#### Transcript N - Purdue Physics

This Week • Forces on an object • Newtons laws Relating force to acceleration • Riding in an elevator What we feel going up and down • Cars and Trains • Reaction /action What makes us walk or a car move Sailing Up Wind 4/4/2017 Physics 214 Fall 2011 1 What causes motion In our everyday life we observe that objects change their state of motion. In fact everything that happens in the Universe results from a change in motion. That is a static inert object does not contribute to any of the things we consider to be useful. The functioning of our body depends on continual change throughout our bodies. These changes are produced by forces and in our everyday life there are just two forces. Gravity acts on mass F = Gm1m2/r2 Electric charge F = kq1q2/r2 4/4/2017 Physics 214 Fall 2011 2 Newtons Second and First Law Second Law The acceleration of an object is directly proportional to the magnitude of the imposed force and inversely proportional to the mass. The acceleration is in the same direction as the force F = ma F and a are vectors unit is a Newton (or pound) 1lb = 4.448N First Law An object remains at rest or in uniform motion in a straight line unless it is acted on by an external force. F = 0 a = 0 so v = constant http://www.physics.purdue.edu/academic_programs/courses/phys214/movies.php (anim0003.mov) (anim0004.mov) 4/4/2017 Physics 214 Fall 2011 3 Force at the earths surface F = GMEm/rE2 rE But F = ma = mg so g = GME/rE2 Famous experiment by Cavendish Measured F = Gm1m2/r2 using two known masses in the laboratory and so measured G Then using earth 4/4/2017 g = GME/rE2 he determined the mass of the Physics 214 Fall 2011 4 Mass and weight Newtons second law enables us to measure relative mass. If we apply the same force to two objects and measure the accelerations then. F = m1a1 and F = m2a2 so m1/m2 = a2/a1 We then need to have one mass as a calibration and a kilogram is the mass of a piece of platinum held in Paris. Since gravity acts proportional to mass then the force near the earths surface is F = mg this is the weight of an object so if we compare F1 = m1g and F2 = m2g then weight 1/weight 2 = m1/m2 4/4/2017 Physics 214 Fall 2011 5 Inertia Inertia = tendency of an object to resist changes in its velocity. Since F = ma and a = Δv/t then Ft = mΔv So if a force acts for a time t the change in velocity will be smaller for larger masses so it is mass that determines inertia. In particular if t is very small and m is large then F can also be large but Δv can still be very small. 4/4/2017 Physics 214 Fall 2011 6 Friction In our everyday world any object which is moving feels a force opposing the motion --- this is friction. An object which is sliding The air resistance on your car These types of friction result in energy being lost and minimizing friction is very important. But Friction is also useful and essential since with no friction a car would not move but just spin it’s wheels a car would not be able to turn a corner we would not be able to walk objects would slide off surfaces unless perfectly horizontal 4/4/2017 Physics 214 Fall 2011 7 Force diagrams When we are analyzing a particular object we have to take into account all the forces acting on the body both in magnitude and direction. The acceleration of the object is equal to F/m in the direction of F where F is the net force acting. As in the example above we know that if we want to move an object that there is a force called friction which opposes what we want to do. In the case shown the 10N force is + and the 2N force - so the net force is 8N and 8 = 5a 4/4/2017 Physics 214 Fall 2011 8 Terminal velocity As an object moves through the atmosphere the air exerts a frictional retarding force which increases with velocity. So an object that is dropped from a great height first accelerates at 9.8m/s2 but this acceleration decreases until the retarding force = mg and at that point the acceleration is zero and the object has it’s terminal velocity. Similarly when a car is moving at constant speed on the highway the forward force produced by the