2.  Mechanics is the branch of physics which concerns itself with forces, and how they affect a body's motion.  Kinematics is the sub-branch of mechanics which studies only a body's motion without regard to causes.  Dynamics is the sub-branch of mechanics which studies the forces which cause a body's motion. Topic 2: Mechanics 2.2 Forces and dynamics Galileo Kinematics Newton Dynamics The two pillars of mechanics Topic 2.1 Topic 2.2

3. 2.2.1Calculate the weight of a body using the expression W = mg. 2.2.2Identify the forces acting on an object and draw free-body diagrams representing those forces. 2.2.3Determine the resultant force in different situations. 2.2.4 State Newton’s first law. 2.2.5 Describe examples of Newton’s first law. 2.2.6 State the condition for translational equilibrium. 2.2.7 Solve problems involving translational equilibrium. Topic 2: Mechanics 2.2 Forces and dynamics

4. Calculate the weight of a body using the expression W = mg. Identify the forces acting on an object and draw free-body diagrams representing those forces.  A force is a push or a pull measured in newtons.  One force we are very familiar with is the force of gravity, AKA the weight.  The very concepts of push and pull imply direction. The direction of the weight is down toward the center of the earth.  If you have a weight of 90 newtons (or 90 N), your weight can be expressed as a vector, 90 N, down.  We will show later that weight has the formula Topic 2: Mechanics 2.2 Forces and dynamics W = mg weight where g = 10 m s -2 and m is the mass in kg

5. Free body diagram Calculate the weight of a body using the expression W = mg. Identify the forces acting on an object and draw free-body diagrams representing those forces. Topic 2: Mechanics 2.2 Forces and dynamics EXAMPLE: Calculate the weight of a 25-kg object. SOLUTION:  Since m = 25 kg and g = 10 m s -2,  W = mg = (25)(10) = 250 N (or 250 n).  Note that W inherits its direction from the fact that g points downward.  We sketch the mass as a dot, and the weight as a vector in a free body diagram: mass force W W = mg weight where g = 10 m s -2 and m is the mass in kg

6.  Certainly there are other forces besides weight that you are familiar with.  For example, when you set a mass on a tabletop, even though it stops moving, it still has a weight. The implication is that the tabletop applies a counterforce to the weight, called a normal force.  Note that the weight and the normal forces are the same length – they balance.  The normal force is called a surface contact force. Identify the forces acting on an object and draw free-body diagrams representing those forces. Topic 2: Mechanics 2.2 Forces and dynamics W N

7. Identify the forces acting on an object and draw free-body diagrams representing those forces.  Tension can only be a pull and never a push.  Friction tries to oppose the motion.  Friction is parallel to the contact surface.  Normal is perpendicular to the contact surface.  Friction and normal are mutually perpendicular.  Friction and normal are surface contact forces. Topic 2: Mechanics 2.2 Forces and dynamics T the tension W N f Contact surface

8. Identify the forces acting on an object and draw free-body diagrams representing those forces.  Weight is drawn from the center of an object.  Normal is always drawn from the contact surface.  Friction is drawn along the contact surface.  Tension is drawn at whatever angle is given. Topic 2: Mechanics 2.2 Forces and dynamics T W N f

9. EXAMPLE: An object has a tension acting on it at 30° as shown. Sketch in the forces, and draw a free body diagram. SOLUTION:  Weight from center, down.  Normal from surface, up.  Friction from surface, parallel. Free body diagram Identify the forces acting on an object and draw free-body diagrams representing those forces. Topic 2: Mechanics 2.2 Forces and dynamics T 30° W N f T W N f

10. Determine the resultant force in different situations.  The resultant (or net) force is just the vector sum of all of the forces acting on a body. Topic 2: Mechanics 2.2 Forces and dynamics EXAMPLE: An object has mass of 25 kg. A tension of 50 n and a friction force of 30 n are acting on it as shown. What is the resultant force? SOLUTION:  Since the weight and the normal forces cancel out in the y- direction, we only need to worry about the forces in the x- direction.  The net force is thus 50 – 30 = 20 n (+x-dir) T W N f 50 n 30 n

