1. Copyright Sautter 2003

2. THE LAWS OF MOTION Isaac Newton, in the 1600s, proposed three fundamental laws of motion which are found to be correct even today! Newton’s First Law of Motion – Inertia – “Objects in motion tend to remain in motion at the same rate (speed) and the in same direction, unless acted on by an outside force” This law says essentially then, that objects keep doing what they have been doing, unless they are forced to change by an external factor. This law explains why for example a car skids on ice when the brakes are applied (the lack of the outside force of friction on the tires) or why it is difficult to negotiate a tight curve at a high rate of speed (the direction of motion tends to remain straight due to inertia).

3. “Objects in motion tend to remain in motion, at the same rate, And in the same direction, unless acted on by an outside force” Outside Force Click here to continue

4. Objects tend to move in the same direction and at the same rate unless acted on by an outside force Constant velocity Outside Force Constant velocity stops Acceleration occurs

5. THE LAWS OF MOTION Newton’s Second Law of Motion – “the acceleration of an object is directly proportional to the force exerted on it and inversely proportional to its mass’. Force means a push or a pull and the second law then says that the harder you push or pull an object, the more rapidly it speeds up or slows down. Cars with large engines (more available force) accelerate more quickly and those with smaller engines! The second law also tells us that large, massive objects are harder to speed up or slow down that small objects. Mass then is the inertial property of matter which means it determines how readily an object maintains its state of motion. As an example, it is much easier to accelerate a sports car than a 10 ton truck!

6. Force Large masses with the same force applied results in small accelerations Small masses with the same force applied results in large accelerations Click here to continue

7. THE LAWS OF MOTION Newton’s Second Law of Motion in abbreviated mathematical form states, F = MA, (force equals mass times acceleration). The units used in expressing force, mass and acceleration vary depending on the measurement system which is used. Three systems are available and the one chosen depends on the units cited in the problem to be solved. (1) MKS – metric units involving meters as displacement units, kilograms as mass units and seconds as time units. MKS force units are newtons (2) CGS – metric units involving centimeters as displacement units, grams as mass units and seconds as time units. CGS force units are dynes. (3) English Units– involving feet as displacement units, slugs as mass units and seconds as time units. English force units are pounds.

8. FORCE

10. THE LAWS OF MOTION Newton’s Third Law of Motion tells us that “for every action there must always be an equal and opposite reaction” What the Third Law says then is if you push on something it must push back equally hard or nothing will happen. As an example, pretend that you are at an ice skating rink, standing on skates at the center of the rink and you attempt to move by pushing on the surrounding air with your hands. No motion occurs because the air cannot push back sufficiently! You do not move. Now, pretend that a friend in next to you on skates and you chose to push on him. You move one way and your friend moves the other. Equal and opposite pushes result in motion occurring!

11. “For every action there must be an equal but opposite reaction” Click here to continue

12. For every action there must always be an equal and opposite reaction. Pushes and pulls occur in pairs.

13. Review of Kinematic Equations S = displacement, t = time V o = original velocity, a = acceleration V i = instantaneous velocity, V ave = average velocity Equations S = V o t + ½ at 2 (displacement vs. time) V i = V o + at (velocity, acceleration & time) a = (V i 2 – V o 2 ) / 2 S (displacement, velocity & acceleration, time not required) V ave = (V 1 + V 2 ) / 2 (average velocity) V ave =  S /  t (average velocity)

14. Solving Force Problems A 5 Kg rifle fires a 9 gram bullet with an acceleration of 30,000 m/s 2. (a) What force acts on the bullet? (b) What force acts on the rifle? ( c) If the bullet is in the rifle for 0.007 seconds, what is its muzzle velocity? (a) F = ma, F = 0.009 kg x 30,000 m/s 2 = 270 nt (grams must be converted to kilograms in order to use the MKS system and get newtons as force unit answer) (b) Newton’s 3 rd Law (Action / Reaction) – the rifle is “shot” in the opposite direction with the same force as the bullet or – 270 nts (c) (muzzle velocity means the velocity at which the bullet leaves the rifle barrel) Using the equation V i = V o + at, V i = o + 30,000 x 0.007 = 210 m/s (V o = 0 since the bullet is not moving before it is fired) a = 30,000 m/s 2

