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AP DYNAMICS NOTES – Newtons Laws 1. 2. 3. 4. 5. 6. Center of Mass Newtons 3 laws Freebody Diagrams Friction Inclined Planes Tension / Atwood machine INTRO: MASS VS WEIGHT (demos: space guy and planets & alien) Mass and Weight are two concepts easily confused. In this unit, we must clearly know the difference between them. Mass – The amount of matter in an object Units – Kg (a gram is tiny, so the standard SI unit is KG) The english system uses slugs (from sluggish) a more massive object is harder to move, or more sluggish. For most of history, there wasn’t a unit for mass, since most people just use weight. English units: Slug (for sluggish) – 1 slug = 14.6 kg Mass doesn’t change, unless you physically add or take away pieces of the object. Weight – The force of gravity acting on an object. Fg = M x a SI units: Newton (N) = 1 kg m/s2 English units – Pound. 1 pound = 4.5 N On earth, a = 9.8 m/s2. However, on other planets acceleration of g is different, so weight can change. Also, in water, you are bouyant, so you weigh less. Weight can change, mass stays constant The only way to change mass is to physically add or subtract matter from an object. Where the confusion begins: On earth, gravity is always 9.8, so weight seems constant. Allot of people confuse mass with weight. This is also compounded by the english system. The unit for mass is the Slug (sluggish). No one really uses a “slug” most people just equate pounds (weight) with mass. On earth, 1 pound of weight = 2.2 kg 6. Center of Mass Center of Mass – The point where all of the mass of an object can be considered to be concentrated. For example, if we have an object, like a boat with people: The center of mass will be the balancing point on the boat. If we could pick the boat up, and balance it, we would need to pick it up at the center of mass, in order for it to balance. Notice, the CM depends on where all of the individual masses (in this case boat and people) are distributed. Equation: where the summation signs mean to add the total If we have a collection of particles, we must place each particle into the equation. For example, if we have 4 individual masses making up the entire object, the equation will be: Now, what if the object was 2 or 3 dimensional? How do we find the center of mass? Find the center of mass in each dimension: Xcm, Ycm, and Zcm Steps in solving Center of Mass problems: 1.) Define the center point and create a coordinate system at this point. Make this step easy on yourself, and pick the center of one of the masses. 2.) Find the position of ALL of the particles in the mass system 3.) Use CM formula to find the Center of Mass in ALL dimensions Example 1: Lets place a mass at the end of a long poll, as shown below. Calculate the center of mass of the system. 1.) Define the center point and create a coordinate system at this point. Make this easy on yourself, lets say the center of the pole is the center point. This will be position (0,0,0) (X,Y,Z) Notice, there is only an X dimension (assume the mass and pole are thin), so we don’t have to bother with Y and Z. 2.) Find the position of ALL of the particles in the mass system There are two particles, the pole (M) and the mass (3M) 1. Center of the poll is located at (0,0,0) 2. Center of the mass is located at (1/2L,0,0) 3.) Use CM formula to find the Center of Mass in ALL dimensions So, the center of mass is (L/8,0,0) Example #2 Find the center of mass of the following object: 1.) Define a center point and create a coordinate system around this point Lets make the center point the center of the rod. 2.) Find the position of ALL of the particles in the mass system 1. The rod is (0,0,0) 2. The larger mass is: (3/2 L, -2L, 0) 3. The smaller mass is: (-3/2 L, -L, 0) 3.) Use CM formula to find the Center of Mass in ALL dimensions There is an X and Y dimension… Find Xcm: Find Ycm: Answer: CM = (3/4 L, -5/6L,0) Do example with masses and weightsw in class CM HW CM Lab _________________________________________________________________________________________ Newton’s 1st Law (demos: tablecloth pull, card and nickels, cart and flying doll) An object in motion will tend to stay in motion, an object at rest will tend to stay at rest, unless acted upon by an unbalanced (external) force. Example. A book sitting on a table has gravity pushing down, but the table stopping it. The forces are balanced. It will stay there until some other force disturbes it, causing motion. A spacecraft launched to another planet: Once it leaves earth, the rockets shut off, and it simply coasts to the other planet. Once in motion, it will stay in constant motion until it hits the other planet (in many cases years). The 1st law is also called the Law of Inertia Interia: The tendency for an object to resist a change in motion. Inertia is directly related to Mass. Bigger Mass = More Inertia This is an easy concept. For example, take two rocks, one boulder, and one pebble. Which one is harder to kick? Obviously, the boubler is MUCH harder to kick, it has more mass, so