11. Determine the resultant force in different situations.  The resultant (or net) force is just the vector sum of all of the forces acting on a body. Topic 2: Mechanics 2.2 Forces and dynamics EXAMPLE: An object has exactly two forces F 1 = 50 n and F 2 = 30 n applied simultaneously to it. What is the resultant force’s magnitude? SOLUTION:  F net =  F = F 1 + F 2 so we simply graphically add the two vectors:  The magnitude is just given by F net 2 = 50 2 + 30 2 so that  F net = 58 n. F1F1 50 n F2F2 30 n F net =  F net force F x,net =  F x F y,net =  F y F net

12. Determine the resultant force in different situations.  The resultant (or net) force is just the vector sum of all of the forces acting on a body. Topic 2: Mechanics 2.2 Forces and dynamics EXAMPLE: An object has exactly two forces F 1 = 50 n and F 2 = 30 n applied simultaneously to it as shown. What is the resultant force’s direction? SOLUTION:  Direction is measured from the x-axis traditionally.  Opposite and adjacent are given directly, so use tangent.  tan  = opp/adj = 30/50 = 0.6 so that  = tan -1 (0.6) = 31°. F1F1 50 n F2F2 30 n F net =  F net force F x,net =  F x F y,net =  F y F net 

13. Determine the resultant force in different situations.  The resultant (or net) force is just the vector sum of all of the forces acting on a body. Topic 2: Mechanics 2.2 Forces and dynamics EXAMPLE: An object has exactly two forces F 1 = 50 n and F 2 = 30 n applied simultaneously to it. What is the resultant force’s magnitude? SOLUTION:  Begin by resolving F 1 into its x- and y-components.  Then F net,x = 44 n and  F net,y = 23 + 30 = 53 n.  F net 2 = F net,x 2 + F net,y 2 so that  F net 2 = 44 2 + 53 2, F net = 70 n. F1F1 50 n F2F2 30 n F net =  F net force F x,net =  F x F y,net =  F y 28° 50cos28  44 n 50sin28  23 n

14. State Newton’s first law.  Newton’s first law is related to certain studies made by Galileo Galilee which contradicted Aristotelian tenets.  Aristotle basically said “The natural state of motion of all objects (except the heavenly ones) is one of rest.”  A child will learn that if you stop pushing a cart, the cart will eventually stop moving.  This simple observation will lead the child to come up with a force law that looks something like this: “In order for a body to be in motion, there must be a force acting on it.”  As we will show on the next slide, both of these statements is false! Topic 2: Mechanics 2.2 Forces and dynamics FALSE

15. State Newton’s first law.  Here’s how Galileo (1564-1642) thought:  If I give a cart a push on a smooth, level surface, it will eventually stop.  What can I do to increase the distance without pushing it harder?  If I can minimize the friction, it’ll go farther.  In fact, he reasoned, if I eliminate the friction altogether the cart will roll forever!  Galileo called the tendency of an object to not change its state of motion inertia. Topic 2: Mechanics 2.2 Forces and dynamics Inertia will only change if there is a force.

16. State Newton’s first law. State the condition for translational equilibrium.  Newton’s first law is drawn from his concept of net force and Galileo’s concept of inertia.  Essentially, Newton’s first law says that the velocity of an object will not change if there is no net force acting on it.  In his words...  In symbols...  The above equation is known as the condition for translational equilibrium. Topic 2: Mechanics 2.2 Forces and dynamics Every body continues in its state of rest, or of uniform motion in a straight line, unless it is compelled to change that state by forces impressed thereon. v = 0 v = CONST FF If  F = 0, Newton’s first law then v = CONST. A body’s inertia will only change if there is a net force applied to it.

