2. Force is a vector quantity with magnitude & direction. e.g. a ball moves because you exerted a force by. If an object changes velocity, then a acted upon it. i.e. a Force causes an object to Types of Forces 1.) : from physical contact. e.g. 2.) : no physical contact. e.g..

3. 1 st Law: (Law of ) “An object will remain in its constant state of motion, whether it is at or with a constant, unless it is acted upon by an external.” i.e. If F Resultant = thena = Inertia: the tendency of an object to remain at or in constant. i.e. has a tendency to resist a. Mass: used to measure Inertia. - inherent property of matter. Units: kilograms (kg)

4. Mass is different to weight. Mass is a quantity. Weight: is a with magnitude & direction. It is the Force of acting on a mass.

5. “The acceleration of an object is directly to the Resultant (Nett) Force and is to its mass.” The direction of the acceleration is in the direction of the Resultant (Nett) Force (F R ). i.e.a α a α  a α so: F R = ΣF = F = Units = kg.m/s 2 = Newtons (N) 1 N = lbs of Force, or 1 lb = N

6. Weight is a Force exerted by the Earth on an object with a mass. It is a vector. Units: Newtons (N) If F = ma then where the Weight vector always points, towards. Therefore, Weight is a vector where the direction is not stated. (Explain 2-way attraction.)

7. “To every action, there’s an equal but opposite.” A book resting on a table experiences a Force of Gravity (W), yet it does not move. (a=0) Why? There must be an equal reactionary force satisfying Newton’s 3 rd Law. The book applies a force on the table, and the table exerts a force onto the book. This force is called the

8. The word normal means to the surface. The book is not moving, therefore  F = (1 st Law), in other words: but in the direction. When an object moves, we now have 3 forces to consider.

9. If an object is not accelerating, then ΣF = 0 and it is in a state of i.e.a = & v =  a = &  This also means that the sum of their components also = 0. i.e. = 0& = 0 Note: When a force is applied to an object via a rope, chain, string etc…, it is then called

10. A diagram can be simplified by only displaying the Force vectors.Diagram | FBD

11. The object in the Figure has a mass of 15kg. The object is suspended by cables as shown. Calculate the tension (T 1 ) in the cable at 30° with the horizontal. Diagram Axes FBD Since the object and its supporting cables are motionless (i.e., in equilibrium), we know that the net force acting on the intersection of the cables is zero. The fact that the net force is zero tells us that the sum of the x-components of T I, T 2,and T 3 is zero, and the sum of the y-components of T I,T 2,and T 3 is zero.

12. Solution:Table Information: T 3 = W = mg = T 1x = = ΣF x = T 1y = = ΣF y = Xy T1T1 T2T2 T3T3 R00

14. Solution to Ch.4 p.107 Q.16:Table Information: T 3 = T 1x = = T 1y = = ΣF x = T 2x = = T 2y = ΣF y = = xy T1T1 T2T2 T3T3 R00

15. We can now solve mass & pulley system problems. E.g. A 10kg block rests on a frictionless table. A rope is attached that goes over a pulley and is connected to a hanging 5kg mass. Calculate the Tension in the rope and the acceleration of the two masses. FBD’s m 1 = 10kg m 2 = 5kg T = ? a = ?

16. Solution: m 1 : m 2 : Make equal:

17. Consider an object on an inclined plane.

18. From the diagram, we can derive that: n = W y n = F G = W x F G = Special Situations when: θ = 0 (horizontal)θ = 90 o (vertical) Cosθ = ∴ n = Sinθ = ∴ F G = F G =

19. Let’s revisit the pulley system problem, but this time, the table will be at an angle of 60 o. DiagramFBD’s m 1 = 10kg m 2 = 5kg θ = 60 o T = ? a = ?

20. Solution: m1:  Fx = ΣF y = m2:  Fy =

21. Substitute: m 1.a + m 1.gSinθ = m 2.g – m 2.a m 1.a + m 2.a = m 2.g – m 1.gSinθ a(m 1 + m 2 ) = g(m 2 – m 1 Sinθ) a = g(m 2 – m 1 Sinθ) (m 1 + m 2 ) a = (9.80m/s 2 )(5kg – 10kg.Sin60 o ) (5kg + 10kg) a = -2.39m/s 2 (why negative?) Substitute back:

22. When a body is in motion, there is resistance to the motion because of its interaction with its surroundings, whether it’s sliding on a surface, or moving through a fluid. (air, water etc…) This resistance is called the: The object remains at rest because the Friction as the applied Force up to a certain point.

23. After this pt, F is too large (>ƒ) and the object moves. This break point is called the: force of So, ifthen the object moves. Once the object is moving, there is still friction, but this time it is called the: force of and this time:

24. Note that Why?

26. So if: then object explains why you still have to exert a Force for a i.e. (Newton’s 1 st Law still holds.)

27. ƒ S and ƒ K are directly proportional to the normal force (n). i.e. and where is the coefficient of and is the coefficient of ƒ must be so Units:

28. Or see Table 4:2 on p.101

29. E.g.At what rate would a 20kg crate accelerate if it is sliding on a wooden floor (μ K = 0.22) and being pushed with a constant Force of 100N?

30. E.g.You are pulling a 10kg box by using a rope attached at an angle of 30 o. If it requires a Tension of 125N to move it at constant velocity, then calculate the coefficient of kinetic friction.

32. E.g. A 25kg wooden block is 2.0m up a 40 o incline. It is attached to a 5kg mass that is hanging over a pulley. The coefficient of Kinetic Friction between the 15kg block and the wood incline is 0.333. Calculate the speed of the block at the bottom of the incline.

33. m 2 = 25kg ΣF y = m 1 = 5kg θ = 40 o x = 2.0m μ K = 0.333 ΣF x =

34. m 1 a + m 1 g = m 2 g.Sinθ – μ K m 2 g.Cosθ – m 2 a m 1 a + m 2 a = m 2 g.Sinθ – μ K m 2 g.Cosθ - m 1 g a(m 1 + m 2 ) = m 2 g.Sinθ – μ K m 2 g.Cosθ - m 1 g a = m 2 g.Sinθ – μ K m 2 g.Cosθ - m 1 g (m 1 + m 2 ) a = g[m 2.Sinθ – μ K m 2.Cosθ – m 1 ] (m 1 + m 2 ) a = (9.80m/s 2 )[(25kg)Sin40 o ) – (0.333)(25kg)Cos40 o – (5kg)] (5kg + 25kg) a = 1.53m/s 2