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Chapter 1: Mechanics Before we start chapter 1, I want to be sure you al recall a few simple rules…. 1. you are expected to know how to work with powers and run your calculator 2. simplifying equations and substituting one equation into another is a common practice in AP physics 3. all students are expected to have a working knowledge of advanced algebra and trigonometry 4. all tests will consist of 3 questions (15 pts. constructed response) from old AP physics tests and 3-5 multiple choice (5 pts. each) and they will have a 40min time limit to complete 5. all homework problems will be worth 5 pts. each assignment and they’re due at the beginning of the hour the next day 6. you will have approximately 20 take home quizzes (valued at 15-25 pts.) throughout the year 7. we will do a chapter about every 6 days throughout the year (30 chapters-38 weeks) 8. sharing/copying answers will lead to zeros for all involved, however cooperative work is encouraged 9. you are expected to have a prior knowledge of mechanics- heat, waves, light, etc. from last year 10. we work only in the SI system- meters, kilograms, seconds, etc. 11. you will be given a formula sheet at the beginning of the year and it will be the only sheet you may use on quizzes/ tests in class 12. I do not care about sig figs- round answers to 3 places 13. UNITS MATTER!! please use them 14. dimensional analysis is the approved method for solving problems converting units 15. if you can do it in your head, cool! 16. if you don’t convert units it is wrong 17. if you don’t use units in your answer you will lose points = proportionality < = less than > = greater than = less than or equal to = greater than or equal to << = much less than = approximately equal to = not only equal to b = sum of values from a b x = change in x = xf – xi the origin is the reference point (R, ) = distance R from the origin at angle in polar coordinates (R, ) AP Problem Solving Strategies 1. 2. 3. 4. 5. 6. 7. read the problem draw a diagram identify the data given and the data needed choose and write out an equation solve the equation(s) showing all work evaluate your answer to see if it’s reasonable check your answer and your math 8. turn it in for a big, beautiful Assignment: pg 18-21 multiple-choice #3,6,7 and problems #3,6,14,16;21,39,45 Chapter 2: Motion in One Dimention 2.1. Displacement The study of the motion of an object and the relationship of this motion to force and mass is called dynamics. Describing motion wihout regard to its causes is called kinematics. In chapter 2 we will focus on kinematics in one dimentional motion (in a straight line). displacement: x which is the difference between the final and starting position of an object distance: how far something moved vector: direction and magnitude of movement/force scalar: magnitude of movement *our book used a bolded small a for an acceleration vector and a bolded small v for a velocity vector 2.2. v = average velocity = x/ t 2.3. v lim t 0 x/ t = instantaneous velocity velocity is speed in a scalar quantity *look at the pg 29 application—auto speedometers Example 2.1. on pg 30 A toy train: find i. v total trip ii. v seconds 0-4 iii. v seconds 4-8 iv. instantaneous v at T= 2 sec v. instantaneous v at T= 5 sec i. v = x/ T = 10cm/12sec = .83cm/sec ii. v = x/ T = 4cm/4sec = 1cm/sec iii. v = x/ T slope = zero, thus v = zero iv. slope at T= 2sec is 4cm/4sec = 1cm/sec *instantaneous v= 1 cm/sec v. slope = zero and instantaneous v = 0cm/sec Why is v equal to the instantaneous v in iv. and v.? a = v/ T = (m/s)/s = m/s2 instantaneous acceleration is a lim T 0 v/ T or the slope of the velocity-time curve at that point go over example 2.2. on pg 33-37 Assignment: multiple choice #1, 2, problems #2, 7, 8, 12, 21, 28 on pg 44 Open books to pg 46 and take 3 minutes in groups of 2. How deep is the well in meters? 