it wants to stay are rest more than the little pebble. Again, lets say that boulder is rolling down the road. How hard would it be to stop? Very hard compared to the pebble. Since it has such a large mass, once in motion it will stay in motion. Example: Sliding a book across a desk. You apply a force to start it in motion. According to the 1st law, it should stay in motion. Why doesn’t it? Because friction from the surface of the desk applies an unbalanced force, which causes it to slow down. What if we slid it on ice? There is less friction, so it would travel further. Wrap-up of 1st law: Law of Inertia So, if an object is at rest, one must apply a force to overcome the objects inertia in order to get it to move. If an object is moving, one must apply a force to overcome the objects inertia in order to stop it’s motion. _______________________________________________________________________________ Newton’s 2nd Law (demos: hovering globe, cart with increasing mass) The 2nd law deals with unbalanced forces. According to 1st law, if forces are balanced, there will not be any acceleration (no change in motion) (do hovering globe) Balanced forces: A = 0 m/s2 Two cases: If V = 0, the object will stay at rest or if V ± 0, the object will stay in constant motion (like a projectile in the X direction) According to the 1st law, if forces are Unbalanced, the object will accelerate (change in motion) Unbalanced forces: A ± 0 m/s2 Newton experimented with unbalanced forces and found a relationship: a proportional 1/m and a proportional Fnet a prop 1/m - makes common sense, a larger mass has more inertia, therefore it won’t change motion as fast (a will be less) a prop F - makes common sense also, a larger force, larger disturbance, will cause a bigger change in motion. (do demo with cart varying masses) Newton combined the two proportions: A prop F/m - turn prop sign to = sign (add constant) constant = 1. a = F/m - Newton’s 2nd law. Often we see it written as F = ma Newton’s 3rd Law: (rope “tug of war” with force readers demo) For every action force there is an equal but opposite reaction force. What this means is forces ALWAYS come in pairs. This may sound counter intuative, but action/reaction pairs are easily seen in nature. Swimmer in water – swimmer pushes arms back, water pushes person foreward Birds – Wings push air down, air pushes bird up Car – the wheels spin, pushing Backward on the road. Road pushes foreward on the car Walking – You push backward on the ground, ground pushes you foreward. Jumping – You push DOWN on ground, ground pushes you upward. Student sitting on chair - Gravity pushes person down, chair pushes back up on person Notice all of these reaction forces are OPPOSITE in direction. If measured, they are also EQUAL in magnitude. Just because the forces are equal, doesn’t mean the accelerations are equal: Example: Moon in orbit – Moon pulls on Earth, Earth pulls back on moon The two forces are equal. However, the moons mass is MUCH smaller, so by newtons 2nd law, the acceleration of the moon falling toward earth is MUCH grater. So the moon orbits the earth, not vice versa (skateboard demo: raft/boat example) This is a really important law, and the basis for constructing “FREE BODY DIAGRAMS” Diagrams used to solve a variety of force/motion problems. (Look at a horse and cart example:) A FREEBODY DIAGRAM only shows the forces acting ON the object. They don’t show any forces that are NOT acting on the object. Example: (Fan Sail Demo) Freebody Diagrams & problem solving In order to sucessfully do “Newton’s laws “ problems, you should first construct a free body diagram of the situation. This makes solving the problem much easier. Steps: 1. 2. 3. 4. 5. 6. Draw a freebody diagram of situation Write a equations for total force in the X directoin and the Y direction. Write down knowns and unknowns Pick appropriate equation(s). Substitute/ Sovle for appropriate variable Plug in #’s and solve By using freebody diagrams, we can do many applicatoins of newtons laws. Two notable ones are Friction and Tension. Friction: Application of Newton’s laws: A good example of where we use freebody diagrams is friction. Friction: A force that resists motion when two objects are in contact (floor and object you are dragging). Friction always acts in the opposite direction of motion. Friction is NOT a conservative force. This means the energy used during the frictional force cannot go back into the object, the energy is used in sound, heat, light, etc… that is given off to the environment. What causes friction? If we look on the microscopic level, even a smooth surface appears rugged. The two surfaces that are in contact rub against each other, the microscopic peaks and valleys catch one another. There are also other reasons for friction, even if the objects are perfectly smooth, interactions with the electrons will cause friction. ___________________________________________________________________________________ There are two types of Friction: Static and Kinetic Static Friction (Fs) – The friction that needs to be