/*! 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 12 A body with a mass \(m\) slides ... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 12

A body with a mass \(m\) slides along the surface of a trihedral prism of mass \(M\), whose upper plane is inclined at an angle \(\alpha\) to the horizontal. The prism rests on a horizontal plane having a vertical wall at the rear edge of the prism to keep it at rest. The normal reaction on the block is (a) \(m g\) (b) \(M g\) (c) \(m g \cos \alpha\) (d) \(m g \sin \alpha \cos \alpha\)

Short Answer

Expert verified
The normal reaction on the block is \(mg \cos \alpha\) (option c).

Step by step solution

01

Understanding the Forces Involved

The body slides along the inclined plane of the prism, which means we need to consider the gravitational force acting on the body and how it relates to the normal force. The gravity force component perpendicular to the inclined plane affects this normal force.
02

Components of Gravitational Force

The gravitational force acting on the body of mass \(m\) is \(mg\). When considering the inclined plane, we decompose this force into two components: one perpendicular to the slope and one parallel. The perpendicular component (which affects the normal force) is given by \(mg \cos \alpha\).
03

Determine the Normal Reaction

The normal reaction is the force exerted by the inclined plane perpendicular to its surface to support the body. Since no additional vertical forces act on the body, the normal force must equal the perpendicular component of gravity, which is \(mg \cos \alpha\). This means that (c) is the correct option.

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Key Concepts

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

Inclined Plane
An inclined plane is a flat surface that is tilted at an angle to the horizontal. This simple machine is often used in physics to study the behavior of objects as they move along a slope. By adjusting the angle of the incline, we can observe different effects on motion and calculate forces involved.

When an object moves along an inclined plane, it doesn't just feel the force of gravity pulling it down. Instead, this force can be divided into two components, which we'll discuss in more detail later.
  • The normal force: this is the force exerted perpendicularly by the plane to support the object.
  • The gravity component: this force is parallel to the inclined plane and causes the object to slide down.
Inclined planes are not just theoretical; they have practical applications as well. They are used in everyday life to make lifting objects easier. For instance, ramps on trucks and buildings help reduce the effort needed to move heavy loads.
Gravitational Force Components
Whenever an object is placed on an inclined plane, gravity acts on it. The force due to gravity can be represented by the formula: \(mg\), where \(m\) is the mass of the object, and \(g\) is the acceleration due to gravity, approximately \(9.81 \, m/s^2\) on Earth.

However, this gravitational force can be split into two separate components:
  • Parallel Component: This component acts down the slope and is responsible for the object's potential sliding down the incline. It is expressed as \(mg \sin \alpha\).
  • Perpendicular Component: This part acts directly into the surface of the plane and is responsible for the normal force or the pressing force against the plane. It is expressed as \(mg \cos \alpha\).
Understanding these components is crucial. The normal force depends directly on the perpendicular component of gravity, balancing it to prevent the object from simply falling off the incline. Calculating these forces can help predict the motion of objects on various slopes.
Forces in Physics
In physics, forces are vectors, meaning they have both magnitude and direction. Understanding how they operate is pivotal in explaining how objects move or remain at rest. When dealing with forces, it's important to note how they interact with each other, particularly on systems like inclined planes.

Let's break down some key points:
  • Normal Force: This is a force perpendicular to surfaces in contact and acts to support objects resting or moving along a surface. In the case of the inclined plane, it acts perpendicular to the slope.
  • Frictional Force: Although not detailed in our specific exercise, it typically acts opposite to motion direction along the surface.
  • Net Force: This is the sum of all forces acting on an object. In a balanced scenario, it results in no acceleration. On an inclined plane without friction, the net force is primarily composed of the gravitational components.
Understanding these forces and how they interplay is essential to solving many problems in physics. Calculating each correctly can help us explain and even predict the behavior of objects in the physical world.

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Most popular questions from this chapter

A small sphere is suspended by a light string from the ceiling of a car and the car begins to move with a constant acceleration \(a\). The inclination of the string to the vertical is (a) \(\tan ^{1}(a / g)\) in the direction of motion (b) \(\tan ^{1}(a / g)\) opposite to the direction of motion (c) \(\tan ^{1}(g / a)\) in the direction of motion (d) tan \({ }^{1}(g / a)\) opposite to the direction of motion

Statement-1 : When a block moves on a rough horizontal surface with some specd, it cventually slows down. Statement-2 : Friction always opposes motion.

The cocflicient of friction between the block \(A\) of mass \(m\) and block \(B\) of mass \(2 m\) is \(\mu-\frac{1}{\sqrt{3}}\). The inclined plane is sinooth. If the system of blocks \(A\) and \(B\) is releascd from rest and there is no slipping between \(A\) and \(B\), then \(\theta<\frac{\pi}{\alpha}\). Find the value of \(\alpha\). Solution When there is no slipping, then both the blocks move together with acceleration \(a=g \sin \theta\), down the plane. I Iorizontal component of this acceleration is \(a_{H}=a \cos \theta\) and vertical component is \(a_{y}=a \sin \theta\), where \(a=g \sin \theta\) \(a_{H}=a \cos \theta=g \sin \theta \cdot \cos \theta\) and \(a_{V}=a \sin \theta=g \sin ^{2} \theta\) \(N=\) normal reaction between \(A\) and \(B\). Equations of motion in horizontal and vertical directions give: \(m g-N=m a_{p}\) or \(N=m g-m a_{v}=m g-m g \sin ^{2} \theta=m g\) \(f=\) friction force \(\mu N \geq m a_{H}\) or \(\mu m g \cos ^{2} \theta \geq m g \sin \theta \cos \theta \quad \Rightarrow \quad \mu \geq \tan \theta \quad\) or \(\theta \leq \tan ^{-1}(\mu)\) \(\Rightarrow \theta<\tan { }^{1}\left(\frac{1}{\sqrt{3}}\right) \Rightarrow \quad \theta<\frac{\pi}{6} \quad \therefore \alpha-6\)

A system consisting of two blocks kept onc over the other rests over a smooth horizontal surface. Somchow it is sct in motion so that the system of blocks acquires a constant velocity. Friction cocfficicnts between the two blocks is \(\mu(\mu \neq O)\) Statement-1 : Aflerwards, friction between the blocks is static in nature and non zero. Statement-2 : 'The lower block is in translational equilibrium.

In case of a central force (a) force is position dependent (b) Torque is zero (c) angular momentum is constant (d) force is conservative
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