2.5. Motion Diagrams Recall from last year….. constant velocity positive acceleration negative acceleration 2.6. One Dimenional Motion with Constant Acceleration because a = ainst when a is constant we will use a= (vf-vi)/T Vf = Vi +aT also… dx = Vi T+ (1/2) aT2 and Vf2 = Vi2 + 2ad *d = x Recall that these are the big three kinematic equations from last year!!! Add: dx = 1/2( Vf + Vi)T Example 2.4. A car with a velocity of 30m/s passes a trooper. The trooper takes off after the car exactly 1sec. after it passes. He moves with a Vi = 0m/s and a = 3m/s2. How long does it take for the trooper to catch the speeding car? speeding car xc = 30m + 30m/s(T) trooper xt = ½ aT2 = ½(3m/s2)T2 Because the cop catches the car… dxc = dxt 30m + 30m/s(T) = ½(3m/s2)T2 Move all to one side…. 0 = 1.5m/s2 T2 – 30m/sT – 30m then factor or use quadratic equation You get a negative value which is not useful (extraneous root) and T= 21.0 sec. Thus the cop catches up in 21sec. Dx = (30m/s)(21sec)= 630m Example 2.7. If 2 skydivers dive out of a plane 3sec. apart connected by a bungie cord would the tension on the cord remain the same, increase, or decrease? *answer on pg 42 Example: A golf ball is released from rest from the top of a tall building. Neglecting air resistance, calculate its position at a) 1sec. b) 2sec. c) 3sec. choose the roof at x=0 and y=0 Vi = 0m/s a = -9.8m/s2 T = 1, 2, 3sec. d = ??? a) v = (-9.8m/s2)(1sec.)= -9.8m/s2 dy = ½ aT2 = ½(-9.8 m/s2)(1sec.)2 = -4.9m b) v = (-9.8m/s2)(2sec.)= -19.6m/s dy = ½ aT2 = ½(-9.8 m/s2)(2sec.)2 = -19.6m c) v = (-9.8m/s2)(3sec.)= 29.6m/s dy = ½ aT2 = ½(-9.8 m/s2)(3sec.)2 = -44.1m Assignment: pg48-49 #26 ,29 ,30, 34 TEST on chapters 1 and 2 will be in 2 days—tomorrow we will review Oh no!!! I hate tests!!! Ahhh!!! Pasco lab free fall, with computers. (twenty minutes at start of class) Chapter 3: Vectors and 2 Dimentional Motion 3.1. Vectors Revisited *remember a scalar has magnitude but no direction…. like a lost freshman!!! *a vector has direction as well as magnitude *an example of a vector quantity is a force *scalars are added using simple arithmetic *vectors are added by parts 3.1. Adding Vectors drawn tip to tail R=A+B A B If you multiply a vector by a scalar you get a vector, same for division ex. 3(vector A) = 3A -3/(vector A) = -3/A *sign shows direction 3.3. Component Vector Addition A= 30N @ 72 B= 60N @ 100 30 cos72 60 cos100 x = R=√( x² + y²) 30 sin72 60 sin100 y= tan-1( y/ x) = Now look at example 3.3 page #58 using vector components. Solving Vectors 1. sketch vectors on an x,y coordinate plane 2. break into x,y components 3. add the squares of the sums of the x and y 4. square root the result from #3 5. draw sum of x and sum of y on the x,y grid to locate the correct quadrant 6. use the appropriate trig function to find 7. add to or subtract from to find angle from 0 Assignment: pg 72 #1, 8, 11, 13, 16, 19 3.4. Velocity in 2 Dimensions recall that v = x/ T now… v = R/ T *you must remember that whenever a velocity changes, a velocity occurs 3.5. Projectile Motion (woohoo, time to throw stuff ) g = 9.8m/s2, air resistance is ignored, and the Earth’s rotation does not affect our motion Ex. Upward from the ground, a projectile is shot at 20 with an instantaneous velocity = 11m/s. Find the time in the air, distance through the air, and the maximum height. Vf = 0m/s at top Vi = 11m/s (sin20) g = 9.8m/s2 Time: Vf = Vi + gT 0 = (11m/s)(sin20)+(-9.8m/s2)(T) T = .384sec. *use 2T to find total time in the air T(2) = 2(.384) = .768sec. Distance: dx = Vi T dx = (11m/s)(cos20)(.786sec.) dx = 7.94m Maximum Height: dy = Vi T+ ½ aT2 ymax = (11m/s)(sin20)(.384sec.)- ½ (9.8m/s2)(.384sec.)2 ymax = .722m Now an Ex. at 0 straight out off of a 100m cliff at 40m/s Vi = 40m/s = Vix * velocity in x direction remains constant * velocity in y direction = 0 at T = 0 -100m dx distance in x direction is dx = Vix (T) distance in y direction is dy = Viy (T)+1/2g (T2) -100 = (0m/s)(T)+1/2(-9.8m/s2)( T2) -100 = -4.9m/s2 ( T2) 20.4sec2 = T2 4.51sec = T solve for T: dx = Vix (T) dx = (40m/s)(4.51sec) dx = 180m 3.6. Relative Velocity Ex. A boat heads due north across a river with an initial velocity of 10km/hr. The river has a velocity of 5km/hr east. Find the relative velocity and angle. 