Physics 231 Topic 3: Forces & Laws of Motion + Uniform Circular Motion Wade Fisher September 10-14 2012 MSU Physics 231 Fall 2012 1 Quick Quiz What aspects of your lectures do you find most useful? A) Discussion of concepts B) Demonstrations C) Problem solving examples D) Opportunity to ask questions E) Opportunity to cover homework problems MSU Physics 231 Fall 2012 2 Recall from Last Lecture Vectors and Scalars Two Dimensional Motion Velocity in 2D Acceleration in 2D Projectile motion Throwing a ball or cannon fire Uniform Circular Motion Centripetal acceleration MSU Physics 231 Fall 2012 3 An example A football player throws a ball with initial velocity of 30 m/s at an angle of 30o degrees w.r.t. the ground. 1) How far will the ball fly before hitting the ground? 2) What about an angle of 60o? 3) At which angle is the distance thrown maximum? Maximum if sin(2) is maximum, so =45o X(=45o) = 91.7 m MSU Physics 231 Fall 2012 4 Another example… Calculate how far the ball goes. MSU Physics 231 Fall 2012 5 Clicker Quiz You enter a circle intersection (roundabout) and travel at 20 mph around the circle at constant velocity. Which direction does your acceleration vector point? A) Along the direction you’re moving B) Towards the center of the circle C) Away from the center of the circle D) In the opposite direction you’re moving E) There is no acceleration MSU Physics 231 Fall 2012 6 Uniform Circular Motion Consider a car at constant speed, constrained to a circular track • Velocity is always determined by the direction the car is facing at a given time Velocity is constantly changing, so this implies an acceleration • But it’s just the direction of the velocity that’s changing! MSU Physics 231 Fall 2012 7 Uniform Circular Motion 𝑣𝑓 − 𝑣𝑖 Δ𝑣 𝑎= = 𝑡𝑓 − 𝑡𝑖 Δ𝑡 Sin( /2 ) = ( s/2 ) / r Sin( /2 ) = ( v/2 ) / v MSU Physics 231 Fall 2012 8 Uniform Circular Motion 𝑣𝑓 − 𝑣𝑖 Δ𝑣 𝑎= = 𝑡𝑓 − 𝑡𝑖 Δ𝑡 ( s/2 ) / r = sin( /2 ) sin( /2 ) = ( v/2 ) / v ( s/2 ) / r = ( v/2 ) / v ( s/t ) v/r = ( v/t ) 𝑎𝑐 = 𝑣 𝑣 𝑟 = MSU Physics 231 Fall 2012 𝑣2 𝑟 ( s/t ) v 9 Key Concepts: Forces & Laws of Nature Force and Mass Newton’s Laws of Motion Inertia, F=ma, Equal/Opposite Reactions Application of Newton’s Laws Force diagrams Component Forces Friction and Drag Kinetic, static, rolling frictions Uniform circular motion Covers chapter 4 in Rex & Wolfson MSU Physics 231 Fall 2012 10 Isaac Newton (1642-1727) “In the beginning of 1665 I found the…rule for reducing any dignity of binomial to a series. The same year in May I found the method of tangents and in November the method of fluxions and in the next year in January had the Theory of Colours and in May following I had the entrance into the inverse method of fluxions and in the same year I began to think of gravity extending to the orb of the moon…and…compared the force requisite to keep the Moon in her orb with the force of Gravity at the surface of the Earth.” “Nature and Nature’s Laws lay hid in night: God said, Let Newton be! And all was light.” Alexander Pope (1688-1744) PHY 231 MSU Physics 231 Fall 2012 11 11 Newton’s Laws First Law: If the net force exerted on an object is zero the object continues in its original state of motion; if it was at rest, it remains at rest. If it was moving with a certain velocity, it will keep on moving with the same velocity. Second Law: The acceleration of an object is proportional to the net force acting on it, and inversely proportional to its mass: F=ma Third Law: If two objects interact, the force exerted by the first object on the second is equal but opposite in direction to the force exerted by the second object on the first: F12=-F21 12 PHY 231 MSU Physics 231 Fall 2012 12 Identifying the forces in a system PHY 231 MSU Physics 231 Fall 2012 13 13 Force, Mass and Weight Forces are quantified in units of Newton (N). 1 N = 1 kgˑm/s2 F = ma A force is a vector: it has direction. But what’s this “mass” thing?? m = 1 kg F = m a = 1 kgˑm/s2 = 1 Newton a = -9.81 m/s2 PHY 231 MSU Physics 231 Fall 2012 14 Two forces that luckily act upon us nearly all the time. Normal Force: elastic force acting perpendicular to the surface the object is resting on. Name: n 1) No net force: remains at rest. 2) Fg = mg = n 3) Fmass-ground = -Fground-mass Gravitational Force Fg = mg (referred to as weight) g = 9.81 m/s2 PHY 231 MSU Physics 231 Fall 2012 15 15 Quiz Question 5 kg 10 kg 5 kg 3 cases with different mass and size are standing on a floor. which of the following is true: a) the normal forces acting on the crates is the same b) the normal force acting on crate b is largest because its mass is largest c) the normal force on the green crate is largest because its size is largest d) the gravitational force acting on each of the crates is the same e) all of the above 16 PHY 231 MSU Physics 231 Fall 2012 16 Some handy things to remember. 