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# Forces

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Explains Newtons Laws of Motion
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### Forces

2. 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). 2
3. 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 3
4. 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 4
5. 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! 5
6. 6. Force 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 6
7. 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. 7
8. 8. FORCE 8
9. 9. 9
10. 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! 10
11. 11. “For every action there must be an equal but opposite reaction” Click here to continue 11
12. 12. For every action there must always be an equal and opposite reaction. Pushes and pulls occur in pairs. 12
13. 13. Review of Kinematic Equations • S = displacement, t = time Vo = original velocity, a = acceleration Vi = instantaneous velocity, Vave = average velocity • Equations • S = Vot + ½ at2 (displacement vs. time) • Vi = Vo + at (velocity, acceleration & time) • a = (Vi 2 – Vo 2) / 2 S (displacement, velocity & acceleration, time not required) • Vave = (V1 + V2) / 2 (average velocity) • Vave = S / t (average velocity) 13
14. 14. Solving Force Problems A 5 Kg rifle fires a 9 gram bullet with an acceleration of 30,000 m/s2. (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/s2 = 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 3rd 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 Vi = Vo + at, Vi = o + 30,000 x 0.007 = 210 m/s (Vo = 0 since the bullet is not moving before it is fired) a = 30,000 m/s2 14
15. 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! 15
16. 16. scale 150 lbs scale 25.6 lbs scale 406 lbs g = 9.81 m/s2 g = 1.67 m/s2 g = 26.6 m/s2 Mass is the same in all cases 16
17. 17. YOU COULD LOSE WEIGHT BY MOVING TO ANOTHER PLANET BUT YOUR MASS WOULD BE THE SAME AND YOU WOULD STILL LOOK THE SAME! 17
18. 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/s2 = 200 slugs. • Now, to find acceleration, Vi = Vo + at or a = (Vi – Vo) / 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 Vo = (20 x 5280) / 3600 = 29.4 ft /sec and Vi = (10 x 5280) / 3600 = 14.9 ft/sec. • a = (14.9 – 29.4)/ 5 = - 2.98 ft/sec2 and F = 200 x (-2.98) = - 596 nt (negative means the force is opposing the motion) • (b) Using S = Vot + ½ at2 we get S = (29.4 x 5) + ½ (-2.98) 52 = 110 feet is the distance traveled during stopping. 20 mph 10 mph time = 5 sec 18
19. 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. 19