Key Areas covered

1. Key Areas covered • Balanced and unbalanced forces. The effects of friction. Terminal velocity. • Forces acting in one plane only. • Analysis of motion using Newton’s first and second laws. Frictional force as a negative vector quantity. • Tension as a pulling force exerted by a string or cable on another object.

2. What we will do today: • Revise Newton’s first two laws. • Use free body diagrams to analyse the forces on an object. • Carry out examples on the above • Carry out calculations using Fun = ma on problems involving lifts.

3. Newton’s First and Second Laws of Motion In your class jotter write down Newton’s first two laws

4. An accelerating object has an unbalanced (resultant) force, F, acting on it in the same direction as the acceleration. If an object is not accelerating then the unbalanced force F = 0.

5. Newton’s First Law • An object will remain stationary or continue at constant velocity unless acted on by an unbalanced force. • i.e. if forces are balanced an object will remain stationary or move at constant velocity.

6. Newton’s Second Law • Motion can exist without forces, but for change of motion to occur, forces must be involved. • i.e. Unbalanced forces cause acceleration.

7. Fun = ma Fun = Unbalanced Force in N m = mass in kg a = acceleration in ms-2 One Newton is the (unbalanced) force that gives a 1 kg mass an acceleration of 1 ms-2. Force is a vector.

8. Solving Force Problems • Draw a free body diagram showing the forces acting on the object and the direction of its acceleration. • Deal with one object at a time. However, when objects are tied together, they behave like a single larger object. • Use F = ma, but remember F is the unbalanced force!!

9. A man pushes a trolley of mass 20 kg along a flat surface of 40 N. If the effects of friction can be ignored, what is the acceleration of the shopping trolley? a = F / m = 40 / 20 = 2 ms-2 Example 1 – Single Force, Single Mass a 40 N 20 kg

10. A guided missile of mass 1000 kg is fired vertically into the air. The missile’s rocket motors provide a thrust of 20 000 N, and there is a drag force of 2 000 N. What is the acceleration of the missile? Fun = 20000 - 9800 - 2000 = 8200 N a = F / m = 8200 / 1000 = 8.2 ms-2 Example 2 – Multiple Force, Single Mass Thrust = 20000 N N A S A a a Drag = 2000 N W = mg = 9800N

11. A ski-tow pulls two skiers, who are connected by a thin nylon rope, along a frictionless surface. The tow uses a force of 70 N and the skiers have masses of 60 kg and 80 kg. a) What is the acceleration of the system? b) What is the tension in the rope? a) a = F / m = 70 / (80 + 60) = 70 / 140 = 0.5 ms-2 b) Tension T is a Force. It is caused by 80 kg person. T = ma = 80 x 0.5 = 40 N Example 3 – Single (External) Force, Multiple Mass 70 N T 80 kg 60 kg

12. 2003 Qu: 3

13. 2005 Qu: 3

14. 2008 Qu: 3

15. 2009 Qu: 4

16. 2010 Qu: 21

17. Lift Problems

18. Tension In all Lift problems, there will always be a tension in the cable supporting it, and a weight (W = mg). There could also be an unbalanced force, Fun. Lift of mass ‘m’ or Fun = ma Weight= mg

19. There are 6 possible lift situations: • Lift at rest Tension = Weight (no Fun) 2) Lift travelling at constant speed Tension = Weight (no Fun) 3) Lift accelerating up Tension = Weight + Fun = mg + ma (Fun up) • Lift decelerating up Tension = Weight – Fun = mg – ma (Fun down) 5) Lift accelerating down Tension = Weight – Fun = mg – ma (Fun down) 6) Lift decelerating down Tension = Weight + Fun = mg + ma (Fun up)

20. A package of mass 4 kg is hung from a spring balance attached to the ceiling of a lift which is accelerating upwards at 3 ms-2. What is the reading on the spring balance? Solution Fsb = force exerted by spring balance [to acc up it has to overcome the weight (mg) and then have Fun (ma)] Fsb = W + Fun Fsb = mg + ma Fsb = (4 x 9.8) + (4 x 3) Fsb = 51.2 N Example Fsb a = 3 ms-2 4 kg W = mg

21. 2006 Qu: 4

22. 2007 Qu: 3

23. Example 2 (…cont) • As the rocket rises its acceleration is found to increase, state a possible reason why. • As it rises fuel is burnt up so mass decreases. • Thinner air results in a reduction in friction. • The value of ‘g’ decreases so weight is less. • (Learn two of these for the exam).

24. Questions • Activity sheets: • Page 22 – 30 • Forces Questions Revisited & Higher Forces • You should now be able to answer all questions in class jotter