90- Choose your coordinate system in a clever way: Define one axis along the direction where you expect an object to start moving, the other axis perpendicular to it (these are not necessarily the horizontal and vertical direction. MSU Physics 231 Fall 2012 17 A first example 1) Draw all Forces acting on the red object 2) If =40o and the mass of the red object equals 1 kg, what is the resulting acceleration (assume no friction). MSU Physics 231 Fall 2012 18 Clicker Question question A ball is rolling from a slope at angle . It is repeated but after decreasing the angle . Compared to the first trial: a) b) c) d) the normal force on the ball increases the normal force gets closer to the gravitational force the ball rolls slower all of the above MSU Physics 231 Fall 2012 19 question END LECTURE 1 MSU Physics 231 Fall 2012 20 Forces seen previously • Gravity: Force between massive objects • Normal force: Elasticity force from supporting surface n = -FgL F = mg cos F = mgsin 90- Fg=mg MSU Physics 231 Fall 2012 21 MSU Physics 231 Fall 2012 22 Decomposing Forces An object has a mass of 30.0 kg and three forces are acting on it as shown in the figure. What is the magnitude and direction of acceleration? 690 N 70o 24o Net Force 300 N mg 139 N 35.2o Force 300 N 690 N Weight Resultant x(N) -122 236 0 114 N y(N) -274 648 -294 80 N Total Force: F = (1142+80.02) = 139 N Direction: = tan-1(Fy/Fx) = 35.2o Acceleration: a = F/m = 139/30.0 = 4.65 m/s2 MSU Physics 231 Fall 2012 23 PHY 231 MSU Physics 231 Fall 2012 24 24 Measuring mass and weight. Given that gearth = 9.81 m/s2, gsun = 274 m/s2, gmoon = 1.67 m/s2, what is the mass of a person on the sun and moon if his mass on earth is 70 kg? And what is his weight on each of the three surfaces? The mass is the same on each of the surfaces On Earth: w = mg = (70 kg)(9.81 m/s2) = 686.7 N On the Moon: w = mg = (70 kg )(1.67 m/s2) = 116.7 N On the Sun: w = mg = (70 kg)(274 m/s2) = 19180 N PHY 231 MSU Physics 231 Fall 2012 25 25 Example Buck Rogers travels from earth to a planet which has a radius that is 2 times larger than that of earth and has a mass 6 times that of earth. By what factor does his weight change relative to earth? Fg~1/r2~1/4 Fg~M~6 Fg~M/r2=6/4=1.5 His weight is 1.5 times larger. PHY 231 MSU Physics 231 Fall 2012 26 26 QUIZ! PHY 231 MSU Physics 231 Fall 2012 27 27 A 2000 kg sailboat is pushed by the tide of the sea with a force of 3000 N to the East. Because of the wind in its sail it feels a force of 6000 N toward to North-West direction. What is the magnitude and direction of the acceleration? MSU Physics 231 Fall 2012 Problem 28 Tension T The magnitude of the force T acting on the crate, is the same as the tension in the rope. Spring-scale You could measure the tension by inserting a spring-scale... MSU Physics 231 Fall 2012 29 Tension T1 T1=T2 ??? T2 M As long as it’s one rope/string/cable, the tension is the same at both ends! MSU Physics 231 Fall 2012 30 Newton’s second law and tension m1 n T No friction. T Fg What is the acceleration of the objects? m2 Fg Object 1: F=m1a, so Object 2: F=m2a, so T = m1a Fg-T = m2a & m2g-T = m2a Combine 1&2 (Tension is the same): m2g - m1a = m2a a = m2g/(m1+m2) MSU Physics 231 Fall 2012 31 Problem What is the tension in the string and what will be the acceleration of the two masses? MSU Physics 231 Fall 2012 32 Example A 1000-kg car is pulling a 300 kg trailer. Their acceleration is 2.15 m/s2. Ignoring friction, find: a) the net force the car b) the net force on the trailer c) the net force exerted by the trailer on the car d) the resultant force exerted by the car on the road MSU Physics 231 Fall 2012 33 Question ceiling A block of mass M is hanging from a string. What is the tension in the string? M a) b) c) d) e) T=0 T=Mg with g=9.81 m/s2 T=0 near the ceiling and T=Mg near the block T=M one cannot tell MSU Physics 231 Fall 2012 34 question END LECTURE 2 MSU Physics 231 Fall 2012 35 EXTRA CREDIT QUIZ M What force must Sparty apply in order to crush the Fighting Irish? a) b) c) d) e) None, the Irish couldn’t fight