Lecture-IV. Contact forces & Newton s laws of motion



Similar documents
Chapter 4. Forces and Newton s Laws of Motion. continued

Chapter 5 Using Newton s Laws: Friction, Circular Motion, Drag Forces. Copyright 2009 Pearson Education, Inc.

circular motion & gravitation physics 111N

Serway_ISM_V1 1 Chapter 4

VELOCITY, ACCELERATION, FORCE

At the skate park on the ramp

Physics 1A Lecture 10C

PHY231 Section 2, Form A March 22, Which one of the following statements concerning kinetic energy is true?

Chapter 9 Circular Motion Dynamics

Physics 11 Assignment KEY Dynamics Chapters 4 & 5

If you put the same book on a tilted surface the normal force will be less. The magnitude of the normal force will equal: N = W cos θ

v v ax v a x a v a v = = = Since F = ma, it follows that a = F/m. The mass of the arrow is unchanged, and ( )

Lecture 6. Weight. Tension. Normal Force. Static Friction. Cutnell+Johnson: , second half of section 4.7

PHY231 Section 1, Form B March 22, 2012

Midterm Solutions. mvr = ω f (I wheel + I bullet ) = ω f 2 MR2 + mr 2 ) ω f = v R. 1 + M 2m

Physics 111: Lecture 4: Chapter 4 - Forces and Newton s Laws of Motion. Physics is about forces and how the world around us reacts to these forces.

LAB 6 - GRAVITATIONAL AND PASSIVE FORCES

Experiment: Static and Kinetic Friction

PHYSICS 111 HOMEWORK SOLUTION #10. April 8, 2013

Columbia University Department of Physics QUALIFYING EXAMINATION

C B A T 3 T 2 T What is the magnitude of the force T 1? A) 37.5 N B) 75.0 N C) 113 N D) 157 N E) 192 N

Physics: Principles and Applications, 6e Giancoli Chapter 4 Dynamics: Newton's Laws of Motion

STATIC AND KINETIC FRICTION

B) 286 m C) 325 m D) 367 m Answer: B

Chapter 6. Work and Energy

5. Forces and Motion-I. Force is an interaction that causes the acceleration of a body. A vector quantity.

Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx Acceleration Velocity (v) Displacement x

W02D2-2 Table Problem Newton s Laws of Motion: Solution

Physics 2A, Sec B00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam

LAB 6: GRAVITATIONAL AND PASSIVE FORCES

Problem Set 5 Work and Kinetic Energy Solutions

Problem Set 1. Ans: a = 1.74 m/s 2, t = 4.80 s

8.012 Physics I: Classical Mechanics Fall 2008

TEACHER ANSWER KEY November 12, Phys - Vectors

Conceptual Questions: Forces and Newton s Laws

AP Physics C. Oscillations/SHM Review Packet

B Answer: neither of these. Mass A is accelerating, so the net force on A must be non-zero Likewise for mass B.

Solution Derivations for Capa #11

Angular acceleration α

AP Physics Applying Forces

3 Work, Power and Energy

Physics 9e/Cutnell. correlated to the. College Board AP Physics 1 Course Objectives

PHYSICS 111 HOMEWORK SOLUTION, week 4, chapter 5, sec 1-7. February 13, 2013

PHY121 #8 Midterm I

F N A) 330 N 0.31 B) 310 N 0.33 C) 250 N 0.27 D) 290 N 0.30 E) 370 N 0.26

D Alembert s principle and applications

Kinetic Friction. Experiment #13

A Determination of g, the Acceleration Due to Gravity, from Newton's Laws of Motion

Two-Body System: Two Hanging Masses

AP Physics Circular Motion Practice Test B,B,B,A,D,D,C,B,D,B,E,E,E, m/s, 0.4 N, 1.5 m, 6.3m/s, m/s, 22.9 m/s

Lecture 17. Last time we saw that the rotational analog of Newton s 2nd Law is

Mechanics 1: Conservation of Energy and Momentum

8. Potential Energy and Conservation of Energy Potential Energy: When an object has potential to have work done on it, it is said to have potential

Lecture 07: Work and Kinetic Energy. Physics 2210 Fall Semester 2014

Spring Simple Harmonic Oscillator. Spring constant. Potential Energy stored in a Spring. Understanding oscillations. Understanding oscillations

WORK DONE BY A CONSTANT FORCE

Lecture L22-2D Rigid Body Dynamics: Work and Energy

When showing forces on diagrams, it is important to show the directions in which they act as well as their magnitudes.

