Newton s Laws http://mathsforeurope.digibel.be/images
Forces and Equilibrium If the net force on a body is zero, it is in equilibrium. dynamic equilibrium: moving relative to us static equilibrium: appears to be at rest An unbalanced force causes an object to accelerate (speed up, slow down, or change direction) Force is a vector!
Types of Forces Contact forces: involve contact between bodies. Normal, Friction Field forces: act without necessity of contact. Gravity, Electromagnetic, Strong, Weak Question: Is there really any such thing as a contact force?
Galileo s Thought Experiment The Physics Classroom, Tom Henderson 1996-2007
Galileo s Thought Experiment This thought experiment lead to Newton s First Law. The Physics Classroom, Tom Henderson 1996-2007
Newton s First Law The Law of Inertia. A body in motion stays in motion in a straight line unless acted upon by an external force.
Newton s Second Law A body accelerates when acted upon by a net external force SF = ma
Newton s Third Law For every action there exists an equal and opposite reaction. If A exerts a force F on B, then B exerts a force of -F on A.
Commonly Confused Terms Inertia: or the resistance of an object to being accelerated Mass: the same thing as inertia (to a physicist). Weight: gravitational attraction (=F g =mg) inertia = mass weight
Sample Question 1 st Law Two forces, F 1 = (4i 6j + k) N and F 2 = (i 2j - 8k) N, act upon a body of mass 3.0 kg as it is moving at constant speed. What can you infer? (no calculation please)
Sample Question 2 nd Law Two forces, F 1 = (4i 6j + k) N and F 2 = (i 2j - 8k) N, act upon a body of mass 3.0 kg. No other forces act upon the body at this time. What can you infer? (no calculation please)
Sample Problem 3 rd Law A tug-of-war team ties a rope to a tree and pulls hard horizontally to create a tension of 30,000 N in the rope. Suppose the team pulls equally hard when, instead of a tree, the other end of the rope is being pulled by another tug-of-war team such that no movement occurs. What is the tension in the rope in the second case? a) 0 N b) 15,000 N c) 30,000 N d) 60,000 N e) Can not be determined
2 nd Law Procedure 1. Identify the body to be analyzed. 2. Select a reference frame, stationary or moving, but not accelerating 3. Draw a force or free body diagram. 4. Set up ΣF = ma equations for each dimension. 5. Use kinematics or calculus where necessary to obtain acceleration. (Do the algebra first, before plugging in numbers! And remember units!)
2 nd Law Problem You hang a 1 kg mass from the ceiling in an elevator. What is the tension in the string when the elevator is accelerating a) Upward at 2 m/s 2? 12N b) Downward at 2 m/s 2? 8N
Sample Problem A 2 kg mass undergoes an acceleration given by a = (3i + 4j) m/s 2. Find the resultant force F and its magnitude.
Normal force Normal force = the force that keeps one object from invading another object. Our weight = the force of attraction of our body for the center of the planet. We don t fall to the center of the planet. The normal force keeps us up!
Normal Force on Flat surface N mg
Normal Force on Ramp N The normal force is always perpendicular to the surface. mg
Normal Force on Ramp N Find the acceleration. SF= ma mgsin = ma gsin = a mg
Tension A pulling force (ropes can t push!). Generally exists in a rope, string, or cable. Arises at the molecular level, when a rope, string, or cable resists being pulled apart.
Tension (static 2D) The horizontal and vertical components of the tension are equal to zero if the system is not accelerating. T 3 mg 2 1 T 1 30 o 45 o 3 15 kg T 2 T 3 SF x = 0 SF y = 0
Pulley problems N SF = ma m 1 g T T m 2 g = (m 1 +m 2 )a m 1 m 2 g m 2
N Pulley problems Tension is determined by examining one block or the other T T SF = (m 1 +m 2 )a m 2 g T+T m 1 gsin = (m 1 +m 2 )a m 1 g m 2 g m 2
N m 1 g Pulley problems Tension is determined by examining one block or the other T T m 2 g SF 2 = m 2 a m 2 g - T = m 2 a SF 1 = m 1 a T-m 1 gsin = m 1 a m 2
Atwood machine T m 1 m 2 T A device for measuring g. If m1 and m2 are nearly the same, slows down freefall such that acceleration can be measured. Then, g can be measured. m 1 g m 2 g SF = ma m 2 g-m 1 g = (m 2 +m 1 )a
Moderate Problem Describe the motion of the 5 kg block. The table and pulley are ideal and frictionless. 20 o 1.0 kg
Friction Friction opposes a sliding motion. Static friction exists before sliding occurs F f s N Kinetic friction exists after sliding occurs F f = k N
Friction on flat surfaces y y x x Draw a free body diagram for a braking car. Draw a free body diagram for a car accelerating from rest.