their way out of a wet paper bag None, without friction the only force is gravity! None, MSU plays fair (especially when there’s no real fight) Final Score H: 6 V:64 Sparty doesn’t appy force, Sparty only hands out humiliating defeat MSU Physics 231 Fall 2012 PHY 231 36 36 EXTRA CREDIT QUIZ Force due to airblower. A small block is put on a frictionless slope. Attached on the block is an air-blower which keeps the block from slipping down the slope. Which of the following is true: a) The force by the airblower on the block equals the total gravitational force on the block b) The force by the airblower on the block is smaller than the total gravitational force on the block c) The force by the airblower on the block is larger than the total gravitational force on the block 37 MSU Physics 231 Fall 2012 PHY 231 37 Tension MSU Physics 231 Fall 2012 38 Quiz! 10 Kg A homogeneous block of 10 kg is hanging with 2 ropes from a ceiling. The tension in each of the ropes is: a) 0 N b) 49 N c) 98 N d) 196 N e) don’t know MSU Physics 231 Fall 2012 39 T 900 T 1kg A mass of 1 kg is hanging from a rope as shown in the figure. If the angle between the 2 supporting wires is 90 degrees, what is the tension in each rope? MSU Physics 231 Fall 2012 40 Friction Friction are the forces acting on an object due to interaction with the surroundings (air-friction, ground-friction etc). Two variants: • Static Friction: as long as an external force (F) trying to make an object move is smaller than fs,max, the static friction fs equals F but is pointing in the opposite direction: no movement! fs,max = sn s = coefficient of static friction • Kinetic Friction: After F has surpassed fs,max, the object starts moving but there is still friction. However, the friction will be less than fs,max! Fk = kn k = coefficient of kinetic friction MSU Physics 231 Fall 2012 41 PHY 231 MSU Physics 231 Fall 2012 42 42 MSU Physics 231 Fall 2012 43 Friction pull 20 kg A person wants to drag a crate of 20 kg over the floor. If he pulls the crate with a force of 98 N, the crate starts to move. What is s? After the crate starts moving, the person continues to pull with the same force. Given k=0.4, what is the acceleration of the crate? MSU Physics 231 Fall 2012 44 General strategy • If not given, make a drawing of the problem. • Put all the relevant forces in the drawing, object by object. • Think about the axis • Think about the signs • Decompose the forces in direction parallel to the motion and perpendicular to it. • Write down Newton’s law for forces in the parallel direction and perpendicular direction. • Solve for the unknowns. • Use these solutions in further equations if necessary • Check whether your answer makes sense. MSU Physics 231 Fall 2012 45 Example Problem A) If s=1.0, what is the angle for which the block just starts to slide? B) The block starts moving. Given that k=0.5, what is the acceleration of the block? MSU Physics 231 Fall 2012 46 puck on ice After being hit by a hockey-player, a puck is moving over a sheet of ice (frictionless). The forces working on the puck are: a) b) c) d) e) No force whatsoever the normal force only the normal force and the gravitational force the force that keeps the puck moving the normal force, the gravitational force and the force that keeps the puck moving PHY 231 MSU Physics 231 Fall 2012 47 47 All the Forces Come Together If a=3.30 m/s2 (the 12kg block is moving downward), what is the value of k? MSU Physics 231 Fall 2012 48 Example 1 kg Is there a value for the static friction of surface A for which these masses do not slide? If so, what is it? A 20o 0.5 kg MSU Physics 231 Fall 2012 49 Airblower again Force due to blower A cart is placed on a slope that makes an angle with the horizontal. An airblower is placed on the cart, creating a force in the direction up the slope. If the cart and blower have a combined weight of 10 kg and =100, what is the force exerted by the blower (ignore friction) is the cart is stationary? MSU Physics 231 Fall 2012 50 Elevator T An object of 1 kg is hanging from a spring scale in an elevator. What is the weight read from the scale if the elevator: a) moves with constant velocity b) accelerates upward with 3 m/s2 c) accelerates downward with 3 m/s2 d) decelerates with 3 m/s2 while moving up MSU Physics 231 Fall 2012 51 For Next Week Rex & Wolfson Chapter 5: Work and Energy Homework Set 3: Covers Chapters 4 Due Wednesday 11PM MSU Physics 231 Fall 2012 52