17. Describe examples of Newton’s first law.  As a memorable demonstration of inertia – matter’s tendency to not change its state of motion (or its state of rest) - consider this:  A water balloon is cut very rapidly with a knife.  For an instant the water remains at rest!  Don’t try this at home, kids. Topic 2: Mechanics 2.2 Forces and dynamics

18. Solve translational equilibrium problems. Topic 2: Mechanics 2.2 Forces and dynamics EXAMPLE: An object of mass m is hanging via three cords as shown. Find the tension in each of the three cords, in terms of m. SOLUTION:  Give each tension a name to organize your effort.  Draw a free body diagram of the mass and the knot.  T 3 is the easiest force to find. Why?  m is not moving so its FBD tells us that  F y = 0 or T 3 – mg = 0 or T 3 = mg. knot 30° 45° T1T1 T2T2 T3T3 mgmg T3T3 FBD, m FBD, knot T2T2 T1T1 T3T3 30° 45° m

19. Solve translational equilibrium problems. Topic 2: Mechanics 2.2 Forces and dynamics EXAMPLE: An object of mass m is hanging via three cords as shown. Find the tension in each of the three cords, in terms of m. SOLUTION: T 3 = mg  Now we break T 1 and T 2 down to components.  Looking at FBD knot we see that  T 1x = T 1 cos 30° = 0.866T 1  T 1y = T 1 sin 30° = 0.500T 1  T 2x = T 2 cos 45° = 0.707T 2  T 2y = T 2 sin 45° = 0.707T 2 knot 30° 45° T1T1 T2T2 T3T3 mgmg T3T3 FBD, m FBD, knot T2T2 T1T1 T3T3 30° 45°

20. Solve translational equilibrium problems. Topic 2: Mechanics 2.2 Forces and dynamics EXAMPLE: An object of mass m is hanging via three cords as shown. Find the tension in each of the three cords, in terms of m. SOLUTION: T 3 = mg  Putting this into the FBD of knot we get: knot 30° 45° T1T1 T2T2 T3T3 mgmg T3T3 FBD, m FBD, knot T2T2 T1T1 T3T3 30° 45° ∑F x = 0 T 2 = 1.225T 1 0.707T 2 - 0.866T 1 = 0 ∑F y = 0 0.707T 2 + 0.500T 1 - T 3 = 0 0.707(1.225T 1 ) + 0.500T 1 = T 3 T 1 = mg/1.366 T 2 = 1.225(mg/1.366) T 2 = 0.897mg

21. Solve translational equilibrium problems. Topic 2: Mechanics 2.2 Forces and dynamics EXAMPLE: A 25-kg mass is hanging via three cords as shown. Find the tension in each of the three cords, in newtons. SOLUTION:  Since all of the angles are the same use the formulas we just derived:  T 3 = mg = 25(10) = 250 n  T 1 = mg/1.366 = 25(10)/1.366 = 180 n  T 2 = 0.897mg = 0.897(25)(10) = 220 n knot 30° 45° T1T1 T2T2 T3T3 FYI  This is an example of using Newton’s first law with v = 0. The next example shows how to use Newton’s first when v is constant, but not zero.

22. Solve translational equilibrium problems. Topic 2: Mechanics 2.2 Forces and dynamics EXAMPLE: A 1000-kg airplane is flying at a constant velocity of 125 m s -1. Label and determine the value of the weight W, the lift L, the drag D and the thrust F if the drag is 25000 n. SOLUTION:  Since the velocity is constant, Newton’s first law applies. Thus  F x = 0 and  F y = 0.  W = mg = 1000(10) = 10000 n and points down.  L = W = 10000 n and points up since  F y = 0.  D tries to impede the aircraft and points left.  F = D = 25000 n and points right since  F x = 0. W L D F

23. 2.2.8State Newton’s second law of motion. 2.2.9Solve problems using Newton’s second law. Topic 2: Mechanics 2.2 Forces and dynamics

24. State Newton’s second law of motion.  Newton reasoned: “If the sum of the forces is not zero, the velocity will change.”  But we know, and he also did, that a change in velocity is an acceleration.  So Newton then asked himself: “How is the sum of the forces related to the acceleration.”  Here is what Newton said: “The acceleration of an object is proportional to the net force acting on it, and inversely proportional to its mass.”  In other words, the bigger the force the bigger the acceleration, and the bigger the mass the smaller the acceleration.  In formula form Topic 2: Mechanics 2.2 Forces and dynamics F net = ma Newton’s second law (or  F = ma ) a = F/m

