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The Calculus connection When Newton first wrote the relationship we call the second law, it was not it the form we use now. Instead, he wrote dp F= dt dp dmv dv F= m ma dt dt dt More than one force can act on an object at once. For example, two people could push on a book at the same time. One person could push toward the left and the other could push toward the right. In this case the two forces would act against each other. What is the NET force? 2 N towards the left If the book had a mass of ½ kg, what would be its acceleration? a = Fnet / m a = 2 / .5 a = 4 m/s2 (left) What if the opposing forces were equal? What is the net Force? What is the acceleration? What if the forces were in the same direction? What is the net force? If the book had a mass of 2 kg, what is its acceleration? a = Fnet / m a = 14 / 2 = 7 m/s2 • What are the unknown forces for the given net force? We usually label forces as negative or positive. Forces upward are positive. Forces downward are negative. Forces to the right are positive. Forces to the left are negative + + - What is the net Force? - 6 N + 18 N = + 12 N What is the acceleration? a = Fnet / m 12 / 3 = 4 m/s2 6N 3 kg 18N What is the net Force? - 6 N – 15 N + 18 N = - 3 N What is the acceleration? a = Fnet / m a = - 3 N / 3 kg = - 1 m/s2 6N 15N 3 kg 18N Weight Weight, Wt. is the gravitational force on an object Weight = mass x gravity W = mg Since weight is a force, it is measured in Newtons, N Remember, “g” on Earth is 9.8 m/s2(10 m/s2) What is the weight of a 42 kg child on Earth? W = mg m = 42 kg W = 42 kg x 10 m/s2 W = 420 N What is the mass of a 15000 N car? W = mg m=W/g m = 15000 N / 10 m/s2 m = 1500 kg Remember, one Newton is not a very big force (about the same as a ¼ pound). So, your weight in Newtons is MUCH bigger than your weight in pounds! In fact, you would have to multiply your weight in pounds by 4.45 to get your weight in Newtons. How much do you weigh in Newton’s? Even if you weigh 550 Newtons, You still wouldn’t be much of a Sumo Wrestler! (that’s only around 120 lbs) A car weighs 1680 N on Earth. What is its mass? W = mg m=W/g m = 1680 N / 10 m/s2 m = 168.0 kg What is the net force on the 50 kg parachutist? -100 N What is his acceleration? a = Fnet / m a = -100 / 50 a = -2 m/s2 What is the net Force? Hold on, there’s another force not drawn! The gravitational force of weight! Wt = mg, (g = 10 m/s2) Wt = 2 kg x 10 m/s2 = 20 N Draw the weight vector also! Now, what is the net Force? Net force = +19 N – 5 N – 20 N = Net Force = - 6 N What is the acceleration? a = Fnet / m = a=-6/2 a = -3 m/s2 It will accelerate downward. 19 N 2 kg 5N mg = 20 N Tension Tension, T, is the force that cables, ropes, and strings pull with. A child pulls up on a string that is holding 2 fish of total mass 5 kg. If he is providing a tension of 60 N, what is the net force on the fish? Fnet = + T – W Fnet = 60 N – 50 N Fnet = 10 N What is the acceleration of the fish? a = Fnet / m a = 10 / 5 a = 2 m/s2 , “sigma” is a Greek letter that is used to signify “the sum of” Quite often, in Newton’s 2nd Law, we write F = ma instead of Fnet = ma A child pulls a 5 kg bucket out of well with a rope. If the bucket accelerates upward at 1.2 m/s2, what is the tension in the rope? m = 5 kg a = 1.2 m/s2 T T=? F = ma T – mg = ma mg T = ma + mg T = 5 x 1.2 + 5 x 9.8 T = 55 N Three kids pull on a 2 kg cat. One child pulls left with a force of 20 N. One child pulls right with a force of 50 N. If the cat accelerates to the left at 3 m/s2, what was the force exerted by the third child, F3? F1 = 20 N F2 = 50 N F = ma -F1 + F2 + F3 = ma F3 = ma + F1 – F2 F3 = 2(-3) + 20 – 50 F3 = -36 N The “Normal” Force, N When an object is pressed against a surface, the surface pushes back. (That’s Newton’s 3rd Law) This “push back” from the surface is called the Normal Force, N The word “normal” in math terminology means “perpendicular” The surface pushes back in a direction that is perpendicular to the surface. Normal force, N Weight = mg If the box is not accelerating, then the Normal force must be balanced by the Weight But… what if you were accelerating up or down? The Normal force would NOT be equal to your weight if you’re accelerating up or down. And… your weight seems to change! Apparent Weight When you ride an elevator, you “feel” heavier or lighter than you actually weigh because of the acceleration of the elevator. Your “apparent weight” is found by taking your REAL weight, mg, and adding the term ma, where “a” is your acceleration Apparent weight = mg + ma A 50 kg woman steps in an elevator that accelerates upward at 1.5 m/s2. What is her REAL weight? mg = 500 N What is her APPARENT weight? Fnet = ma N – mg = ma Her APPARENT weight is what she feels like she weighs, given by the Normal force, N. N = ma + mg N = 50 x 1.5 + 500 Apparent weight = 575 N N mg Do falling objects REALLY accelerate toward the Earth at 9.8 m/s2? No, because of air resistance. Air resistance is a force