Problem Set #8 Solutions

Determination of Acceleration due to Gravity

CHAPTER 6 WORK AND ENERGY

SOLID MECHANICS TUTORIAL MECHANISMS KINEMATICS - VELOCITY AND ACCELERATION DIAGRAMS

Kinetic Friction. Experiment #13

Physics 201 Homework 8

FRICTION, WORK, AND THE INCLINED PLANE

Ideal Cable. Linear Spring - 1. Cables, Springs and Pulleys

Review Chapters 2, 3, 4, 5

There are four types of friction, they are 1).Static friction 2) Dynamic friction 3) Sliding friction 4) Rolling friction

Chapter 6 Circular Motion

AP Physics - Chapter 8 Practice Test

Newton s Laws of Motion

Objective: Equilibrium Applications of Newton s Laws of Motion I

Weight The weight of an object is defined as the gravitational force acting on the object. Unit: Newton (N)

Recitation Week 4 Chapter 5

AP Physics: Rotational Dynamics 2

Experiment 5 ~ Friction

Chapter 18 Static Equilibrium

11. Rotation Translational Motion: Rotational Motion:

3600 s 1 h. 24 h 1 day. 1 day

University Physics 226N/231N Old Dominion University. Getting Loopy and Friction

Lab 7: Rotational Motion

Fric-3. force F k and the equation (4.2) may be used. The sense of F k is opposite

AP Physics C Fall Final Web Review

Review D: Potential Energy and the Conservation of Mechanical Energy

ENERGY CONSERVATION The First Law of Thermodynamics and the Work/Kinetic-Energy Theorem

COEFFICIENT OF KINETIC FRICTION

PHYS 101 Lecture 10 - Work and kinetic energy 10-1

SOLID MECHANICS DYNAMICS TUTORIAL CENTRIPETAL FORCE

Newton s Law of Motion

Name: Partners: Period: Coaster Option: 1. In the space below, make a sketch of your roller coaster.

Chapter 6 Work and Energy

Physics 211 Lecture 4

Chapter 3.8 & 6 Solutions

Physics 125 Practice Exam #3 Chapters 6-7 Professor Siegel

Simple Harmonic Motion

Classical Physics I. PHY131 Lecture 7 Friction Forces and Newton s Laws. Lecture 7 1

Forces. Definition Friction Falling Objects Projectiles Newton s Laws of Motion Momentum Universal Forces Fluid Pressure Hydraulics Buoyancy

Answer, Key Homework 7 David McIntyre Mar 25,

Chapter 4: Newton s Laws: Explaining Motion

Name Class Period. F = G m 1 m 2 d 2. G =6.67 x Nm 2 /kg 2

Transcription:

Lecture-IV Contact forces & Newton s laws of motion

Contact forces: Force arises from interaction between two bodies. By contact forces we mean the forces which are transmitted between bodies by short-range atomic or molecular interactions. Examples: push, pull, tension of a string, the normal force, the force of friction, etc. The origin of these forces can be explained in terms of the fundamental properties of matter. However, our approach will emphasize the empirical properties of these forces and the techniques for dealing with them in physical problems, with only brief mention of their microscopic origins.

Tension in a string: A string consists of long chains of atoms characteristic of the particular material found in the string. When a string is pulled, we say it is under tension. The long chains of molecules are stretched, and inter-atomic forces between atoms in the molecules prevent the molecules from breaking apart. A detailed microscopic description would be difficult and unnecessary for our purpose. Instead, a macroscopic model would be developed for the behavior of strings under tension. Consider a block of mass M in free space pulled by a string of mass m. A force F is applied to the string, as shown. What is the force that the string transmits to the block? String is inextensible: a S =a M. By Newton's third law: F 1 =F 1 The forceon the block is less than F; the string does not transmit the full applied force. If the mass of the string is negligible compared with the block,f 1 =Ftogoodapproximation.

A string is composed of short sections interacting by contact forces. Each section pulls the sections to either side of it, and by Newton's third law, it is pulled by the adjacent sections. The magnitude of the force acting between adjacent sections is called tension.thereisnodirectionassociatedwithtension.inthesketch,thetensionatais FandthetensionatBisF'. Although a string may be under considerable tension, if the tension is uniform, the net string force on each small section is zero and the section remains at rest unless external forces act on it. If there are external forces on the section, or if the string is accelerating, the tension generally varies along the string.