engine is balanced by the frictional force of the air. Fair Ftire Ff mg v Mg – Ff =ma Freaction – Fair = ma Freaction 4/4/2017 Physics 214 Fall 2011 9 Reaction action: Newtons Third Law For every force that acts on a body there is an equal and opposite reaction force. Ff F F Ff You pull on the block the block pulls back on you The floor exerts a frictional force holding the block back The block exerts a frictional force on the floor trying to move the floor to the right. The block accelerates providing F > Ff 4/4/2017 Physics 214 Fall 2011 10 Force analysis To analyze the motion of an object we need to draw a diagram and put in all the forces that are acting on the object. We will only deal with problems that have an acceleration along a single axis. The net force along an axis perpendicular to this axis is zero Ff F F Ff We normally choose + in the direction of the acceleration. We now can use the equations F = ma v = v0 + at d = v0t + ½ at2 v2 = v02 + 2ad 4/4/2017 Physics 214 Fall 2011 11 Reaction action forces N For the book Nt = mg mg For the table Nearth = mg + mtableg mtableg mg Ne For a stack of plates or a brick wall and even a mountain each layer has to support the weight of everything higher. So for a stack of 48 plates the force on the second plate from the top is mg. For the bottom plate it is 47mg but of course each plate has a net force of mg to balance it’s own weight. 4/4/2017 Physics 214 Fall 2011 12 Forces in an elevator W = mg = true weight with no acceleration N = apparent weight N – mg is the net force taking + to be up N – mg = ma is the equation of motion If N > mg a is positive and the apparent weight is > than the true weight If N < mg a is negative and the apparent weight is less than the true weight + N g mg IRRESPECTIVE OF THE DIRECTION OR MAGNITUDE OF THE VELOCITY 4/4/2017 Physics 214 Fall 2011 13 Connected objects T Both objects have the same acceleration The coupling pulls back on the 4kg mass but accelerates the 2kg mass. 30 – 8 – T = 4a T – 6 = 2a Or 30 – 8 – 6 = 6a a = 16/6 m/s2 T = 68/6N If I have a freight train with 100 cars each of mass m then for the coupling between the engine and the first car T1 = 100ma And between the last two cars T99 = ma or 100 times less 4/4/2017 Physics 214 Fall 2011 14 Connected objects 6mg 7mg In the example shown each brick has a net upward force = mg down. But each brick has to support all the bricks above it. So the 7th brick has a downward force on it’s top surface of 6mg and an upward force of 7mg on it’s bottom surface. 4/4/2017 F F3 = 3ma F = 4ma 4 In the example shown of 7 equal mass objects being accelerated F = 7ma. The tension in each coupling pulls back on the mass ahead and accelerates the mass behind. So F4 =4ma but the net force on block 4 is 4ma – 3ma = ma. So the net force on each block is ma but the tension in each coupling reduces by ma as one goes down the chain. Physics 214 Fall 2011 15 Acceleration of a car Fair Ftire Freaction – Fair = ma Freaction Generally Fair is proportional to v2 so the difference between 55mph and 80mph is a factor of 2.16. This means you use 2.16 more gasoline to cover the same distance Travel from Indianapolis to Lafayette a distance of 60 miles Car does 30miles/gallon at 55mph gas costs $4/gallon At 55mph use 2 gallons cost 8$ time = 65.45 minutes At 80 mph use 4.32 gallons cost $17.28 time = 45 minutes Cost of saving 20.45 minutes = $9.28 4/4/2017 Physics 214 Fall 2011 16 Summary of Chapter 4 Forces are responsible for all physical phenomena Gravitation and the electromagnetic force are responsible for all the phenomena we normally observe in our everyday life. Newton’s laws F = ma where F is net force v = v0 + at d = v0t + ½ at2 d = ½(v + v0)t v2 = v02 + 2ad Every force produces an equal and opposite reaction Weight = mg where g = 9.8m/s2 locally Apparent weight in an elevator depends on the acceleration a up weight is higher a down weight is lower If your weight becomes zero it’s time to worry because you are in free fall!! 4/4/2017 Physics 214 Fall 2011 17 Sailing Up Wind The force on the sail balances out the force