5km/hr 10km/hr 52+102 = R2 11.2km/hr = R = tan-1(5/10) = 26.6 The boat travels at 11.2km/hr at an angle of 26.6. Assignment: pg 73 #20, 24, 26, 29, 32, 41 Extra Credit pg 78 #60, due on test day, next Tuesday covering chapters 1-3 all Chapter 4 4.1. Concept of Force *When two objects touch and one “forces” another to move it is called contact force. *Field forces are forces that interact without contacting each other. Gravity attractions, strong and weak nuclear forces, and electromagnetism are the four fundamental forces and are all examples of field forces. 4.2. Newton’s First Law (Law of Inertia) *Objects at rest remain at rest unless acted upon by an outside force. The same applies for objects in motion which must experience an external net force. *When the net external force on an object equals 0 the object’s velocity (if moving) will be constant. *The tendency of an object to resist a change of motion is its inertia. The greater the mass, the higher its inertia. 4.3. Newton’s Second Law (f =ma) *Force is directly proportionate to the product of mass and acceleration. *The unit of force is kg(m/s2) = 1 Newton 1g(cm/s2) = 1 dyne 105 dynes = 1N = .225lb *Weight is a force…. fw = mg the acceleration here is gravity where a = g = 9. 8m/s2 4.4. Newton’s Third Law If two objects interact, the force exerted by one object is equal and opposite to the force exerted by the object. Forces come in opposite or equal pairs and the action force is equal and opposite of the reaction force. If there is no movement because of these interactions they are in equilibrium. 4.5. Applications of Newton’s Laws The normal force pushes back upward, opposite of gravity. To do force problems always do the following steps: 1. Draw a free body diagram 2. Label all x and y forces 3. Solve as a system of equations Ex. T1 T2 y1 = T1sin37 y2 = T2sin53 37 x1 = T1cos37 53 x2 = T2cos53 Fw=100N (a neg Y) Find the tension on each cable… x direction: x1 = x2 in equilibrium, the light is not moving left of right Fx = -T1(cos37)+ T2(cos53) = 0 T2 = T1(cos37/ cos53) = 1.33T1 = T2 y direction: Fy = T1sin37+ T2sin53-100N = 0 T1sin37+(1.33T1)(sin53) = 100N T1 = 60.1N Put into 1.33T1 = T2 (1.33)(60.1) = T2 79.9N = T2 T3 = 100N (look at picture) When would T1= T2? (when 1 = 2) Ex. 2 A 77N sled is pulled up a 30 incline (frictionless) by a child. a) find the force exerted by the child on the rope b) find the force exerted on the sled by the hill sled 30 y x Fn Ft Fn = normal force is perpendicular to the plane Ft = tension Fg = force of gravity (aka fw = weight) 30 Fg = 77N = 30 fx = T-77N(sin30) = 0 T = 38.5N fy = N-77N(cos30) = 0 N = 66.7N Since the hill pushes up at 66.7N the sled pushes down with 66.7N. Why is the downward force of 66.7N < Fw at 77N? (The sled is on a 30 incline off of Fn.) Ex.3 Moving a Box A person pulls with a constant force of 20N a box on a dolly with a weight of 300N. a) find the acceleration of the system b) how far will it move in 2sec? (assume Vi = 0 and friction = 0) w = mg m = w/g = 300N/9.8m/s2 = 30.6kg f = ma a = f/m = 20N/30.6kg = .654m/s2 Because force is constant the acceleration is also constant. Vi = 0 dx = Vi T+ ½ aT2 dx = ½ aT2 dx = ½ (.654m/s2)(2sec.2) dx = 1.31m Assignment: pg 105-106 #11, 12, 17, 18, 2, 6, 10 T = 2sec. 4.6. Force of Friction *recall two types of friction: static and kinetic *Fs > Fk for all surfaces because to get an object moving you must overcome surface adhesions *all objects have a coefficient of friction for both static and kinetic—see chart on pg 97 *turn to pg 98 and go over examples 4.5. and 4.6. fx = µk .3 on a 4kg object 4kg Find the acceleration of the two objects and the tension on the string. N 7kg 4kg Fk Ft mg Top Object: fx = T – Fk = (4kg)(a) fy = N – (4kg)(g) = 0 because Fk = μkN and (4kg)(g) = 39.2N Fk = (.3)(39.2N) = 11.8N therefore T = Fk + (4kg)(a) = 11.8 + (4kg)(a) = equation #1 now do the 7kg object… fy = (7kg)(g) – T = (7kg)(a) or (7kg)(9.8m/s2) – (7kg)(a) = T = equation #2 68.6N – 7a = T (equation #2) -11.8N + 4a = T (equation #1) 56.8N – 11a = 0 56.8N = 11a a = 5.16m/s2 put into equation #1 2 11.8N + 4kg(5.16m/s ) = T = 32.4N Assignment: pg 105-106 #7, 15, 20, 25, 34, 38, 68 Chapter 5 Work and Energy “S” = distance work – an object undergoes a displacement of distance “S” along a straight line while acted on by a constant force “F” that makes an angle of with “S” W = Fcos ”S” Work is a scalar with units force times length equals a Newton meter (aka Joule (J)) If you lift something up you do positive work. Put it down and you do negative work. Together they equal zero, so by picking up 100lbs, holding it up for 30min, and dropping it back down, you have done zero work!!! Ex. A force of 50N is exerted at an angle of 30 to move an object a distance of 30 meters. Calculate the work done by the force. W = Fcos S = (50N)(cos30)(30m) = 1300J 5.2. W = FS = maS = m (Vf2-Vi2)/2 = ½ m Vf2-1/2 m Vi2 thus… Wnet = KEf - KEi = KE (the work done on an object is equal to the change in its kinetic energy) Ex. A 1400kg car has a net force of 4500N applied on it. If the car starts from rest on a flat track, find its kinetic energy and velocity after 100m. Vi = 0 Vf = ? F = 4500N m = 1400kg s = 100m Wnet = FS = (4500N)(100m) = 4.5x105 J since this Wnet = KE 1/2mv2 = 4.5x105J v = (4.5x10^5J ) /(1/ 2)(1400kg) = 25.4m/s 5.3. Potential Energy (Gravitational Potential Energy) PE = mgy Wgravity = PEi – PEf *pick a reference level to assign zero, usually your starting point ex. A 60kg skier drops from point A to point B (10m). Find PE at A and the difference at B (using B as zero). PEA = mgy = (60kg)(9.8m/s)(10m) = 5880J PE = PEA since B= 0 5.4. Conservation and Non-conservation of Forces The work done on an object by a conservative force depends only on the initial and final positions of the object. It is a conservative force if the work it does on an object moving through any closed path is zero. Gravity is conservative Wg = mgyi – mgyf note that wg = 0 only if yi = yf Non-conservative - any force that leads to a dissipation of mechanical energy. (thermal pollution) Section 5.5. Conservation of Mechanical energy Wconservative = PEi – PEf remember… Wnet = PEi - PEf and… W = KE = KEf – KEi so… KEi + PEi = KEf + PEf thus… ½mvi2+ mgyi = ½mvf2 + mgyf Turn to page 129 and we will go over example #5.5. Potential energy stored in a spring F = kx = Hooke’s Law and the work done by an applied force on a spring is… W = Fx = ½kx2 This equates to the PE = ½kx2 now stored in the spring. Since PE is proportional to x2, PE is always positive. This new form of energy is included as another term in our equation for conservation of mechanical energy. (KE + PEg + PEs)i = (KE + PEg + PEs)f where PEg = gravitational potential energy Ex. A .5kg block is pressed against a spring with K=80N/m for a distance of 2cm, then released. Find Vf of the block at x = 0. (KE + PEg + PEs)i = (KE + PEg + PEs)f zero PEgi = PEgf no height involved PEsi = KEf ½kxi2 = ½mvf2 cancel ½ from both sides… kxi2 =mvf2 (80N/m)(.02m)2 = (.5kg)Vf2 .25m/s = Vf Assignment: pg 141 #1, 4, 7, 13, 15, 17, 19 5.6. Non-conservative forces and the Work—Kinetic Energy Theorem Since all the work done by non-conservative forces equals the change in mechanical energy of the system we use… Wnc = (KEf + PEf) – (KEi + PEi) Ex. A 20kg child slides down an irregularly curved water slide with a height 6m. If the child starts at the top with an initial velocity of zero, what is the kid’s speed at the bottom? (assume no friction) yf = 0 yi = 6m (20kg)(9.8m/s2)(6m) = ½(20kg)(Vf)2 10.8m/s = Vf (you could also have used Vf2 = Vi2 + 2ad because friction = 0 5.7. Conservation of Energy in General Recall that energy can never be created or destroyed, only transformed. The total energy in the universe is constant. 