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Newton’s 2 nd Law

Newton’s 2 nd Law. particle. e.g.1 A pebble of mass 0 ·2 kg is dropped from the top of a cliff. Find the acceleration of the pebble as it falls, assuming that the pebble falls in a straight line and there is a constant resistance of 0·06 newtons. Solution:.

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Newton’s 2 nd Law

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  1. Newton’s 2nd Law

  2. particle e.g.1 A pebble of mass 0·2kg is dropped from the top of a cliff. Find the acceleration of the pebble as it falls, assuming that the pebble falls in a straight line and there is a constant resistance of0·06 newtons. Solution: Tip: When we are given mass rather than weight, it’s a good idea to use W = mg to replace W on the diagram. 0·2g W

  3. particle e.g.1 A pebble of mass 0·2kg is dropped from the top of a cliff. Find the acceleration of the pebble as it falls, assuming that the pebble falls in a straight line and there is a constant resistance of0·06 newtons. Solution: 0·06 a We don’t substitute for g at this stage since we need to see both mass and weight clearly in the diagram. 0·2g

  4. particle e.g.1 A pebble of mass 0·2kg is dropped from the top of a cliff. Find the acceleration of the pebble as it falls, assuming that the pebble falls in a straight line and there is a constant resistance of0·06 newtons. Solution: 0·06 N2L: a Resultant force = mass  acceleration 0·2g Beware! On the left-hand side of the equation we have force, so we need weight. On the right-hand side we have mass. We always resolve in the direction of the acceleration.

  5. 1·9 0·2 particle e.g.1 A pebble of mass 0·2kg is dropped from the top of a cliff. Find the acceleration of the pebble as it falls, assuming that the pebble falls in a straight line and there is a constant resistance of0·06 newtons. Solution: 0·06 N2L: a Resultant force = mass  acceleration a 0·2g 0·2  - 0·06= 0·2g 0·2 9·8- 0·06=0·2a   =a a = 9·5ms-2 

  6. Connected Particles and Newton’s 3rd Law The following example shows a pulley problem. It covers the methods used to find the forces and acceleration of the system, including the force on the pulley.

  7. Connected Particles and Newton’s 3rd Law The following example shows a pulley problem. It covers the methods used to find the forces and acceleration of the system, including the force on the pulley.

  8. A B e.g.2. A particle, A, of mass 0·6kg, is held at rest on a smooth table. A is connected by a light, inextensible string, which passes over a smooth fixed pulley at the edge of the table, to another particle, B, of mass 0·4kg hanging freely. The string is horizontal and at right angles to the edge of the table. A is released. Find (a) the magnitude of the reaction of the table on A, (b) the acceleration of the system (c) the tension in the string, and (d) the force on the pulley due to the particles.

  9. a R A: mass 0·6kg B: mass 0·4kg T A T 0·6g a B 0·4g Solution: N2L: Resultant force = mass  acceleration (a) Find R. - 0·6g R =0·6g  A: R = 0 - - - - (1) R =5·88 newtons ( 3 s.f. )  (b) Find a. A accelerates horizontally, so there is no vertical component of acceleration. 0·6a - - - - (2) = A: T - - - - (3) = - 0·4g 0·4a T B:

  10. a = 0·4 9·8  a R A: mass 0·6kg B: mass 0·4kg T A T 0·6g a R =5·88N B 0·4g 0·6a - - - - (2) = (b) Find a. T = - 0·4g 0·4a T - - - - (3) (2) + (3): a = 0·4g a = 3·92ms-2  (c) Find T. Substitute in (2): T= 0·6  3·92 T= 2·35 newtons ( 3 s.f. ) 

  11. T T 3·92ms-2 a= T= 2·35N T T a R A: mass 0·6kg B: mass 0·4kg T A T 0·6g a B 0·4g (d) Find the force on the pulley due to the particles. Since the 2 forces are equal, the resultant lies between them, at an equal angle to each. 45 What can you say about the direction of the resultant? F F=Tcos45 + Tcos45 = 2Tcos45  F= 3·33newtons ( 3 s.f. ) Reminder: Use the value of T from your calculator so that your answer will be correct to 3s.f.

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