on the keel and leaves a component of force “against the wind”. Then F = ma and the boat will increase speed until the component of the wind force equals the drag force The sail and keel forces are like lift forces on an airplane wing An example is when the boat is moving perpendicular to the wind the force of the winds on the sails remains constant. The sails are set at about 450 to the direction of motion and the wind. The boats equilibrium speed is determined by the resistance to the motion. If the resistance can be made small the boat speed can be very high. Sail iceboats have very little resistance and reach speeds in excess of 90mph with wind speeds of 30mph 4/4/2017 Physics 214 Fall 2011 18 1C-04 Bocce Ball Tracks Two balls are released with the same initial velocity. One travels a flat path the other goes down a slope increasing it’s speed and then climbs a slope and slows down before reaching the end. The balls roll to minimize friction. V V V Which ball will reach the end first ? 1 5 2 3 4 The time for the yellow ball is just d/v. The time for the blue ball can be computed in 5 parts. Parts 1 and 5 are identical for both balls but the blue ball is faster for part 3 and qualitatively one would expect the blue ball to arrive first. 4/4/2017 Physics 214 Fall 2011 19 1F-02 Stack of Washers This is a demonstration of Inertia where a washer can be removed from a stack if the blow is fast. Why does it work less well as the stack gets shorter ? Strike the stack quickly. So the friction will be very short-lived and the stack will not gain speed before the force is gone. THIS TENDENCY TO RESIST CHANGES IN THEIR STATE OF MOTION IS DESCRIBED AS INERTIA. MASS IS A MEASURE OF THE AMOUNT OF INERTIA. SO FOR A FIXED FRICTIONAL FORCE ACTING FOR A SHORT TIME. THE BIGGER THE MASS, THE LESS IT WILL MOVE. 4/4/2017 Physics 214 Fall 2011 20 1F-07 Table Cloth Jerk Can the table cloth be removed without breaking any dishes ? THERE IS A FORCE ACTING ON THE DISHES, BUT IT LASTS FOR A VERY SHORT TIME. COMBINED WITH THE RELATIVELY LARGE MASS OF THE DISHES, THIS FORCE IS OVER SO QUICKLY AND IS SO SMALL THAT THE DISHES HARDLY MOVE. 4/4/2017 Physics 214 Fall 2011 21 1F-03 Egg drop Is it possible to get the eggs in the beakers without touching them ? IF THE PAN IS HIT SHARPLY A FORCE WILL ACT ON THE EGGS FOR A VERY SHORT TIME AND THEY WILL NOT MOVE HORIZONTALLY. THE PAN HAS TO BE HIT HARD ENOUGH SO THAT HAS MOVED OUT OF THE WAY BEFORE THE EGGS DROP ANY APPRECIABLE DISTANCE 4/4/2017 Physics 214 Fall 2011 22 1H-02 Fan Cart (Action-Reaction) Can a fan attached to a cart propel the cart? What if the sail is removed? Reaction Action In which direction will it move ? What if the sail is canted at an angle ? INTERNAL FORCES IN A SYSTEM CANCEL EACH OTHER WHEN THE SYSTEM AS A WHOLE IS CONSIDERED. SO IF THE SAIL IS PERPENDICULAR THE FAN DOES NOT MOVE. IF THE SAIL IS REMOVED THE FAN MOVES IN THE OPPOSITE DIRECTION TO WHICH IT BLOWS AIR. THE FAN WOULD NOT MOVE IN OUTER SPACE. 4/4/2017 Physics 214 Fall 2011 23 1H-03 CO2 Rocket Imagine that you are sitting in a cart with a pile of bricks. ROCKET PROPULSION ! How could you use the bricks to get yourself and the cart to move ? What would happen if you throw a brick out of the cart ? Then you throw out another . What if you throw smaller bricks faster and more frequently ? Now, if the bricks were the size of molecules. . . What happens when the fire extinguisher rapidly “throws” out CO 2 molecules ? THIS IS DIFFERENT THAN THE FAN CART. CO2 IS EXPELLED AT HIGH VELOCITY AND IN Because the repulse TERMS OF FORCES THE REACTION FORCE CAUSES THE CART TO MOVE IN THE OPPOSITE DIRECTION. THE QUANTITY CALLED MOMENTUM IS CONSERVED AND THE MASS OF THE CO2 X AVERAGE SPEED = TOTAL MASS OF THE CART X AVERAGE VELOCITY. THIS IS HOW A ROCKET IS ACCELERATED AND IT WORKS IN OUTER SPACE. 