25. State Newton’s second law of motion.  Looking at the form  F = ma note that if a = 0 then  F = 0.  But if a = 0, v = CONST.  Thus Newton’s first law is just a special case of his second. Topic 2: Mechanics 2.2 Forces and dynamics F net = ma Newton’s second law (or  F = ma )

26. Solve problems using Newton’s second law. Topic 2: Mechanics 2.2 Forces and dynamics F net = ma Newton’s second law (or  F = ma ) EXAMPLE: An object has mass of 25 kg. A tension of 50 n and a friction force of 30 n are acting on it as shown. What is its acceleration? SOLUTION:  The vertical forces W and N cancel out.  The net force is thus 50 – 30 = 20 n (+x-dir).  From F net = ma we get 20 = 25a so that a = 20 / 25 = 0.8 m s -2. T W N f 50 n 30 n

27. Solve problems using Newton’s second law. Topic 2: Mechanics 2.2 Forces and dynamics F net = ma Newton’s second law (or  F = ma ) PRACTICE: Use F = ma to show that the formula for weight is correct.  F = ma.  But F is the weight W.  And a is the freefall acceleration g.  Thus F = ma becomes W = mg.

28. Topic 2: Mechanics 2.2 Forces and dynamics EXAMPLE: A 1000-kg airplane is flying in perfectly level flight. The drag D is 25000 n and the thrust F is 40000 n. Find its acceleration. SOLUTION:  Since the flight is level,  F y = 0.  F x = F – D = 40000 – 25000 = 15000 n = F net.  From F net = ma we get 15000 = 1000a a = 15000 / 1000 = 15 m s -2. W L D F Solve problems using Newton’s second law. F net = ma Newton’s second law (or  F = ma )

29. Topic 2: Mechanics 2.2 Forces and dynamics Solve problems using Newton’s second law. F net = ma Newton’s second law (or  F = ma ) EXAMPLE: A 25-kg object has exactly two forces F 1 = 40. n and F 2 = 30. n applied simultaneously to it. What is the object’s acceleration? SOLUTION:  Begin by resolving F 1 into its x- and y-components.  Then F net,x = 36 n and  F net,y = 17 + 30 = 47 n.  F net 2 = F net,x 2 + F net,y 2 so that  F net 2 = 36 2 + 47 2, F net = 59 n.  Then from F net = ma we get 59 = 25a a = 59 / 25 = 2.4 m s -2. F1F1 40 n F2F2 30 n 25° 40cos25  36 n 40sin25  17 n

30. Topic 2: Mechanics 2.2 Forces and dynamics Solve problems using Newton’s second law. F net = ma Newton’s second law (or  F = ma ) EXAMPLE: A 25-kg object resting on a frictionless incline is released, as shown. What is its acceleration? SOLUTION:  Begin with a FBD.  Break down the weight into its components.  Since N and mg cos 30° are perpendicular to the path of the crate they do NOT contribute to its acceleration.  F net = ma mg sin 30° = ma a = 10 sin 30° = 5.0 m s -2. 30° 6.0 m mg N 60  30  mg cos 30  mg sin 30 

31. Topic 2: Mechanics 2.2 Forces and dynamics Solve problems using Newton’s second law. F net = ma Newton’s second law (or  F = ma ) EXAMPLE: A 25-kg object resting on a frictionless incline is released, as shown. What is its speed at the bottom? SOLUTION:  We found that its acceleration is 5.0 m s -2.  We will use the timeless equation to find v so we need to know what s is.  From trigonometry, we have opposite and we want hypotenuse so we use sin  = opp/hyp. Thus  s = hyp = opp/sin  = 6 / sin 30° = 12 m. v 2 = u 2 + 2as = 0 2 + 2(5)(12) v = 11 m s -1. 30° 6.0 m s u = 0 v = ? a = 5