overcome in order to start an object in motion (Friction of a stationary object) If you apply a small force to an object, it will not move. You need to apply enough force to overcome its static friction. Kinetic Friction (Fk)- The firction that needs to be overcome in order to keep an object in motion. (Friction of a moving object) Static friction is ALWAYS greater than Kinetic friction. The reason why related to newton’s 1st law of motion. An object at rest wants to stay at rest, so one needs to apply an extra force to overcome it’s inertia An object in motion wants to stay in motion, so one does not need ot apply an extra force to keep it moving. So, in the presence of friction, once an object starts moving, it requires less force to keep it moving. How do we measure friction? What causes two surfaces to push on each other? Gravity. Gravity pushes down, and the Normal force from the surface pushes back up. Therefore, the Normal force is the root cause of friction. Fk is proportional to (Fn) To get proportion to an equal sign, Fk = (constant)(Fn) The constant is a number that tells how much friction a surface has. For example, Ice has a small constant, while Rubber has a large constant. This constant is called the Coefficient of friction. The symbol is µ - mu – it is unitless For kinetic friction, the coefficent of friction si µk For static friction, the coefficent of friction si µs So, the equation for kinetic friction is Fk = µk Fn For static friction, the equation is Fs = µs Fn Again, µs is always greater than µk *** Coefficients can not be greater than 1. 1 is the maximum friction. Some common values of coefficients of kinetic and static friction: Surfaces µ µ (static) (kinetic) Steel on steel 0.74 0.57 Glass on glass 0.94 0.40 Metal on Metal (lubricated) 0.15 0.06 Ice on ice 0.10 0.03 Teflon on Teflon 0.04 0.04 Tire on concrete 1.00 0.80 Tire on wet road 0.60 0.40 Tire on snow 0.30 0.20 Synovial Fluid (joints) 0.01 0.003 Solving Friction problems: 1. 2. 3. 4. It is important to first draw a freebody diagram of the situation Next, pull out your 2 freebody equations Solve the equations for the appropriate variables Do substitutions for friction and gravity if necessary Hints: Usually you need to find FN, use your freebody equation in the Y direction to find it. Remember, any time you are looking for Force of Gravity (the weight) Fg, Fg = m ag Examples: 1.) A 15 kg object is being dragged with a force of 70 N at a constant rate along a level surface. What is the coefficient of kinetic friction between this object and the ground? 2.) A 15 kg object is being dragged with a force of 70 N at an angle of 30 degrees a constant rate along a level surface. What is the coefficient of kinetic friction between this object and the ground? 3.) Draw a freebody diagram of the following situation in order to determine the force applied in the x direction. An object of mass 10 kg is accelerating at a rate of .2m/s2 along a level surface along the x direction. k =.2 ____________________________________________________________________________ Inclined Planes: If we look at an inclined plane, and draw a freebody diagram of An object sliding down it, we get the following illustration: Notice, Fg is straight, all the others are crooked. Instead of using trig twice to rotate the FN and Fk, we can take a short cut and rotate the X and Y axis. Now, if we look at the diagram, only Fg is crooked. We just have to use trig for Fg. Lets make a triangle and solve Fgx and Fgy Notice Fgy is perpendicular to the plane hyp of the inclined plane, and Fgx is parallel to the hyp of the inclined plane. This means the inclined plane triangle and the Fg triangle is a similar triangle, which also means the two thetas are the same (as shown in the diagram) From our Fg diagram, we can use trig: Fgx = Fg sin θ Fgy = Fg cos θ Now that we know Fgx and Fgy we can redraw our original freebody diagram using Fgx and Fgy, so now all of the force vectors are aligned with the X and Y axis. Once we have this diagram set up, and the equations for Fgx and Fgy, we can now write a freebody equation for the X and Y directions, and solve for what the problems ask for. Special Case: If we have an inclined plane, set at a low angle, the object will not move, friction will hold it back. Now, if we raise the angle of the inclined plane, to a high angle, the object will accelerate down, Fgx will be greater than Fk. What if we change the angle so it is between the two above cases, where the angle is small enough so that it doesn’t accelerate, but large enough so that is moves. This means Fgx = Fk, and the object will travel at a constant velocity. This is a special angle. There is only 1 angle, a threshold angle, in which the object will just start sliding at a constant velocity. Any higher, it accelerates, and lower it stops. At this special angle, Fgx = Fk Example: What is the maximum angle a plane can be tilted so that it does not move? Solution: If we think carefully about this problem, we actually want to find the angle at which the object just starts to move, that