Forces Revision Newton’s First Law. An object will remain at rest or continue to move with constant velocity unless acted on by an unbalanced force. 22/09/2018

Forces Revision Newton’s Second Law. When an unbalanced force acts on an object it will accelerate. The rate of this acceleration is given by; Where; Fun = Unbalanced Force (N) m = mass (kg) a = acceleration (ms-2) 22/09/2018

Examples of Newton’s First Law; A car being driven at constant velocity ~ forces balanced are the drive force of the car and friction. A ship floating in the water ~ forces balanced are the weight of the ship and the upthrust caused by the water. A parachutist who has reached “Terminal Velocity”. 22/09/2018

Examples of Newton’s Second Law; A car moving off from rest – Drive Force >> Friction. A ship sinking – Weight >> Upthrust A parachutist who is slowing down after he/she opens the parachute. Air Resistance >> Weight 22/09/2018

More About Parachutists The graph below shows how the velocity of a parachute jumper varies with time. Velocity (ms-1) Time (s) Freefall 1st Terminal Velocity Parachute Opens 2nd Terminal Velocity lands 22/09/2018

Explanation of Graph Freefall – Weight >> Air Resistance but Air Resistance increases the faster the parachutist falls. 1st Terminal Velocity - Weight = Air Resistance, no unbalanced force so constant velocity. 22/09/2018

Explanation of Graph Parachute Opens – Air Resistance >> Weight but Air Resistance decreases as parachutist slows down. Landing - Weight << Upwards Force due to ground so accelration to rest. Note – We will learn why parachutists bend their knees on landing when we do Impulse. 22/09/2018

Definition of The Newton One newton (N) is defined as the unbalanced force required to accelerate a mass of 1 kilogram (kg) with an acceleration of 1 metre per second per second (ms-2) 1N Acceleration = 1ms-2 1kg 22/09/2018

Example 1 A rocket of mass 10 000kg is required to lift off with an acceleration of 3ms-2 . Find; a) Find the size of thrust the engines must produce. b) In reality will this acceleration increase, decrease or remain constant as the rocket climbs? 22/09/2018

Thrust Unbalanced Force = Thrust + Weight Mass = 10 000kg weight = mg = 10 000 x 9.8 = 98 000N 22/09/2018

Fun = Thrust + Weight Also; So we now have; 22/09/2018

This of course assumes a constant Thrust force from the engines. In reality the rate of acceleration will increase as the mass of the rocket will decrease as it burns fuel. The unbalanced force will therefore also increase as F = T – W and weight is getting smaller. So we have a = F/m with F getting bigger and m getting smaller the higher it goes. This of course assumes a constant Thrust force from the engines. 22/09/2018

Example 2 A gardener pulls a roller of mass 50kg at an angle of 45 with a force of 141.42N. If the force of friction on the roller is a constant 90N calculate the acceleration. 141.42 N 90 N 45 Mass = 50kg 22/09/2018

141.42 cos 45 N 90N Mass = 50kg (141.42 cos 45) – 90 = 10N 22/09/2018

Example 3 – Accelerating Systems Wile Coyote (mass 35kg) decides to lower, over a cliff, an anvil (mass 25kg) using a pulley wheel and rope onto Road Runners head. However Wile forgets to take off his frictionless roller blades and finds himself being accelerated towards the edge of the cliff as the anvil falls. Find the tension force in the rope. 22/09/2018

Pulley 25kg 35kg 22/09/2018

The unbalanced force is provided by the weight of the anvil. The system has a mass of 60kg (anvil + Wile). System mass = (35kg (Wile) + 25kg (anvil)) = 60kg Fun = W = mg (anvil) = 25 x 9.8 = -245N 22/09/2018

Firstly, calculate the acceleration of the system Firstly, calculate the acceleration of the system. This is found by considering the TOTAL mass of the system and the accelerating force , gravitational force acting on the anvil. 22/09/2018

The tension is the force that causes Wile to accelerate at 4.08 m s-2. To find the tension in the rope we need to consider the forces around a single part of the system, either will provide the answer. Consider Wile first; Fun (tension) Mass = 35kg a = 4.08ms-2 The tension is the force that causes Wile to accelerate at 4.08 m s-2. Fun = m x a = 35 x 4.08 = 142.8 N 22/09/2018

The same answer can be obtained by considering the anvil, this is slightly more complicated; Tension a = -4.08ms-2 Mass = 25kg Weight = mg = 25x9.8 = -245N 22/09/2018

Example 4 – Choo Choo Type A = 4kg B = 3kg C = 2kg Driving Force = 27N 3N 3N 3N Frictional force Find the tension in the link between A and B. 22/09/2018

Step 1; we find the acceleration of the system. 9kg 22/09/2018

Step 2; consider the link between A and B Tension 6N (Friction) 5kg a = 2ms-2 22/09/2018

Example 5 – The Blocks 10kg 4kg 56N Find the magnitude of the force exerted by the 4kg block on the 10kg block. 22/09/2018

Step 1; we find the acceleration of the system. 14kg 22/09/2018

Step 2; consider forces on the 10kg block Fun 10kg a = 4ms-2 22/09/2018