Newton’s laws The lift problem Case 1 : Lift at rest

Slides:

Advertisements

Similar presentations

Newton’s Universal Law of Gravitation – Practice Problems

Advertisements

 The force that act on the object are balanced in all direction.  The force cancel each other, so that the resultant force or net force is zero.  Newton’s.

Someone of mass 100 kg is standing on top of a vertical cliff with only an old rope that he knows will support no more than 500 N. His plan is to slide.

Ropes and Pulleys.

More on Newton’s 3 rd Law. Conceptual Example 4-4: What exerts the force to move a car? Response: A common answer is that the engine makes the car move.

 Friction opposes motion  Friction is dependent on the texture of the surfaces  Friction is dependent on normal force motionfriction.

Newton’s 2 nd Law Of Motion By Mr. Yum. Newton’s 2 nd Law Newton’s 2 nd Law is defined as: Newton’s 2 nd Law is defined as: F = m x a Where, F = force,

Apparent Weight. Acceleration of Gravity  Objects that fall to the Earth all experience an acceleration.  The acceleration due to gravity is g = 9.8.

ELEVATOR PHYSICS.

Universal Law of Gravitation, Normal Force, Apparent Weight SPH3U.

Aim: How can we explain the motion of elevators using Newton’s 2 nd law? Do Now: What is the acceleration of this object? m = 20 kg F = 150 N F = 100 N.

Newton’s 2 nd Law. Force on Object Objects acted on by a net unbalanced force will accelerate in the direction of the force This means they will speed.

Science Starter! John weighs 735 N on Earth. (a) What is John’s mass in kilograms? (b) What would John’s mass be on Mars where g = 3.8 m/s 2 ? (c) What.

Newton’s Laws of Motion 1. If the sum of all external forces on an object is zero, then its speed and direction will not change. Inertia 2. If a nonzero.

Newton’s Second Law In this chapter we investigate the effect of a net force on a mass. Newton’s Second Law is: Whenever an unbalanced force acts on a.

Physics Unit Four Forces that Affect Motion. Force A push or a pull. Measured in newtons with a spring scale. 1 newton (N) = 1 kg m/s 2 An apple weighs.

A Force Caused by Gravity.  If we apply Newton’s 2 nd Law to objects accelerating due to gravity F G = mg  F G – force of gravity, commonly called.

1 Higher Still Multiple Choice Mechanics: Forces.

What is a Force? A force is a push or a pull causing a change in velocity or causing deformation.

Forces Newton’s Second Law.

Sir Isaac Newton Newton’s Laws of Motion Newton’s 1st Law of Motion -An object at rest, will remain at rest, unless acted upon by an unbalanced.

Objectives  Describe how the weight and the mass of an object are related.  Differentiate between actual weight and apparent weight.

Vertical Acceleration – Ex kg A 5.0 kg mass hangs on a string with a tension of 65 N. What is the acceleration of the mass?

Newton's Laws of Motion 1. Newton 1 st law of motion 2. Newton 3 rd law of motion 3. Newton 2 nd law of motion.

Newton’s Second Law Aims: To know Newton’s second law. To be able to use it to solve problems.

Problems Involving Forces

The tendency of objects to resist change in their state of motion is called inertia  Inertia is measured quantitatively by the object's mass.  Objects.

Apparent Weight The weight of an object is the force of gravity on that object. Your sensation of weight is due to contact forces supporting you. Let’s.

Weight & Normal Force Weight  The force of gravity on an object.

 Friction – force that opposes motion  Caused by microscopic irregularities of a surface  The friction force is the force exerted by a surface as an.

Aim: More Law of Acceleration Solve for the normal force on the following diagram: 10 kg 2 N N mg ΣF = 0 2 N + N – mg = 0 N = mg – 2 N N = (10 kg)(10 m/s.

SECTION Madison Evans, Mitchell Peters, Ryder Briggs.

Forces in One Dimension Chapter 4 Physics Principles and Problems Zitzewitz, Elliot, Haase, Harper, Herzog, Nelson, Nelson, Schuler and Zorn McGraw Hill,

LESSON OBJECTIVE Practise Newton’s 2 nd law A car is using a tow bar to pull a trailer along a straight, level road. There are resisting forces R acting.

Weight vs. Apparent Weight Physics 11. Elevator:  When you enter the elevator and press the button, you feel the normal amount of your weight on your.

More on Newton’s 3 rd Law. Conceptual Example: What exerts the force to move a car? Response: A common answer is that the engine makes the car move forward.