5.8. Power Power is the transfer of energy over a specific time period. P = w/ T = F( S)/ T = F V units are in Watts = 1W = 1J/s = 1kgm2/s3 1hp = 550ftlb/s = 746W T = tension = a force T-F-Mg = 0 A 1000kg elevator carries a max load of 800kg. A constant friction force of 4000N retards its upward motion. What minimum power, in kilowatts and horsepower, must the motor deliver to lift a fully loaded elevator at 3m/s? T-F-Mg = 0 = T = F +Mg = 4000N+ (1800kg)(9.8m/s2) = 21600N using P = FV = TV = (21600N)(3m/s) = 64800W = 64.8kW = 86.9hp Assignment: pg142 Day 1 21, 25, 26, 27, 28 Day 2 35, 41, 43, 44, 54 Chapter 6 6.1. Momentum and Impulse P = mV P = momentum m = mass V = velocity Momentum is a vector quantity. The rate of change of the momentum of a particle is equal to the resultant force on that particle. The impulse of a force equals the change in momentum of the particle on which the force acts. Under the Impulse Approximation, it’s assumed that one force (or one of the forces) acting on a particle is of a short time duration but of much greater magnitude than any of the other forces. *Newton’s 2nd law applies here… F = P/ T = change in momentum/time interval during momentum change F = P/ T = MVf - MVi/ T = M(Vf-Vi)/ T but the impulse momentum theorem states F T = P = MVf-MVi graphically, the impulse is the area under the curve on a force-time curve = WORK? Why? Force Time (sec.) Example #5 on pg 170… A1500kg car moving at 15m/s collides with a utility pole and is brought to a rest in .3sec. Find the average force exerted on the car during the collision. F = (1500kg)(0m/s-15m/s)/.3sec = -75000N The negative sign means the force is in the opposite direction of the car’s velocity. Go over example 6.2. on page 153. 6.2. Conservation of Momentum If two particles of masses M1 and M2 form an isolated system then the total mass of that system will remain constant throughout the collision. The result of this is known as the conservation of momentum. M1V1i+ M2V2i = M1V1f+ M2V2f This only holds true when the system is closed (no external forces) Ex. #9 on page 170 Looking at the graph you will see that the 2kg particle has a force that varies with time. Find the impulse of the force. total impulse = total area under curve = area (A) + area (B) + area (C) = ½ (2sec)(4N) + (1sec)(4N) + ½ (2sec)(4N) = 12N = 12kg m/s 6.3. Collisions Inelastic collisions: momentum is conserved but energy is not (when two objects collide and stick together the collision is perfectly inelastic). The deformation of objects is what causes the loss of kinetic energy. Elastic collisions: both momentum and kinetic energy are conserved, like pool balls and air molecules at a normal temperature (no deformations). Most of these are limiting in cases. Even billiard balls dent so some deformation, however small, does occur. For perfectly inelastic collisions (stick together), use m1v1i + m2v2i = (m1 + m2)vf We can also use KE = ½ mv2 Assignment: pg 170 #4, 10, 15, 17 6.4. Glancing Collisions Until now we have only look at straight-line collisions, and nothing else. Glancing collisions (colliding masses) rebound at an angle relative to the line of motion of incident mass. example #35 on pg 172… A 2000kg car moving east at 10m/s collides with a 3000kg car moving north. They stick together and move as a unit after the collision at a 40º angle north of east at 5.22m/s. Find Vi for the 3000kg car. since momentum is conserved… Pf = Pi so the sum of the vector components of x and y for both cars are the same before and after the collision. and since the 3000kg car (car #2) was moving in the y direction, we only need to look at the y components. finding Vi of car #2 we get … P1 + P2 = P1f + P2f MVi car#1 + MVi car#2 = MVf car#1 + MVf car#2 but car #1 had P = 0 in the y direction sooooooo… 0 + (3000kg)(Vi) = 5000kg(5.22m/s)(sin40º)/3000kg Vi = 5.59m/s Assignment: pg 171 #21, 25, 29, 31 Test Tomorrow!!! Chapters 4, 5 & 6 – pick partners today Chapter 7 7.1. Angular Speed and Acceleration Most equations in chapters 7 and 8 require you to use radians, not degrees!!! s R = s/R