32. 2.2.10 Define linear momentum and impulse. 2.2.11 Determine the impulse due to a time- varying force by interpreting a force-time graph. 2.2.12 State the law of conservation of linear momentum. 2.2.13 Solve momentum and impulse problems. Topic 2: Mechanics 2.2 Forces and dynamics

33. Define linear momentum and impulse.  Linear momentum, p, is defined to be the product of an object’s mass m with its velocity v.  Its units are obtained directly from the formula and are kg m s -1. Topic 2: Mechanics 2.2 Forces and dynamics p = mv linear momentum EXAMPLE: What is the linear momentum of a 4.0-gram NATO SS 109 bullet traveling at 950 m/s? SOLUTION:  Convert grams to kg (jump 3 decimal places left) to get m =.004 kg.  Then p = mv = (.004)(950) = 3.8 kg m s -1.

34. Define linear momentum and impulse.  From F net = ma we can get F net = m∆v/∆t F net = ∆p/∆t  This last is just Newton’s second law in terms of change in momentum rather than mass and acceleration. Topic 2: Mechanics 2.2 Forces and dynamics p = mv linear momentum F net = ∆p/∆t Newton’s second law (p-form) EXAMPLE: A 6-kg object has its speed increase from 5 m s -1 to 25 m s -1 in 30 s. What is the net force acting on it? SOLUTION:  F net = ∆p/∆t = m(v – u)/∆t = 6(25 – 5)/30 = 4 n.

35. Define linear momentum and impulse.  If we manipulate Newton’s second law (p-form) to isolate the change in momentum we get: F net = ∆p/∆t F net ∆t = ∆p  We call the force times the time the impulse J. Topic 2: Mechanics 2.2 Forces and dynamics F net = ∆p/∆t Newton’s second law (p-form) EXAMPLE: A baseball has an average force of 12000 n applied to it for 25 ms. What is the impulse imparted to the ball from the bat? SOLUTION:  J = F net ∆t = 12000(25  10 -3 ) = 300 n s. J = F net ∆t = ∆p impulse FYI  The units for impulse are the units of force (n) times time (s).

36. Determine the impulse due to a time-varying force by interpreting a force-time graph.  The impulse is the area under an F vs. t graph. Topic 2: Mechanics 2.2 Forces and dynamics EXAMPLE: A bat striking a ball imparts a force to it as shown in the graph. Find the impulse. SOLUTION:  Break the graph into simple areas of rectangles and triangles.  A 1 = (1/2)(3)(9) = 13.5 n s  A 2 = (4)(9) = 36 n s  A 3 = (1/2)(3)(9) = 13.5 n s  A tot = A 1 + A 2 + A 3  A tot = 13.5 + 36 + 13.5 = 63 n s. J = F net ∆t = ∆p impulse 0 9 3 6 0 510 Force F/n Time t/s

37. State the law of conservation of linear momentum.  Recall Newton’s second law (p-form):  If the net force acting on an object is zero, we have  F net = ∆p/∆t 0 = ∆p/∆t 0 = ∆p  In words, if the net force is zero, then the momentum does not change – p is constant.  In symbols Topic 2: Mechanics 2.2 Forces and dynamics F net = ∆p/∆t Newton’s second law (p-form) If F net = 0 then p = CONST conservation of linear momentum

38. 2.2.14 State Newton’s third law. 2.2.15 Discuss examples of Newton’s third law. Students should understand that when two bodies interact, the forces they exert on each other are equal and opposite. Topic 2: Mechanics 2.2 Forces and dynamics

39. State Newton’s third law. Discuss examples of Newton’s third law.  In words “For every action force there is an equal and opposite reaction force.”  In symbols  In the big picture, if every force in the universe has a reaction force that is equal and opposite, the net force in the whole universe is zero!  So why are there accelerations all around us? Topic 2: Mechanics 2.2 Forces and dynamics F AB = - F BA F AB is the force on body A by body B. F BA is the force on body B by body A. Newton’s third law