7.3 Newton’s third law of motion Action and reaction The wall exerts an equal but opposite force on your hand. uThe pair of forces is called the action-reaction.

Section 2: Weight and Drag Force

Chapter 4.1 Notes Resistance (is futile!). ► Newton’s 1st law - Every object in motion stays in motion; Every object at rest stays at rest unless acted.

What is a force? An interaction between TWO objects. For example, pushes and pulls are forces. We must be careful to think about a force as acting on one.

Application of Forces Objective: To apply Newton’s Laws of motion to analyze accelerated motion as it applies to a vertically accelerated object and.

Apparent Weight. Apparent Weight of an object is the reading on a ___________ scale when that object is placed on it.

Force is part of an interaction

Key Areas covered Balanced and unbalanced forces. The effects of friction. Terminal velocity. Forces acting in one plane only. Analysis of motion using.

Vertical Acceleration – Ex. 1

Physics Support Materials Higher Mechanics and Properties of Matter

Aim: How can we apply Newton’s Second Law?

Apparent Weight.

Balanced and unbalanced forces

Newton’s Second Law 1.

Weight Fg = mg Fg=weight m=mass g=acceleration due to gravity Ex: Calculate the weight of a 12 kg toy cart. What is the difference between weight and.

Forces 2nd Law.

FORCES AND NEWTON’S LAWS OF MOTION

Notes 2.2: Newton’s 2nd Law of Motion

Newton‘s 3rd Law.

Simple applications Of 1st & 2nd Laws.

Newton’s Laws of Motion

Accelerating Frames.

Forces in One Dimension

Newton’s Second Law 1.

A 700 N person stands on a bathroom scale in an elevator

Engineering Mechanics

Weight, Mass, and the Dreaded Elevator Problem

Newton’s Second Law If all forces are in balance, object is at equilibrium and does accelerate Newton’s second law applies when forces are unbalanced;

A block of mass m resting on a horizontal

ELEVATOR –WEIGHT and NEWTON’S SECOND LAW

Apparent Weight.

Mechanics Chapter 3 Vertical Motion.

Using Newton’s Laws.

Presentation transcript:

Newton’s laws The lift problem Case 1 : Lift at rest Imagine a mass of m kg hung from a spring balance in a lift. The tension in the spring is given by the measurement on the scale of the balance. Case 1 : Lift at rest T F = ma : T = mg m Tension = weight mg

Case 2 :Lift accelerating upwards The lift problem Case 2 :Lift accelerating upwards T F = ma : T – mg = ma m  T = m( g + a ) a apparent weight = T = mg + ma  mass apparently weighs more mg

Case 3 :Lift accelerating downwards The lift problem Case 3 :Lift accelerating downwards T F = ma : mg - T = ma m  T = mg - ma a apparent weight = T = mg - ma  mass apparently weighs less mg

Case 4 :Lift moving up or down with constant speed The lift problem Case 4 :Lift moving up or down with constant speed T F = ma : T -mg = 0 m  T = mg a T = weight mg

Example 1 A lift and its passengers have a total mass of 500 kg. Find the tension in the cable supporting the lift if (i) the lift is at rest (ii) the lift is moving at constant speed (iii) the lift is accelerating upwards at 0.6 ms-2 (iv) the lift is accelerating downwards at 0.4 ms-2 Solution (i) At rest T – mg = 0 T – 500 x 9.8 = 0 T = 4900 N

(ii) Ascending at constant speed Constant speed or at rest will have the same tension. i.e. T = 4900 N (iii) Accelerating upwards T – mg = ma T – 500 x 9.8 = 500 x 0.6 gives T = 5200 N (iv) Accelerating downwards T – mg = m(-a) T – 500 x 9.8 = 500 x (-0.4) gives T = 4700 N

Example 2 A man of mass 80 kg stands o the floor of a lift that is accelerating at 2 ms-2. Find the normal reaction exerted on the man by the lift floor if the lift is (i) going up (ii) going down Solution (i) R – 80g = 80a : R – 784 = 160  R = 944 (ii) R – 80g = 80(-a) : R – 784 = -160  R = 624

Example 3 A dynamo of mass 1200 kg is placed in a cage of mass 600 kg, which is raised vertically by a cable from a crane. The tension in the cable is 18 000 N. Find the acceleration of the cage, and the contact force between the cage and the dynamo. 18 000 N For the cage-and-dynamo a ms-2 T – mg = ma :18000 – 17640 = 1800a So a = 0.2 17640 N For the dynamo alone R – 11760 = 1200 a 0.2 ms-2 R – 11760 = 1200 x 0.2 So R = 12000 N R 11760 N