Vector and scalar quantities

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Vector and scalar quantities

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Vector and scalar quantities A scalar quantity is defined only by its magnitude (or size) for example: distance, speed, time. It is easy to combine two or more scalar quantities e.g. 2 metres + 3 metres = 5 metres! A vector quantity is defined by stating both its magnitude (or size) and direction for example: displacement, velocity, force The direction must be taken into account when combining two or more vector quantities e. 2 metres + (-3 metres) = -1 metre

Introduction to Forces II Sharon Tripconey

Some Basic Principles Newton s first law (N1L) Every particle continues in a state of rest or uniform motion in a straight line unless acted on by a resultant external force. This means that for a particle to be in equilibrium* it must be the case that there is no resultant force acting on it (*dynamic or static equilibrium) Newton s second law (N2L) When a force acts on a particle, the change in momentum is proportional to the force. For constant mass, F = ma Newton s third law (N3L) When one object exerts a force on another there is always a reaction that is equal in magnitude and opposite in direction to the applied force. This means that we expect forces to be found in pairs

Draw a force diagram

W = Weight of the person Assumptions: Each object is a particle in equilibrium The scales and the brush have zero mass W

W Weight of the person F F S Force exerted on the brush Contact force (brush and floor) N Contact force (person and scales) F R Contact force (scales and floor) N R W N W S R S Hint: Think about pairs of forces (N3L)

Problem: Explain why the reading goes down when I press on the floor with a brush Consider the forces acting on the person F has the same magnitude as the force exerted on the brush N is the contact force (the reading on the bathroom scales) + W is the weight of the person (which is constant) F + N W = 0 F + N = W

Newton s Second Law In a given direction F = ma Example A toy train of mass 0.5 kg moves on a horizontal straight track. There is a driving force of 0.4 N and a resistance to motion of 0.35 N. Find the acceleration. 0.35 N 0.5 kg 0.4 N a m s -2

Newton s Second Law 0.35 N 0.5 kg 0.4 N a m s -2 N2 L Resultant force = ma 0.4 0.35 0.5a 0.05 = 0.5a a 0.1 The acceleration is 0.1 ms -2 in the direction of the 0.4 N force.

What happens if. Two identical objects are connected to the ends of a light inelastic string which passes over a fixed pulley as shown What happens if the system is released from rest?

Motion in a lift A boy of mass 50 kg is standing in a lift of mass 200 kg. The lift is raised by a vertical cable. Find the reaction of the floor on the boy and the tension in the cable when the acceleration is 1 g upwards. 4 boy (50kg) R lift (200kg) T lift and boy (250kg) T ¼ g ¼ g ¼ g 50g 200g R 250g

Motion in a lift For the boy, N2L R 50g 50 g so R 62.5g 1 4 For the lift and boy, N2L, T 250g = 250 1 4 g so T = 312.5g For the lift, N2L, T R 200g 200 g so T 312.5g 1 4

Mathematical modelling A model is a representation of a real situation. A real situation will invariably contain a rich variety of detail and any model of it will simplify reality by extracting those features which are considered to be most important. Modelling is at the heart of the subject of Mechanics

Modelling assumptions Common assumptions are: the body is a particle value of g is constant the string is light and inextensible the pulley is light and smooth air resistance is negligible friction is negligible friction obeys the law F µ R

Modelling Cycle Real-world problem Simplifying assumptions Mathematical model (equations etc) Experiment Prediction Analysis and solution

How can we support students with their understanding of mathematical modelling in Mechanics? Incorporate simple practical demonstrations into our teaching; many students quickly see something of the principle even if their depth of understanding is not great As teachers, knowledge of common student misconceptions helps us to plan lessons that will reduce the likelihood of their development and increase the likelihood that misconceptions already formed will be corrected. An approach of confronting hard things is usually more successful than that of avoiding them for as long as possible!

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