4/4/2017 Physics 214 Fall 2011 24 1H-04 Hero's Engine A glass bulb emits steam from small nozzles What happens when the Glass Bulb begins to emit steam ? Reaction = Bulb Spins Action = Ejects Steam Same Principle causes a Lawn Sprinkler to Turn. THE REACTION FORCE TO THE EJECTION OF MASS CAUSES THE OBJECT TO SPIN. THIS IS THE SAME AS THE CO2 ROCKET IN THAT MATERIAL IS EXPELLED AT HIGH VELOCITY 4/4/2017 Physics 214 Fall 2011 25 1J-04 Scale Paradox 1 A Scale Measures the Force acting on it NOW, What is the reading on the scale ? What is the reading on the scale ? WALL mg mg mg T T T T mg T = mg mg T = mg mg mg THE TENSION IN THE CORD IS THE SAME FOR BOTH CASES. THE SCALE MEASURES THE TENSION IN THE CORD. FOR EXAMPLE THE TENSION IN A ROPE IS THE SAME IF TWO PEOPLE PULL ON EACH END WITH FORCE F OR IF ONE PERSON PULLS WITH FORCE F TO A ROPE TIED TO A WALL. 4/4/2017 Physics 214 Fall 2011 mg 26 Questions Chapter 4 Q8 A 3-kg block is observed to accelerate at a rate twice that of a 6-kg block. Is the net force acting on the 3-kg block therefore twice as large as that acting on the 6-kg block? Explain. The net force is the same Q9 Two equal-magnitude horizontal forces act on a box as shown in the diagram. Is the object accelerated horizontally? Explain. -F F No the net force is zero Q10 Is it possible that the object pictured in question 9 is moving, given the fact that the two forces acting on it are equal in size but opposite in direction? Explain. Yes, constant velocity 4/4/2017 Physics 214 Fall 2011 27 Q18 The acceleration due to gravity on the moon is approximately one-sixth the gravitational acceleration near the earth’s surface. If a rock is transported from the earth to the moon, will either its mass or its weight change in the process? Explain. It’s mass will not change but it’s weight wil be 6 times less Q22 The engine of a car is part of the car and Ftire cannot push directly on the car in order to accelerate it. What external force acting on the car is responsible for the acceleration of the car on a level road surface? Explain. Fair Freaction It’s the reaction force between the tires and the road Q23 It is difficult to stop a car on icy road surface. It is also difficult to accelerate a car on this same icy road? Explain. Because of a lack of friction the wheels will skid or spin 4/4/2017 Physics 214 Fall 2011 28 Q25 When a magician performs the tablecloth trick, the objects on the table do not move very far. Is there a horizontal force acting on these objects while the tablecloth is being pulled off the table? Why do the objects not move very far? Explain. Yes but the force acts for a very short time and the objects start to move, then when the cloth is gone friction stops them. Q30 Two masses, m1 and m2, connected by a string, are placed upon a fixed frictionless pulley as shown in the diagram. If m2 is larger than m1, will the two masses accelerate? Explain. Yes m2 will fall and m1 will rise 4/4/2017 Physics 214 Fall 2011 • m m 2 1 29 Q31 Two blocks with the same mass are connected by a string and are pulled across a frictionless surface by a constant force, F, exerted by a string (see diagram). A. Will the two blocks move with constant velocity? Explain. B. Will the tension in the connecting string be greater than, less than, or equal to the force F? Explain. F A. They will accelerate F = ma B. The tension will be less 4/4/2017 Physics 214 Fall 2011 30 Q33 If you get into an elevator on the top floor of a large building and the elevator begins to accelerate downward, will the normal force pushing up on your feet be greater than, equal to, or less than the force of gravity pulling downward on you? Explain. + N g N – mg = ma but a is negative so N is smaller than mg The only force pulling you down is gravity so if you are accelerating down the force due to gravity must be larger than the reaction force N ( N is apparent weight) 4/4/2017 Physics 214 Fall 2011 mg a 31 Ch 4 E4 A 2.5kg block is pulled with a force of 80N and friction is 5N a) What is the acceleration? 5N 2.5 kg Net force = 75 N 4/4/2017 80 N a = 75/2.5 = 30 m/s2 Physics 214 Fall 2011 32 Ch 4 E6 A 6kg block is being pushed with a force P and has an acceleration of 3.0m/s2 a) What is the net force? b) If P is 20N what is Ff? P a = 3m/s2 6 kg Ff a) F = ma = 6 x 3 = 18 N b) F = P – Ff 4/4/2017 Ff = 2N Physics 214 Fall 2011 33 Ch 4 E14 A 4kg rock is dropped and experiences air resistance of 15N a) What is the acceleration? 