Friction on a ramp Sliding down Sliding up
Friction is always parallel to surfaces. A 1.00 kg book is held against a wall by pressing it against the wall with a force of 50.00 N. What must be the minimum coefficient of friction between the book and the wall, such that the book does not slide down the wall? F F f N F G (Ans: 0.20)
Centripetal Force Inwardly directed force which causes a body to turn; perpendicular to velocity. Centripetal force always arises from other forces, and is not a unique kind of force. Sources include gravity, friction, tension, electromagnetic, normal. ΣF = ma a = v 2 /r ΣF = m v 2 /r
Highway Curves R z Friction turns the vehicle r Normal force turns the vehicle
Sample problem Find the minimum safe turning radius for a car traveling at 60 mph on a flat roadway, assuming a coefficient of static friction of 0.70.
Conical Pendulum T = 2p L cos g θ L r F T For conical pendulums, centripetal force is provided by a component of the tension. mg
Sample problem Derive the expression for the period of a conical pendulum with respect to the string length and radius of rotation. T = 2p L cos g
Non-uniform Circular Motion Consider circular motion in which either speed of the rotating object is changing, or the forces on the rotating object are changing. If the speed changes, there is a tangential as well as a centripetal component to the force. In some cases, the magnitude of the centripetal force changes as the circular motion occurs.
Sample problem You swing a 0.25-kg rock in a vertical circle on a 0.80 m long rope at 2.0 Hz. What is the tension in the rope a) at the top and b) at the bottom of your swing?
Non-constant Forces Forces can vary with time. Forces can vary with velocity. Forces can vary with position.
Drag Forces Drag forces slow an object down as it passes through a fluid. act in opposite direction to velocity. are functions of velocity. impose terminal velocity.
Drag Force in Free Fall F D F D F D when F D equals mg, terminal velocity has been mg mg mg mg reached
Drag Force in Free Fall F D = bv + c v 2 for slow moving objects F D = b v b and c depend on size and shape of object viscosity of the fluid F D mg for fast moving objects F D = c v 2 c = 1/2 D r A Where D = drag coefficient r = density of fluid A = cross-sectional area
Sample Problem: Slow moving objects F D = bv Show that v T = mg/b mg SF = ma mg - bv = ma mg - bv v T T mg = b = 0 0
Slow-moving objects F D = bv What happens when a 0? mg - bv = ma mg - bv = m dv dt mg This is called a differential equation because it relates a variable v to its derivative dv/dt.
Solving the Diff Eq mg - bv = dt t dt = = = mdv mg m - b m - bv m mg - bv ln( mg dv dt dv - bv) - ln( mg) Making it pretty - bt m - bt m e e v (-bt / m (-bt / m = = ln( mg - bv) - ln( mg) mg - bv = ln mg mg b = mg - bv mg = 1- b mg v ( (-bt / m 1- e
Sample Problem: Slow Moving Objects v(t) = (mg/b)(1 e bt/m ) F D = bv What is the velocity at t = 0? What is the velocity at t? mg
Sample Problem: Fast moving objects Show that v T = (2mg / DrA) 1/2 F D = 1/2DrAv 2 SF mg = - ma 1 2 DrAv 2 = ma 0 mg v 2 T = - 1 2 1 2 DrAv mg DrA 2 T = 0 mg v T = 2mg DrA
Sample Problem: Fast moving objects Derive an expression for the velocity of the object as a function of time F D = 1/2DrAv 2 SF = ma mg mg - - 1 2 1 2 DrAv DrAv 2 2 = = ma m dv dt mg Another differential equation because it relates a variable v to its derivative dv/dt. (we re not going to solve it!)
Problems
Force Problem The following applied forces cause a mass to accelerate, as indicated. What is the object s mass? a) 10 kg b) 11 kg c) 12 kg d) 15 kg e) 20 kg Applied force (N) Applied Force (N) 1200 1000 800 600 400 200 0 Acceleration (m/s 2 ) 200 10 300 20 600 50 1100 100 F=ma y = 10x + 100 0 50 100 acceleration (m/s^2)
Problem #1 Assume a coefficient of static friction of 1.0 between tires and road. What is the minimum length of time it would take to accelerate a car from 0 to 60 mph?
Problem #2 Assume a coefficient of static friction of 1.0 between tires and road and a coefficient of kinetic friction of 0.80 between tires and road. How far would a car travel after the driver applies the brakes if it skids to a stop?
Sample problem Derive the expression for the period best banking angle of a roadway given the radius of curvature and the likely speed of the vehicles.
Sample problem A 40.0 kg child sits in a swing supported by 3.00 m long chains. If the tension in each chain at the lowest point is 350 N, find a) the child s speed at the lowest point and b) the force exerted by the seat on the child at the lowest point.
Sample problem A 900-kg automobile is traveling along a hilly road. If it is to remain with its wheels on the road, what is the maximum speed it can have as it tops a hill with a radius of curvature of 20.0 m?
Sample problem Consider a force that is a function of time: F(t) = (3.0 t 0.5 t 2 )N If this force acts upon a 0.2 kg particle at rest for 3.0 seconds, what is the resulting velocity and position of the particle?
Sample problem Consider a force that is a function of time: F(t) = (16 t 2 8 t + 4)N If this force acts upon a 4 kg particle at rest for 1.0 seconds, what is the resulting change in velocity of the particle?