/*! 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 34 A body of mass \(m\) is kept sta... [FREE SOLUTION] | 91Ó°ÊÓ

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A body of mass \(m\) is kept stationary on a rough inclined plane of inclination \(\theta .\) The magnitude of force acting on the body by the inclined plane is (A) \(m g\) (B) \(m g \sin \theta\) (C) \(m g \cos \theta\) (D) \(m g \sqrt{1+\cos ^{2} \theta}\)

Short Answer

Expert verified
The force exerted by the inclined plane on the body is \(F = mg \sin \theta\). Hence, the correct answer is (B) \(mg \sin \theta\).

Step by step solution

01

Draw a force diagram

For better understanding, we begin by drawing a force diagram representing the body on the inclined plane. Mark the gravitational force acting downward (mg), the normal force perpendicular to the inclined plane (N), and the force exerted by the inclined plane on the body (F). The angle between the inclined plane and the horizontal is \(\theta\).
02

Resolve gravitational force into components

Represent the gravitational force mg as two components: one perpendicular to the inclined plane (mg cosθ) and one parallel to the plane (mg sinθ).
03

Balance forces acting on the body

As the body is stationary, the forces acting on the body must be balanced. So, the component of the force acting parallel to the inclined plane must be equal to the force exerted by the inclined plane on the body. This can be expressed as: \(F = mg \sin \theta\)
04

Choose the correct option

According to our analysis, the force exerted by the inclined plane on the body is: \(F = mg \sin \theta\) Hence, the correct answer is (B) \(mg \sin \theta\).

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Key Concepts

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

Gravitational Force
Gravitational force is a fundamental concept in physics, describing the pull an object experiences due to the mass of another object. In our exercise, this force acts on a body resting on an inclined plane. The gravitational force, represented by \(mg\), acts vertically downward where \(m\) is the mass of the object and \(g\) is the acceleration due to gravity.
On an inclined plane, the gravitational force can be resolved into two useful components that help us understand how they affect the object's motion: one is perpendicular to the plane (\(mg \cos \theta\)), and the other is parallel (\(mg \sin \theta\)).
Understanding gravitational forces is essential as it explains why objects do not slide down an inclined plane unless a sufficient perpendicular force is applied to counteract the friction, or parallel forces exceed static frictional forces.
Force Resolution
Force resolution involves breaking down a single force into multiple components. These components are typically perpendicular to each other, and their vector sum equals the original force.
In the case of an inclined plane, we resolve the gravitational force \(mg\) into two components because it provides a clearer understanding of how the forces are working in different directions.
  • One component is parallel to the plane (\(mg \sin \theta\)) and is responsible for trying to pull the object downwards along the plane.
  • The other component is perpendicular to the plane (\(mg \cos \theta\)) and contributes to the normal force that balances the perpendicular forces.
Force resolution is crucial in analyzing such problems because it allows us to consider the effects of forces in different directions independently, facilitating better understanding and solution of the problem.
Newton's Laws of Motion
Newton's laws of motion are pronounced in understanding inclined plane problems. They provide the framework for analyzing how forces interact and influence the motion of objects. In this scenario, Newton's first and second laws are particularly relevant.

By Newton's first law, an object at rest will remain at rest unless acted upon by an unbalanced force. For the body on the inclined plane to remain stationary, the forces along the plane must balance. This is achieved when the gravitational force component parallel to the plane \(mg \sin \theta\) is balanced by the frictional force or any other force acting in the opposite direction.

Newton's second law relates the net force acting on an object to its mass and acceleration (\(F = ma\)). In this scenario, because the object is stationary, the acceleration is zero and hence the net force is also zero, confirming that all forces acting along the plane and perpendicular to it are balanced. Understanding these interactions helps us determine that the magnitude of force exerted by the inclined plane is \(F = mg \sin \theta\), ensuring the object remains stationary.

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Most popular questions from this chapter

A block rests on a rough inclined plane making an angle of \(30^{\circ}\) with the horizontal. The co-efficient of static friction between the block and the plane is \(0.8\). If the frictional force on the block is \(10 \mathrm{~N}\), the mass of the block (in \(\mathrm{kg}\) ) is (Taking \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) ) (A) \(2.0\) (B) \(4.0\) (C) \(1.6\) (D) \(2.5\)

Three identical blocks of masses \(m=2 \mathrm{~kg}\) each are drawn by a force \(F=10.2 \mathrm{~N}\) with an acceleration of \(0.6 \mathrm{~m} / \mathrm{s}^{2}\) on a surface, then what is the tension (in N) in the string between \(B\) and \(C\), if there is no friction between the surface and the blocks \(A\) and \(B\) (A) \(9.2\) (B) \(3.4\) (C) 4 (D) \(9.8\)

A girl of mass \(50 \mathrm{~kg}\) stands on a measuring scale in a lift. At an instant, it is detected that the reading reduces to \(40 \mathrm{~kg}\) for a while and then returns to original value. It can be said that (A) The lift was in constant motion upwards (B) The lift was in constant motion downwards (C) The lift was suddenly started in downward motion (D) The lift was suddenly started in upward motion

A block of \(10 \mathrm{~kg}\) is pulled by a constant speed on a rough horizontal surface by a force of \(19.6 \mathrm{~N}\). The co-efficient of friction is (A) \(0.1\) (B) \(0.2\) (C) \(0.3\) (D) \(0.4\)

A constant retarding force of \(50 \mathrm{~N}\) is applied to a body of mass \(20 \mathrm{~kg}\) moving initially with a speed of \(15 \mathrm{~ms}^{-1}\). How long does the body take to stop? (A) \(2 \mathrm{~s}\) (B) \(4 \mathrm{~s}\) (C) \(6 \mathrm{~s}\) (D) \(8 \mathrm{~s}\)

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