40. State Newton’s third law. Discuss examples of Newton’s third law. Topic 2: Mechanics 2.2 Forces and dynamics EXAMPLE: Consider a door. When you push on a door with 10 n, because of Newton’s third law the door pushes on your hand with the same 10 n, but in the opposite direction. Why does the door move, and you don’t? SOLUTION:  Even though the forces are equal and opposite, they are acting on different bodies.  Each body acts in response only to the force acting on it.  The door can’t resist F AB, but you CAN resist F BA. your action A B F AB A F BA the door’s reaction

41. State Newton’s third law. Discuss examples of Newton’s third law. Topic 2: Mechanics 2.2 Forces and dynamics EXAMPLE: Consider a baseball resting on a tabletop. Discuss all of the forces acting on the baseball, and their reactions. SOLUTION:  Acting on the ball is its weight F BE and the normal force on the ball caused by the table N BT.  The reactions are F EB and N TB.  Note that F BE (the weight force) and N BT (the normal force) are acting on the ball.  N BT (the normal force) acts on the table.  F EB (the weight force) acts on the earth. F EB F BE N BT N TB

42. State Newton’s third law. Discuss examples of Newton’s third law.  We define a system as a collection of more than one body, mutually interacting with each other. Topic 2: Mechanics 2.2 Forces and dynamics EXAMPLE: Three billiard balls interacting on a pool table constitute a system.  The action-reaction force pairs between the balls are called internal forces.  For any system all internal forces cancel!

43. State Newton’s third law. Discuss examples of Newton’s third law.  We define a system as a collection of more than one body, mutually interacting with each other. Topic 2: Mechanics 2.2 Forces and dynamics EXAMPLE: Three billiard balls interacting on a pool table constitute a system. Label and discuss all of the internal forces.  The internal force pairs only exist while the balls are in contact with one another.  Note that a blue and a red force act on the white ball. It responds only to those two forces.  Note that a single white force acts on the red ball. It responds only to that single force.  Note that a single white force acts on the blue ball. It responds only to that single force.

44. State Newton’s third law. Discuss examples of Newton’s third law.  We define a system as a collection of more than one body, mutually interacting with each other. Topic 2: Mechanics 2.2 Forces and dynamics EXAMPLE: Three billiard balls interacting on a pool table constitute a system. Describe the external forces.  External forces are the forces that the balls feel from external origins.  For billiard balls, these forces are the balls’ weights, the cushion forces, and the queue stick forces.

45. State the law of conservation of linear momentum.  In light of Newton’s third law and a knowledge of systems, the conservation of linear momentum can be refined for a system of particles.  Since in a system all of the internal forces sum up to zero, F net can only be the sum of the external forces. Thus  In other words, internal forces cannot change the momentum of a system of particles – ever! Topic 2: Mechanics 2.2 Forces and dynamics If F net = 0 then p = CONST conservation of linear momentum If F net,ext = 0 then p = CONST conservation of linear momentum - system

46. Solve momentum and impulse problems. Topic 2: Mechanics 2.2 Forces and dynamics EXAMPLE: A 12-kg block of ice at rest has a fire cracker inside a hole drilled in its center. When it explodes, the block breaks into 2 pieces, one of which travels at +16 m s -1 in the x-direction. What is the velocity of the other 8.0 kg piece? SOLUTION:  Make before and after sketches.  The initial velocity of the two is 0.  From conservation of momentum we have  p = CONST which means p 0 = p f. Since p = mv,  (8+4)(0) = 8v + 4(16)  so that v = -8 m s -1. 8 4 8 4 16 v If F net,ext = 0 then p = CONST conservation of linear momentum - system

47. EXAMPLE: A 730-kg Smart Car traveling at 25 m s -1 (x-dir) collides with a stationary 1800-kg Dodge Charger. The two vehicles stick together. Find their velocity immediately after the collision. SOLUTION:  Make before and after sketches.  p 0 = p f so that (730)(25) = (730 + 1800)v 18250 = 2530v v = 18250/2530 = 7.2 m s -1. Solve momentum and impulse problems. Topic 2: Mechanics 2.2 Forces and dynamics If F net,ext = 0 then p = CONST conservation of linear momentum - system 7301800 250 730 +1800 v