4 kg 15 N Mg F = 4 x 9.8 – 15 F = ma = 24.2 N a = 24.2/4 = 6.05m/s2 4/4/2017 Physics 214 Fall 2011 34 Ch 4 E16 g N A vertical force of 6N presses on a book. a) What is the gravitational force? b) What is the normal force ? 6N 0.4 kg a) Gravitational Force = mg = 3.92 N b) Upward Force = 6 + 3.92 = 9.92 N 4/4/2017 Physics 214 Fall 2011 35 Ch 4 E18 m = 60 KG A 60kg person is in an elevator With an upward acceleration of 1.2m/s2 a) What is the net force? b) What is the gravitational force? c) What is the normal force? a) Net Force a = 1.2 m/s2 mg F = Ma = 60 x 1.2 = 72 N b) mg = 60 x 9.8 = 588 N c) N = 588 + 72 = 660 N 4/4/2017 Physics 214 Fall 2011 36 Ch 4 CP4 A 60kg crate is lowered from a height of 1.4m and the tension is 500N a) Will the crate accelerate? b) What is the acceleration? c) How long to reach the floor? d) How fast does the crate hit the floor? a) Net Force = 60 x 9.8 – 500 = 588 – 500 g 60 kg = 88 N b) Will accelerate down a = 88/60 = 1.47 m/s2 c) d = 1/2 at2 t = 1.38s d) v = v0 + at v = 2.03 m/s 4/4/2017 500 N Physics 214 Fall 2011 37 Ch 4 CP6 A 60kg person accelerating down at 1.4m/s2 a) What is the true weight? b) What is the net force? c) What is N? d) What is the apparent weight? e) a, b, c, d with 1.4m/s2 up? a) True weight = mg N 1.4 m/s2 Mg = 60 x 9.8 = 588 N b) Net Force = Ma = 84 N c) N = 588 – 84 = 504 N d) 504 N e) ↑1.4 m/s2 Net Force = 84 N↑ 4/4/2017 N = 588 + 84 = 672 N Physics 214 Fall 2011 W = 672 N 38 Review Chapters 1 - 4 - d + x Units----Length, mass, time SI units m, kg, second Coordinate systems Average speed = distance/time = d/t Instantaneous speed = d/Δt Vector quantities---magnitude and direction Magnitude is always positive Velocity----magnitude is speed Acceleration = change in velocity/time =Δv/Δt Force = ma Newtons 4/4/2017 Physics 214 Fall 2011 39 Conversions, prefixes and scientific notation giga 1,000,000,000 109 billion 1 in 2.54cm mega 1,000,000 106 million 1cm 0.394in kilo 1,000 103 thousand 1ft 30.5cm centi 1/100 10- hundredth 1m 39.4in thousandth 1km 0.621mi 1mi 5280ft 1.609km 1lb 0.4536kg g =9.8 1kg 2.205lbs g=9.8 0.01 3.281ft 2 milli micro 1/1000 1/1,000,000 0.00 1 1/106 103 10- millionth 6 nano 1/1,000,000,000 1/109 109 4/4/2017 billionth Physics 214 Fall 2011 40 Speed, velocity and acceleration v = Δd/Δt a = Δv/Δt The magnitude of a is not related to the magnitude of v the direction of a is not related to the direction of v 2 3 4 1 v = v0 + at constant acceleration d = v0t + 1/2at2 d,v0 v,a can be + or d = 1/2(v + v0) t v2 = v02 + 2ad 4/4/2017 Physics 214 Fall 2011 41 One dimensional motion and gravity v = v0 + at d = v0t + 1/2at2 v2 = v02 + 2ad d = ½(v + v0)t + g = -9.8m/s2 + 4/4/2017 At the top v = 0 and t = v0/9.8 At the bottom t = 2v0/9.8 Physics 214 Fall 2011 42 Equations v = v0 + at d = v0t + 1/2at2 d = ½(v + v0)t v2 = v02 + 2ad Sometimes you have to use two equations. v0 = 15m/s v = 50m/s What is h? v = v0 + at 50 = 15 + 9.8t t = 3.57 s v0 ` h = v0t + 1/2at2 g h v 4/4/2017 h = 15 x 3.57 + 1/2x9.8x3.572 = 116m h = ½(15 + 50) x 3.57 = 116m Physics 214 Fall 2011 43 Projectile Motion axis 1 axis 2 v1 = constant and d1 = v1t vv = v0v + at and d = v0vt + 1/2at2 v1 g 9.8m/s2 h v R Use + down so g is + and h is + h = v0vt + 1/2at2 v0v = 0, t2 = 2h/a R = v 1t v = v0v + at 4/4/2017 Physics 214 Fall 2011 44 Complete Projectile v0v v1 9.8m/s2 v1 v1 v0v highest point the vertical velocity is zero vv = v0v + at so t = v0v/9.8 h = v0vt + 1/2at2 end t = 2v0v/9.8 and R = v1 x 2v0v/9.8 and the vertical velocity is minus v0v 4/4/2017 Physics 214 Fall 2011 45 Newton’s Second and First Law Second Law F = ma unit is a Newton (or pound) First Law F = 0 a = 0 so v = constant Third law For every force there is an equal and opposite reaction force N Weight = mg mg Ff F F Ff F = ma 4/4/2017 v = v0 + at d = v0t + ½ at2 d = ½(v + v0)t v2 = v02 + 2ad Physics 214 Fall 2011 46 Examples + T N g 30 – 8 – T = 4a T – 6 = 2a 30 – 8 – 6 = 6a mg N – mg = ma a + N > mg a – N < mg 4/4/2017 Physics 214 Fall 2011 47