threshold angle. Anything before that angle, and the object will not move. Draw your diagram: Write your freebody equations: X: Fgx – Fk = 0 Fgx = Fk (there is no acceleration) Y: Fn –Fgy = 0 Fn = Fgy Look at X equation: Fg sin θ = µkFn Fg sin θ = µk Fg cos θ Fg’s cancel out… sin θ = µk cos θ We are looking for theta, so put thetas on one side: sin θ / cos θ = µk In Trig, sin/cos = tan , therefore: Tan θ = µk This is the equation for the threshold angle, the angle at which a stationary object will just start to move. Any higher angle, the object accelerates, and lower angle the object stops. ___________________________________________________________________________________ TENSION Tension is a reaction force applied by a stretched string (rope or a similar object) on the objects which stretch it. The direction of the force of tension is parallel to the string, away from the object exerting the stretching force. So if an object hangs from a rope due to gravity, the gravitational force on the object points downward, and there is an equal tension force in the rope point upward, making the net force on the object equal to zero. Ftot = Fa – Ft = ma Tensile Strength - measures the stress required to pull something such as rope, wire, or a structural beam to the point where it breaks. Tension: Equilibrium Problems – Sum of all forces = 0 , No Acceleration Key to solving. Break the tensions into the X and Y components, then solve. Examples: Find the weight of the sign: The tension is 30.0 N and the angle is 45 degrees. Thus, sine (45 degrees) = (Fvert) / (30.0 N). The proper use of algebra leads to the equation: Fvert = (30.0 N) • sine (45 degrees) = 21.2 N Each cable pulls upward with 21.2 N of force. Thus, the sign must weigh twice this - 42.4 N The weight of the sign is 42.4 N. ____________________________ M = 50 kg, Fint Tension on rope. Since the mass is 50 kg, the weight is 490 N. Since there is only one "upwardpulling" cable, it must supply all the upward force. This cable pulls upwards with approximately 490 N of force. Thus, sin (30 degrees) = (490 N ) / (Ftens). Proper use of algebra leads to the equation Ftens = (490 N) / [ sine 30 (degrees) ] = 980 N. The tension is 980 Newtons. ___________________________ Mass = 50 kg, find tension on Each cable. Since the mass is 50.0 kg, the weight is 490 N. Each cable must pull upwards with 245 N of force. Thus, sine (45 degrees) = (245 N ) / (Ftens). Proper use of algebra leads to the equation Ftens = (245 N) / [sine (45 degrees)] = 346 N. The tension is 346 Newtons. Atwood Machine: Atwood’s Machine is a simple device. It is a pulley with two masses attached to it by one string. Obviously, the larger mass will fall down, and the smaller mass move up. If the masses are equal, they will balance, and not move. An example of an Atwood’s machine is shown to the right. In this example, M1 > M2. ** Important concept to know: Since there is 1 string, the Tension on the string from Mass 1 IS THE SAME as the Tension on the string from Mass 2. It is the SAME string. T1 = T2 ** Notice, their accelerations are the same, it’s the same string! If you accelerate one side of the string, the other will accelerate the same. Using Newton’s laws (specifically constructing freebody diagrams), we can calculate the accelerations of the masses. From the example pictured above, let’s calculate acceleration of the masses. STEP 1: Draw a freebody diagram: Notice, it’s the same diagram as above, just with the direction of motion added so we don’t get confused. STEP 2: Make your freebody equations: In this case, there are 2 objects, M1 and M2, so we need equations for M1 AND M2. M1: M1 is moving down, we will make the direction of motion positive. It is also accelerating, therefore: Fg1 – T = ma - Fg1 is mass times acceleration – Fg1 = M1 ag We can plug that in: M1 ag – T = M1a M2: M2 is moving up, lets make Up positive: T – M2 ag = M2a STEP 3: Solve for acceleration: We now have two equations. We want to solve for acceleration (a). Lets combine equations. Easiest way is to substitute T, since both have T: M1: T = –M1a + M1 ag T = M1 ag – M1a M2: T = M2a + M2 ag Both are solved for T, lets substitute: M2a + M2 ag = M1 ag – M1a Solve for (a), so get (a) to one side: M2a + M1a = M1 ag – M2 ag Pull out (a) a (M2 + M1) = M1 ag – M2 ag Divide out M2 and M1: a = (M1 ag – M2 ag) / (M2 + M1) We can make it look easier by pulling out gravity: a = ag (M – M2) / (M2 + M1) This is the acceleration of the blocks. **If we wanted to find the velocity of the blocks after a certain time or distance, just use the Kinematic equations, now that you know (a). ** If M2 was greater than M1, just switch M1 and M2 around in the equation. What if we wanted T instead of a? Now that you know the equation for a: a = ag (M – M2) / (M2 + M1) We can substitute that back into one of the original equations for T. Ex: T = M1 ag – M1a Substitute (a) in: T = M1 ag – M1 (ag (M – M2) / (M2 + M1)) You can stop there, or you can simplify it: To simplify, both terms have M1 ag, so pull it out: T = M1 ag (1 – ((M – M2) / (M2 + M1)))