15. Force, Weight & Gravity Weight and mass are related but they are not the same! Weight requires gravity while mass exists independent of gravity. Your weight in outer space would be zero, your mass would not be zero but the same as it is on earth or anywhere for that matter. As the mass of an object increases, its weight increases proportionally if gravity is present. Greater mass gives greater weight. Although, weight in Europe is measured in kilograms, it is technically incorrect! Weight being a force should be measured in newtons. The exact meaning of weight measured in kilograms is “kilograms of force” meaning the mass of an object times Earth’s gravity. In the English system, pounds in the correct force unit and weight values given in pound units are correct. English mass units are slugs!

16. scale 150 lbs scale 25.6 lbs scale 406 lbs g = 9.81 m/s 2 g = 1.67 m/s 2 g = 26.6 m/s 2 Mass is the same in all cases

17. YOU COULD LOSE WEIGHT BY MOVING TO ANOTHER PLANET BUT YOUR MASS WOULD BE THE SAME AND YOU WOULD STILL LOOK THE SAME!

18. Solving Force Problems (a) How much force is needed to reduce the velocity of a 6400 lb truck from 20 mph to 10 mph in 5 seconds? (b) What is its stopping distance? (a) F = ma, we must first find mass. The weight is 6400 lbs. Wt = mass x gravity therefore, m = w/g, mass = 6400 lbs /32 ft/s 2 = 200 slugs. Now, to find acceleration, V i = V o + at or a = (V i – V o ) / t Velocities are given in mph and we need ft / sec. To convert mph to ft / sec we multiply by 5280 ft / mile and divide by 3600 seconds in an hour to get V o = (20 x 5280) / 3600 = 29.4 ft /sec and V i = (10 x 5280) / 3600 = 14.9 ft/sec. a = (14.9 – 29.4)/ 5 = - 2.98 ft/sec 2 and F = 200 x (-2.98) = - 596 nt (negative means the force is opposing the motion) (b) Using S = V o t + ½ at 2 we get S = (29.4 x 5) + ½ (-2.98) 5 2 = 110 feet is the distance traveled during stopping. 20 mph 10 mph time = 5 sec

19. Solving Force Problems Solving problems in physics involving forces often uses the idea of net force. The net force on a object is the vector sum of all forces acting on that object and the net force gives the object its acceleration. As an example, if I push a chair across the floor with a force of 25 newtons and the force of friction opposing the motion is 10 newtons, the net force accelerating the chair is 25 + (-10) or 15 newtons in the direction that I am pushing the chair. When the applied forces are acting at angles other than straight line motion, vector addition methods must be used to determine the net force on the object.

20. Frictional Forces Friction is a force that is commonly encountered in physics problems. FRICTION is a force which ALWAYS OPPOSES THE MOTION OF AN OBJECT Friction is related to two factors. (1) the force holding the surfaces in contact with each other (normal force) and (2) the nature of the surfaces (different materials cause different frictional forces) The normal force holding surfaces in contact is simply the weight of an object on the horizontal. As the the angle of inclination changes the normal force is reduced until at 90 degrees none of the weight holds the surfaces in contact and the normal force is zero. The trig function which describes this relationship is the cosine, Normal force = weight x cosine of the angle of inclination. (F normal = w x cos  and since weight is mass x gravity, F normal = mg cos  ). The nature of the surfaces in contact is described by the coefficient of friction with a symbol called “mu” (  ). Force of friction = coefficient of friction x normal force F friction =  w cos  =  m g cos 

21. Solving Force Problems A force of 20 nts will just set a 50 kg box in motion. What is the coefficient of friction between the box and the floor? F friction =  w cos  =  m g cos   = F friction / (m g cos  ) since the box just moves the force acting on the box and the force of friction are just about equal, therefore F friction = 200 nts  = 200 / (50 x 9.8 x cos 0 0 ) = 0.410 The coefficient of friction is a ratio which has no units! 50 kg 200 nts friction

