbf{F_2} + \mathbf{F_3}\) is no longer being counteracted, which means the net force on the particle is now \(\mathbf{R}\).
04

Calculating the Resulting Acceleration

According to Newton's second law, the acceleration \(\mathbf{a}\) of the particle is given by \(\mathbf{a} = \frac{\mathbf{F}}{m}\), where \(\mathbf{F}\) is the net force. Thus, \(\mathbf{a} = \frac{F_2 + F_3}{m}\), as the net force is now just \(F_2 + F_3\).
05

Finding the Corresponding Answer

Evaluating the given options, none exactly match \(\frac{F_2 + F_3}{m}\). However, understanding the scenario, \(\mathbf{F_1} = -\mathbf{R}\), implies in magnitude, \(\mathbf{R} = F_1\), equates \(F_2 + F_3 = F_1\) as vectors. Thus, the acceleration \(\mathbf{a}\) = \(\frac{F_1}{m}\) matches option a.

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!

Key Concepts

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

Equilibrium of Forces
When a particle remains stationary despite having multiple forces acting upon it, this is because the forces are in equilibrium. Equilibrium occurs when the sum of all forces acting on an object is zero. In the given problem, the forces are balanced, meaning they cancel each other out precisely.
For equilibrium:
  • There should be no net force acting on the object.
  • The total sum of all forces in all directions must equal zero.
Since the forces are in equilibrium, the conditions are perfectly met without any unbalanced force to cause motion.
Understanding equilibrium helps clarify situations where forces seem to act but result in no movement, showing the power of balance.
Vector Addition of Forces
Vector addition is crucial when multiple forces act on an object, especially in different directions. The vector nature of forces means each force has both a magnitude and a direction.
In the exercise, forces \( F_2 \) and \( F_3 \) are perpendicular, and their vector addition creates a resultant force \( \mathbf{R} \). This is a classic application of vector addition, where the Pythagorean theorem helps find the magnitude:
  • The resultant force \( \mathbf{R} \) can be calculated as \( \mathbf{R} = \mathbf{F_2} + \mathbf{F_3} \).
  • Magnitude of \( \mathbf{R} \) can be found using \( R = \sqrt{F_2^2 + F_3^2} \).
Vector addition allows us to combine forces in different directions into a single force vector.
Understanding how to add vectors gives insight into predicting the system's behavior when forces are applied.
Newton's First Law
Newton's first law of motion, often called the law of inertia, states that an object at rest will stay at rest unless acted upon by an external force. This principle explains why the particle in the exercise stays stationary when three forces, \( F_1 \), \( F_2 \), and \( F_3 \), act upon it but balance each other out.
Key considerations:
  • When the forces are balanced, and no net force exists, the object remains at rest or continues with a constant velocity.
  • An unbalanced force is required to change the existing state of motion.
In the given problem, the balance of the three forces illustrates Newton's first law, showing no changes in motion or direction occur without an unbalanced force.
Newton's Second Law
Newton's second law of motion connects the concepts of force, mass, and acceleration. It states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass: \( \mathbf{a} = \frac{\mathbf{F}}{m} \).
In the context of the problem:
  • When force \( F_1 \) is removed, the net force is the resultant of \( F_2 + F_3 \).
  • The resulting acceleration \( \mathbf{a} \) is \( \frac{F_2 + F_3}{m} \).
Since the net force now equals the magnitude of what \( \mathbf{F_1} \) was countering, the particle moves with acceleration \( \frac{F_1}{m} \).
Newton鈥檚 second law helps us understand how removing or adding forces directly influences an object's acceleration.

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

A vehicle is moving with a velocity \(v\) ou a curved road of width \(b\) and radius of curvature \(R\). For counteracting the centrifugal force on the vehicle, the difference in elevation required in between the outer and inner edges of the rod is a. \(v^{2} b / R g\) b. \(v b / R g\) c. \(v b^{2} / R g\) d. \(v b / R^{2} g\)

A box of mass \(8 \mathrm{~kg}\) is placed on a rough inclined plane of inclination \(45^{\circ}\). Its downward motion can be prevented by applying an upward pull \(F\) and it can be made to slide upwards by applying a force \(2 F\). The coefficient of friction between the box and the inclined plane is a. \(\frac{1}{2}\) b. \(\frac{1}{\sqrt{2}}\) c. \(\frac{1}{2 \sqrt{2}}\) d. \(\frac{1}{3}\)

A person is drawing himself up and a trolley on which he stands with some acceleration. Mass of the person is more than the mass of the trolley. As the person increases his force on the string, the normal reaction between person and the trolley will a. increase b. decrease c. remain same d. cannot be predicted as data is insufficient

A block is lying on the horizontal frictionless surface. One end of a uniform rope is fixed to the block which is pulled in the horizontal direction by applying a force \(F\) at the other end. If the mass of the rope is half the mass of the block, the tension in the middle of the rope will be a. \(F\) b. \(2 F / 3\) c. \(3 F / 5\) d. \(5 F / 6\)

A block of mass \(M\) is pulled along a horizontal frictionless surface by a rope of mass \(m\). Force \(P\) is applied at one end of rope. The force which the rope exerts on the block \(\mathrm{s}\) a. \(\frac{P}{(M-m)}\) b. \(\frac{P}{M(m+M)}\) c. \(\frac{P M}{(m+M)}\) d. \(\frac{P M}{(M-m)}\)
See all solutions

Recommended explanations on Physics Textbooks

Translational Dynamics

Read Explanation

Dynamics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Medical Physics

Read Explanation

Electrostatics

Read Explanation

Rotational Dynamics

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.