Dangling rope: A uniform rope of mass Mand length Lhangs from the limb of a tree. Find the tension a distance x from the bottom. At the bottom of the rope the tension is zero, while at the top the tension equals the total weight of the rope Mg.

Whirling rope: AuniformropeofmassMandlengthLispivotedatoneendandwhirlswithuniform angular velocity ω. What is the tension in the rope at distance r from the pivot? Neglect gravity. θˆ 2 a = ( r r θ ) rˆ + ( r θ + 2 r θ ) ˆ θ; r dr ˆr 2 r = r = 0, dm θ = 0 a = rω rˆ 2 M dt ( r) = rω dm; dm = ρdr = dr L T(r) Mω 2 L/2 r

Pulleys: When a pulley is used to change the direction of a rope under tension, there is a reaction force on the pulley. The force on the pulley depends on the tension and the angle through which the rope is deflected. 2T sinθ 0 Thus the element exerts an inward radial force of magnitude T θ on the pulley.

The normal force and friction: Rest your hand on a table; the atoms that form the molecules that make up the table and your hand are in contact with each other. If you press harder, the atoms are also pressed closer together. The electrons in the atoms begin to repel each other and your handispushedintheoppositedirectionbythetable. According to Newton s Third Law, the force of your hand on the table is equal in magnitudeandoppositeindirectiontotheforceofthetableonyourhand. The total force of the table acting on yourhand,, is called the contact force. This force has both a normal component to the surface,, called the normal force, and a tangential component to the surface,, called the friction force. and then are not independent forces.

Normal force and weight: Hold an object in your hand. There are two forces acting on the object. One force is the gravitation force between the earth and the object, and is given by, where is the gravitational acceleration. The other force on the object is the contact force between your hand and the object. Since we are not pushing the block horizontally, this contact force on your hand points perpendicular to the surface, and hence has only a normal component, By Newton s 2 nd law:, Though in this particular example, the normal force is equal in magnitude to the gravitational force and opposite in direction, they are NOT third Law pair. The normal force and the gravitation force are two completely different forces.

Friction: Friction arises when the surface of one body moves, or tries to move, along the surface of a second body. Themaximumvalueofthefrictionisfoundtobe f friction = µn wherenisthenormalforceand µisthecoefficientoffriction. When a body slides across a surface, the friction force is directed opposite to the instantaneous velocity and has magnitude µn. The force of sliding friction is slightly lessthantheforceofstaticfriction,butforthemostpartweshallneglectthiseffect. For two given surfaces, the force of sliding friction is independent of the area of contact. M M f f

Frictionisindependentoftheareaofcontactbecausetheactualareaofcontacton anatomicscaleisaminutefractionofthetotalsurfacearea. Friction occurs because of the interatomic forces at these minute regions of atomic contact. Non rigid bodies, like automobile tires, are more complicated. A wide tire is generally better than a narrow one for good acceleration and braking. Frictional force is also independent of relative velocity between two surfaces. This is approximately true for a wide range of low speeds, as the speed increases This is approximately true for a wide range of low speeds, as the speed increases and air friction is encountered, it is found that the friction not only depends on the speed, but upon the square and sometimes higher powers of the speed.

The spinning terror: Therefore, or or

Block on Block A block of mass m is kept on another block of mass M (M>m) and the assembly is placed on a surface. The coefficients of static and kinetic frictions between the lower block and the surface are µ 1s, µ 1k respectively and those between the surface of the two blocks are µ 2s, µ 2k respectively. The lower block is pulled by a force F which is increased from zero till the blocks are in contact. Plot the accelerations of both the blocks as functions of F. µ µ 2s, µ 2k m 2s > µ 1s µ 1s, µ 1k M No motion: M max 1s ( ) F = f = µ M + m g µ 2k > µ 1k Starts sliding: ( + ) = µ ( + ) F a M m F M m g ( M + m) 1k k F a = µ g, with a M = µ µ g For m: max ( ) 1k f 1s 1k f = µ mg, or a = µ g m max 2s m 2s The corresponding force is: ( + ) = k ( + ) = ( µ + µ )( + ) = µ g M m F µ M m g, or 2s 1 F M m g F 2s 1k 1 ma = µ mg, or a = µ g m 2k m 2k

For M: ( ) a M = F µ M + m g µ mg M 1k 2k F g or am = 1k ( M m) 2km M M µ + + µ g with am, F = µ ( ) 1 2s M + m µ 2km M