22. Solving Force Problems A common problem and laboratory experiment involving forces uses the Atwood’s Machine. It is simply a pulley with two mass suspended on each side. When released the masses accelerate at different rates depending on the masses used. The experiment may be done in two different ways. In one it can be used to predict and verify the acceleration of the system or it may be use to verify the acceleration due to gravity. The calculations shown on the next slide are used to predict the acceleration of the system. An important factor to remember when doing an Atwood’s Machine problem is that both masses are accelerated do to the net force acting on the system.

23. 2 Kg 3 Kg Acceleration 1.96 m/s 2 pulley F 2 = 2 x 9.8 = 19.6 nt F 3 = 3 x 9.8 = 29.4 nt F net = F 3 – F 2 = 29.4 – 19.6 = 9.8 nt F net = ma, mass = 2 + 3 = 5 kg 9.8 nt = 5 kg x a, a = 1.96 m/s 2

24. Solving Force Problems Additional calculations may be applied to the Atwood’s Machine problem, that is to tend the tension in the cord connecting the masses. Doing a problem such as this requires the use of the Free Body Diagram concept. Using the Free Body Diagram requires us to select a point within the system and examine it with only the forces acting at that specific point. Free Body Diagrams are used to simplify otherwise complex systems dealing with one point within the system at a time!

25. Free Body Diagram Weight 2 x -9.8 = -19.6 nt Tension of The cord Force net = mass x acceleration Tension + weight = mass x acceleration T + (-19.6) = 2 x (+1.96) (positive means upwards) T = 3.92 + 19.6 = 23.52 nts (tension in cord) 2 Kg 3 Kg Point to be examined a = 1.96 m/s 2

26. Solving Force Problems Inclined plane are often solved using forces. The forces acting on a body resting on an incline are that of gravity (the weight of the object), the normal force of the plane supporting the object on its surface and the the force of friction. A free body diagram may be constructed for inclined plane situations using the center of mass of the object as the point of force interaction. Center of mass refers to a point within the object around which the mass of the body is equally distributed. It is a simplifying technique by which the entire mass of the body can be considered to be present at a single point rather than working with each and every point of mass within the object. For a symmetric body of uniform density, the center of mass is the geometric center of the body.

27.  Normal force Force parallel Upward force of plane Friction force weight W N P  Vector diagram for Inclined Plane W = weight vector P = parallel force vector N = normal force vector P = W sin N = W cos   Center of Mass

28. Solving an Inclined Plane Problem A box slides down an plane 8 meters long inclined at 30 0 with a coefficient of friction of 0.25. (a) what is the acceleration of the box? (b) what is its velocity at the bottom of the incline? (a) F net = ma, w = mg, m = w /g, F net = (w/g) a 0.5 w – 0.17 w = (w/9.8) a, a = 2.77 m/s 2 (b) V o = 0 (the box starts at rest), a = (V i 2 – V o 2 ) / 2 S V i =( 2 S a + V o ) 1/2, V i = ( 2 x 8 x 2.77 + 0 2 ) 1/2, V i = 6.7 m/s  Normal force Force parallel Upward force of plane Friction force weight Center of Mass F Parallel = w sin 30 0 F friction =  F normal F friction = 0.25 x w cos 30 0 F net = P – F friction F net = 0.5 w – 0.17 w F Parallel = W sin  F Normal = W cos 

29. Solving Force Problems In addition to Atwood’s Machine problems and inclined plane problems another common force problem involves elevators. When a person stands on a scale in a motionless environment (acceleration is zero) the scale reading gives his normal weight (the effect of gravity on his mass that is mass x gravity - remember w = mg) When a person stands on a scale in a moving environment the scale still gives his normal weight if the motion is at a constant velocity that is acceleration is still zero. If, however, the system is accelerated upwards or downwards, the scale reading changes and his normal weight is not the scale reading !