that pushes up on an object as it falls. The faster you fall… The greater the air resistance. Eventually, the air resistance pushing up on you is just as large as your weight that is pulling down on you!! The faster the man falls, the more air resistance pushes up on him. Eventually, there will be just as much air resistance pushing up on him as his weight pulling him down. What will be the NET force acting then? What will be his acceleration? Once the air resistance pushing up is as large as the weight pushing down, the NET force acting on you is ZERO! If the net force is zero, what is your acceleration? ZERO! This doesn’t mean you stop in mid air. But it does mean that you stop accelerating! You still continue to fall towards the Earth, but you don’t pick up any more velocity- you continue to fall towards the Earth at the same velocity. This speed is called your “terminal velocity” You will reach your terminal velocity when Air resistance = your weight Which one will have a faster terminal velocity? You don’t reach terminal velocity until the air resistance grows to as large a force as your weight. The more massive skydiver will have a faster terminal velocity and hit the ground at a faster speed The “Spring Force” If an object is attached to a spring and then pulled or pushed, the spring will exert a force that is proportional to the displacement of the object but opposite in direction. Fs = -kx Where k is called the “spring constant” or the “force constant” and describes how stiff the spring is. The units for k are N/m Friction, f • A force that always opposes motion • Depends on two things: the roughness of the surfaces and how hard they are pressed together. f = mN m , mu- the “coefficient of friction” tells how rough the surfaces are. N, the Normal force tells how hard the surfaces are pressed together Example: How large is the frictional force between 2 surfaces if the coefficient of friction is 0.2 and the Normal force is 80 N? f = mN f = 0.2 x 80 f = 16 N There are two kinds of friction: “static friction” (not moving) must be overcome to initiate motion. “kinetic friction” must be overcome while an object is moving Static friction > Kinetic friction You pull on a box with an applied force of 30 N. The coefficient of friction is 0.4. If the mass of the box is 2 kg, what is its acceleration? 1. Draw the box and all FOUR forces acting on it. 2. Write what you know and don’t know. 3. Write the equations, Fnet = ma and f = mN 4. Calculate the Normal force and the friction force. 5. Calculate the value of the net Force and then the acceleration. FA = 30 N m = 2 kg m = 0.4 a=? Fnet = ma f = mN f = mN N = mg = 2 x 10 = 20N f = mN = 0.4(20) = 8N Horizon: Fnet = FA - f 30N – 8N = 22N a = Fnet / m a = 11 m/s2 Normal force, N FA= 30 N Weight = mg FA = 25N m = 2 kg q = 43 m = 0.4 a=? Fx = ma f = mN Fx = - mN + FA cos q = ma Fy = N – mg - Fsin q = ma = 0 N = mg + Fsin q 36.65 N Normal force, N (-0.4(36.65) + 25cos 43 ) / 2= a a = 1.81 m/s2 f q FA Weight = mg FA = 15 N m = 2 kg q = 43 m = 0.4 a=? Fx = ma f = mN Fx = - mN + FA cos q = ma Fy = N – mg + Fsin q = ma = 0 N = mg – Fsin q 9.77 N Normal force, N FA (-0.4(9.77) + 15cos 43) / 2 = a a = 3.53 m/s2 f q Weight = mg If the box is moving at constant speed, there is no acceleration, Therefore the net force must be zero… so the horizontal forces must cancel each other. f mN = Fcos q Normal force, N FA q If you push hard enough to just get the box moving, the acceleration is zero in that Weight = mg case also, but the friction is static, not kinetic. Inclines: With friction, case 1: at rest or sliding down Draw the weight vector, mg Draw the Normal force vector. Draw the components of the weight vector, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle. Write Newton’s Second Law for the box. F = ma What is friction? N f - mgsin q = ma mmgcos q - mgsinq = ma f mgcos q - gsinq = a q f = mN What is N? N = mgcos q f = mmgcos q mg q Inclines: With friction, case 2: pushed downward Draw the weight vector, mg Draw the Normal force vector. Draw the components of the weight vector, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle. Write Newton’s Second Law for the box. F = ma What is friction? N f - mgsin q – FA = ma mmgcos q - mgsinq – FA = ma f FA q mg q f = mN f = mmgcos q Inclines: With friction, case 3: pushed upward Draw the weight vector, mg Draw the Normal force vector. Draw the components of the weight vector, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle. Write Newton’s Second Law for the box. F = ma N FA - mgsin q – f = ma FA - mgsinq - mmgcos q = ma FA What is friction? f = mN f = mmgcos q q mg f q Friction along an incline An object placed along an incline will eventually slide down if the incline is elevated high enough. The angle at which it slides depends on how rough the incline surface is. To find the angle where it Since it doesn’t move: slides: mgsinq = mmgcosq Therefore: Tan qmax = mmax, the coefficient of static friction Or The angle qmax = tan -1 mmax q