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

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Chapter 4: Problem 10

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 force exerted by the base of the prism on the horizontal plane is (a) \(M g\) (b) \(m g\) (c) \(M g+m g \sin ^{2} \alpha\) (d) \(M g+m g \cos ^{2} \alpha\)

Short Answer

Expert verified
(d) \( M g + m g \cos^2 \alpha \)

Step by step solution

01

Understanding the Forces

The prism is resting on a horizontal plane, and a body slides down along the inclined plane of the prism. The prism does not move horizontally because of the vertical wall at the rear edge. Hence, the normal force exerted by the prism on the horizontal plane comes from the weight of the prism and the components of the weight of the sliding body.
02

Analyzing the Body on the Incline

The weight component of the body perpendicular to the inclined plane is given by the force of gravity on the mass, which is \[ m g \cos \alpha \]. This force acts normal to the inclined plane and contributes to the normal force exerted by the prism on the horizontal plane.
03

Considering the Reaction Forces

The horizontal force component that acts because of the sliding body is \[ m g \sin \alpha \]. The prism has a vertical wall, ensuring it's stationary horizontally, so this force is balanced internally without causing an overall horizontal force on the ground.
04

Calculating the Normal Force on the Ground

The normal force exerted by the ground on the prism consists of the gravitational force of the prism \( M g \) plus the component of the gravitational force of the body acting perpendicular to the incline \( m g \cos^2 \alpha \). Hence, the total force the horizontal plane exerts on the surface is:\[ M g + m g \cdot \cos^2 \alpha \].
05

Selecting the Correct Option

Among the given options, the expression \( M g + m g \cos^2 \alpha \) corresponds to option (d).

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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, slanting surface, or slope. When an object is placed on this surface, the inclination or slant causes gravitational force to break into components. One component acts perpendicular to the plane, while the other acts parallel, causing the object to potentially slide downwards.
Inclined planes make it easier to move objects to higher or lower positions without directly lifting them.
In our exercise, the inclined plane affects how the gravitational force on the sliding body (of mass \(m\)) is distributed:
  • The gravitational force contributes to both sliding motion down the incline and pressing action into the plane, forming components \(m g \sin \alpha\) and \(m g \cos \alpha\) respectively.
  • The angle of the incline (\(\alpha\)) significantly determines the amount of force in each component.
Understanding this distribution is key to analyzing forces at play in problems involving inclined planes.
Gravitational Force
Gravitational force is one of the most fundamental forces in physics. It's the force of attraction between two objects with mass. The gravitational force on an object on Earth is commonly represented as \(m g\), where \(m\) is the mass of the object and \(g\) is the acceleration due to gravity (approximately \(9.81 \text{ m/s}^{2}\) on Earth).
In this exercise, the gravitational force acts on both the sliding body and the prism itself, playing a crucial role:
  • For the sliding body, gravitational force causes it to exert pressure on the inclined plane, represented by \(m g \cos \alpha\).
  • It also pulls the prism and the body downward, exerting a weight of \(M g\) and \(m g\) on their respective surfaces.
These forces are vital in calculating other forces in the system, especially the normal force exerted by the ground.
Forces in Physics
Forces in physics refer to any interaction that changes the motion of an object. An object generates force on another when they interact, either by contact or at a distance. Examples include gravitational, electrostatic, and magnetic forces.
In our inclined plane scenario, several forces need consideration:
  • The gravitational force: Acts downward on both the prism and the mass sliding on it.
  • The normal force: Acts perpendicular to surfaces at contact. It supports weight components and balances forces perpendicularly.
  • Frictional forces: These may exist on inclined planes, although not mentioned in the given exercise, they typically oppose motion.
Forces are vector quantities. They have magnitude and direction, necessitating the use of vector addition to solve related physics problems. Understanding these forces and their interplay aids in analyzing systems like the inclined plane and calculating actual physical scenarios effectively.

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Most popular questions from this chapter

An inextensible string \(A B\) is tied to a block \(B\) of negligible dimensions and passes over a small pullcy \(C\) : so that the frec end \(A\) hangs \(h_{1}\) unil above the ground on which the block \(B\) rests. In this initial position shown in figure, the frec cnd \(A\) is \(h\) unit below \(C\). If now the end \(A\) moves horizontally with a velocity \(u\), obtain an cxpression for the velocity of the block at any time \(t\). Solution In time \(t\), the end \(A\) moves to the position \(A_{1}\). So that \(A A_{1}=u l .\) The block \(B\) moves upwards to the position \(B_{1}\). Let \(B B_{1}=y\). 'Ihen length \(A_{1} C=h+y \quad\) ln the \(\Delta A C A_{1},(h+y)^{2}=h^{2}+(u i)^{2}\) \(\begin{array}{llll}\text { or } h^{2}+y^{2}+2 h y=h^{2}+u^{2} t^{2} & \therefore & y^{2}+2 h y-u^{2} t^{2}=0 & \ldots(1)\end{array}\) \Lambdafter solving equation (1), we get \(y--h+\sqrt{h^{2}+u^{2} t^{2}}\) Ihis is the equation for the displacement \(y\) of the block. Velocity of the block: \(v-\frac{d y}{d t}-\frac{d}{d t}\left[h \mid\left(h^{2} \mid u^{2} t^{2}\right)^{1 / 2}\right]-\frac{1}{2}\left(h^{2} \mid u^{2} t^{2}\right)^{-1 / 2} \times 2 u^{2} t\) or \(\quad v-\frac{u^{2} t}{\left(h^{2}+u^{2} t^{2}\right)^{1 / 2}}\)

Three blocks \(A, B\) and \(C\) of masses \(2 \mathrm{~kg}, 3 \mathrm{~kg}\) and \(4 \mathrm{~kg}\) are placed as shown. Coefficient of friction between \(A\) and \(B\) is \(0.5\) and that between \(B\) and \(C\) is \(0.1\). The surface is frictionless. The maximum force \(F\) that can be applied horizontally onto \(A\) such that the three blocks move together is (a) \(12.22 \mathrm{~N}\) (b) \(13 \mathrm{~N}\) (c) \(11.25 \mathrm{~N}\) (d) None

A t /=0\(, a variable force varying with respect to time as \)F=6 t-2 t^{2}\( starts acting on a \)2 \mathrm{~kg}\( body which is initially at rest \)(F\( is in Newton and \)t\( is in sec.). When the body comes to rest again (we can see that at \)t=0, F=0\( ) the force is withdrawn. What is the time when the velocity of the body is maximum? (a) \)2 \mathrm{~s}\( (b) \)3 \mathrm{~s}\( (c) \)3.5 \mathrm{~s}\( (d) \)4.5 \mathrm{~s}$

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 value of the tension produced in the sting is (a) \(m g\) (b) \(m \sqrt{g^{2}-a^{2}}\) (c) \(m \sqrt{g^{2}+a^{2}}\) (d) \(m \sqrt{g^{2}+2 a^{2}}\)

Statement-1 : \(\Lambda\) body can be pulled with the least effort if it is pulled at an angle equal to the angle of friction from the surface. Statement-2 : If coefficient of static friction is \(\mu\), the angle of friction is tan \({ }^{1}(\mu)\).
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