30. Solving Force Problems If the system is accelerating upwards, the scale must support the weight of the individual and also provide additional force to accelerate the person upwards. The scale reading therefore is greater than the normal weight of the person. If the system is accelerating downwards, the scale must provide a reading less than the normal weight since if is provided an upward force equal to the weight the net force would be zero and no acceleration would occur ! The extreme example of this situation would be when both the person and the scale are in free fall. In this case, no upward force is supplied by the scale and its reading is zero. The acceleration of both the person and scale then is that of gravity (- 9.8 m/s 2 )

31. Upward force of scale Downward force of weight Net force = 0 Acceleration = 0 Scale reads normal weight Upward force of scale Downward force of weight Net force = up Acceleration = + Scale reads > normal weight Upward force of scale Downward force of weight Net force = down Acceleration = - Scale reads < normal weight

32. Scale Reading 100 lbs Scale Reading 125 lbs GOING UP Acceleration = + 8 ft/s 2 Scale Reading 75 lbs GOING DOWN Acceleration = - 8 ft/s 2 Net force = MA T + W = MA T = tension in elevator cable W = weight

33. Solving Force Problems An 800 nt man stands on a scale in an elevator. What is the scale reading when (a) it is ascending a constant velocity of 3 m/s (b) descending at the same rate ( c) ascending with acceleration of 0.8 m/s 2 (d) descending with acceleration of 0.8 m/s 2 ? Upward force of scale Downward force of weight Net force = 0 Acceleration = 0 Scale reads normal weight Questions (a) and (b) In any unaccelerated system (constant velocity) the scale reading equals normal weight, therefore in parts (a) and (b) the scale reads 800 nts

34. Elevator Problem (cont’d) Part ( c) Ascending a = 0.8 m/s 2 Upward force of scale (P) Downward force of weight Net force > 0 (up) Acceleration = + Scale reads > normal weight Force net = upward push of scale (+) + weight downward (-) Force net = P + (-800) Force net = mass x acceleration Weight = mass x gravity -800 = m (-9.8), mass = (-800 / - 9.8) P + (- 800) = (- 800 / -9.8) (0.80) P = ( 81.6) (0.80) + 800 P = + 865 nt The scale reading is greater than the normal weight reading of 800 nts and upward acceleration occurs

35. Elevator Problem (cont’d) Part ( c) descending a = - 0.8 m/s 2 Upward force of scale (P) Downward force of weight Net force < 0 (down) Acceleration = - Scale reads < normal weight Force net = upward push of scale (+) + weight downward (-) Force net = P + (-800) Force net = mass x acceleration Weight = mass x gravity -800 = m (-9.8), mass = (-800 / - 9.8) P + (- 800) = (- 800 / -9.8) ( - 0.80) P = ( 81.6) (- 0.80) + 800 P = + 735 nt The scale reading is less than the normal weight reading of 800 nts and downward acceleration occurs

37. A force which gives a 2.0 kg object an acceleration of 1.6 m/s 2 would give an 8.0 kg object what acceleration ? (A) 0.2 m/s 2 (B) 0.4 m/s 2 (C) 1.6 m/s 2 (D) 6.4 m/s 2 A car slows form 50 ft/s to 15 ft/s in 10 seconds when the brakes exert a force of 200 pounds. What is the weight of the car ? (A) 57 lbs (B) 560 lbs (C) 1830 lbs (D) 22,400 lbs The coefficient of friction between the tires of a car and the road is 0.80. If the car is to stop in 3.0 seconds what can be its maximum speed? (A) 2.4 m/s (B) 2.6 m/s (C) 7.8 m/s (D) 23.5 m/s A skier stands on a 5 degree hill The coefficient of friction is 0.1 Does the skier slide down the hill ? (A) yes (B) no (C) it is impossible to tell A cable supports an elevator which is 2000 Kg. The tension in the cable lifting the elevator is 25 KN. What is the acceleration ? (A) 2.7 m/s 2 (B) 0.4 m/s 2 (C) 1.6 m/s 2 (D) 8.4 m/s 2 Click here for answers