Since the angle through which the disk rotates is = 2- 1, we can define angular displacement as . Angular speed as (omega) is the rotation of a ridged object through that angular displacement in the time interval T. sooo… = 2- 1/T2-T1 = / T then we could say that instantaneous angular speed is the limit of the average speed Ex. on pg 210 #1 Tires on a compact car with a diameter of 2ft are warranted for 60000 miles. a) Determine the angle through which a tire will rotate under warranty. b) How many revolutions of the tire is the answer is the answer from part a? a) = s/R = (60000mi/1ft)(5280ft/ 1mi) = 3.2x108 radians b) = 3.2x108rad = 1rev/2 rad = 5x107 revolutions 7.2. Rotational Motion Under Constant Acceleration (Notice the Comparison?) v = x/ T and = / T Additionally, notice the similarities between… a = v/ T and = / T (average acceleration for angular motion) Ex. A bike wheel rotates with a constant angular acceleration of 3.5rad/sec2. If i = 2rad./sec. at T = 0, through what angle does the wheel rotate in 2 sec? = i T+1/2 T2 = (2rad/s)(2sec) + 1/2(3.5rad/s2)(2sec) 2 = 11rad = 630 degrees at t= 2 sec f = i + t = 2 rad/s + 3.5 rad/s2(2sec) = 9 rad/s also you can use = v/R where v = velocity tangent to radians which solves into velocity tangent … Vt = R and acceleration tangent is… at = R ( = angular acceleration) *the tangential acceleration of a point on a rotating object equals the distance of that point from the axis of rotation multiplied by the angular acceleration. #Ex 11 on pg. 210 A rotating wheel requires 3sec to rotate 37 revolutions. Its angular velocity at the end of 3sec is 98rad/sec. What is the constant angular acceleration of the wheel? = /t = (37rev)(2 rad/1rev)/3sec = 77.5rad/sec 7.4. Centripetal Acceleration Driving around a circular track at a constant speed you will still have acceleration. Why? Because you are changing direction (vector change). An acceleration going around a circle is called centripetal acceleration. (means center seeking) ac = Vt2/R = R2 2/R = R 2 In circular motion the centripetal acceleration is directed inward towards the center of the circle and has a magnitude of either V2/R or R 2. Vc ac Assignment: page 210 #2, 3, 5, 7, 12, 16, 17 7.5. Forces Causing Centripetal Acceleration ac = Vt2/R F = mac = mVt2/R Example #21 on page 211 A sample of blood is placed in centrifuge with a radius of 15cm. The mass of the blood is 3x10-16 kg (red corpuscle) and the magnitude of the force required to make it settle out of the plasma is 4x10-11 N. At how many revolutions per second should the centrifuge be operating??? m = mass R = radius of centrifuge Fc = 4x10-11 N = mVt2/R = mR 2 so… = (Fc/mR)1/2 = (4x10-11 N/(3x10-16 kg)(.15m))1/2 = 942.8rad/sec. = 150rev/sec. Skip 7.6 7.7 Newton’s Universal Law of Gravitation F = G(m1m2/R2) R = distance between objects G = 6.67x10-11 (N m2/kg2) for any object on the earth’s surface use… F = G(mEm/RE2) using two equations we get… V = (Gm/R)1/2 This explains why dark matter exists. Distant stars are too far away from other stars (masses) to move as fast as they do without the existence of an unseen mass (a.k.a. dark matter). Yes it’s vector time!!!!! Ex. Two .3kg balls are placed on a table. One is .4m north and the other is .3m east of a third .3kg ball. Find the resultant force on the third ball m2 m1 m3 The force on m1 by m2 will be upward. The force on m1 by m3 will be to the right. Find the force of each and add the vectors. F2on1 = G(m1m2/R2)= (6.67x10-11(Nm2/R2))((.3kg)(.3kg)/(.4m2))= 3.75x10-11N F3on1 = G(m1m2/R2)= (6.67x10-11(Nm2/R2))((.3kg)(.3kg)/(.3m2))= 6.67x10-11N Fr = (f2on1)2+( f3on1)2 = (3.75x10-11)2+( 6.67x10-11)2 = 7.65x10-11N @29˚ tan-1 = 3.75/6.67 = 29˚ 7.8. Gravitational Potential Energy (again) Recall that PE = mgh only works for objects near the earth’s surface. For objects far from the surface use PE = -G(mEm/R2) where R= the distance of the satellite from the center of the earth. do #36 in class Assignment: pg #212 #15, 18, 31, 32, 33, 35, 37 satellite