1. NEWTON’S SECOND LAW

2. Newton’s Second Law states: The resultant force on a body is proportional to the acceleration of the body. In its simplest form: F = m a This is often referred to as the equation of motion. This can be applied in a number of situations:

3. Example 1: A body of mass 5 kg is acted on by 3 forces; F 1 = 3i + j, F 2 = 7i – 2j and F 3 = 5i + 6j. Find the magnitude and direction of the acceleration of the body. The resultant force, F = F 1 + F 2 + F 3 F = F = m a Using: 5a5a a = 3 1 a θ The magnitude of the acceleration, a = a = 3.16 ms -2. 1313 tan θ = 18.4º θ = The direction of motion is 18.4 º to the horizontal as shown.

4. The resultant force, F = + + = Example 2: A particle P of mass 0.2 kg is acted on by a force of 3N acting due North, a force of 7N acting due East, and a force of 8N at 43º to the horizontal as shown. Find the magnitude of the acceleration, and the direction of motion, giving the answer as a bearing. 3 43º 8 7 θ 1.149 2.456 F The magnitude of the force, F = F = 2.711 tan θ = 2.456 1.149 64.9º The bearing is 90 + θ =155º (To nearest degree). 2.711 = 0.2 a a = 13.6 ms -2. F = m a Using: θ =

5. Alternatively, note that, since F = F = m a Using: 0.2 a θ 5.745 12.28 a a = = 13.6 ms -2. tan θ = 12.28 5.745 θ = 64.9º a = The bearing is 90 + θ = 155º

6. Example 3: A particle of mass 0.5 kg is acted on by a force 3i – j. If the particle has velocity 2i ms -1 initially, find the velocity of the particle 5 seconds later. F = F = m a Using: 0.5a This is constant acceleration,so “ v = u + at ” + 5 The velocity is ms -1 a = v =

7. Example 4: A stone of mass 4 kg, sliding horizontally on ice decelerates uniformly from a speed of 2 ms -1 to rest, moving a distance 8 m. Find the resistive force acting on the stone. v = u = a = s = t = a N 4g R 2 0 8 Using “ v 2 = u 2 + 2as ” 0 = 4 + 2(a)(8) – 4 = 16a “F = ma” – R = 4a a = – 0.25 – R = 4(– 0.25) – R = – 1 R = 1 Resistive force = 1N The force diagram is: The motion is constant acceleration.

8. Example 5: A particle of mass 2 kg is sliding down a rough slope inclined at 35º to the horizontal, against a resistive force of 8 N. If the particle starts from rest, find its speed after 1.5 seconds. 35° a N 2g 8 2g sin35 – 8 =2a “F = ma” a = 0 1.621 v = u = a = s = t = 1.5 Using “ v = u + at ” v = 0 + (1.621) (1.5) v = 2.43 1.621 The speed is 2.43 ms -1. The force diagram is:

9. Summary of key points: This PowerPoint produced by R.Collins ; Updated Sep. 2008 In it’s simplest form, Newton’s Second Law states: F = m a This is often referred to as the equation of motion. Problems of this type often lead to constant acceleration questions. v = u + a t s u t + a t2a t2 1212 = v 2 = u 2 + 2as So the following are also needed: