* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Identifying Forces: Non
Survey
Document related concepts
Jerk (physics) wikipedia , lookup
Equations of motion wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Modified Newtonian dynamics wikipedia , lookup
Center of mass wikipedia , lookup
Relativistic mechanics wikipedia , lookup
Fictitious force wikipedia , lookup
Classical mechanics wikipedia , lookup
Seismometer wikipedia , lookup
Centrifugal force wikipedia , lookup
Fundamental interaction wikipedia , lookup
Renormalization group wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Classical central-force problem wikipedia , lookup
Centripetal force wikipedia , lookup
Transcript
Physics 207 – Lecture 7 Lecture 7 Goals: Solve 1D and 2D problems with forces in equilibrium and non-equilibrium (i.e., acceleration) using Newton’ 1st and 2nd laws. Distinguish static and kinetic coefficients of friction Differentiate between Newton’s 1st, 2nd and 3rd Laws Assignment: HW4, (Chapters 6 & 7 (1st part) due 2/17, 9 am, Wednesday) For Thursday, Read Chapter 7 1st Exam Thursday, Feb. 17 from 7:15-8:45 PM Chapters 1-7 Physics 207: Lecture 7, Pg 1 Identifying Forces: Non-contact All objects having mass exhibit a mutually attractive force (i.e., gravity) that is distance dependent At the Earth’s surface this variation is small so little “g” (the associated acceleration) is typically set to 9.80 or 10. m/s2 FB,G Physics 207: Lecture 7, Pg 2 Page 1 Physics 207 – Lecture 7 Contact (i.e., normal) Forces Certain forces act to keep an object in place. These have what ever force needed to balance all others (until a breaking point). FB,T Physics 207: Lecture 7, Pg 3 No net force No acceleration r r r ∑ F = Fnet = ma = 0 ∑ Fx = 0 ∑ Fy = 0 y (Force vectors are not always drawn at contact points) FB,T Normal force is always ⊥ to a surface ∑ Fy = −mg + N = 0 N = mg FB,G Physics 207: Lecture 7, Pg 4 Page 2 Physics 207 – Lecture 7 No net force No acceleration r r r ∑ F = Fnet = ma = 0 If zero velocity then “static equilibrium” If non-zero velocity then “dynamic equilibrium” This label, static vs. dynamic, is observer dependent Forces are vectors r r r r r r ∑ F ≡ Fnet = ma = F1 + F2 + F3 + K Physics 207: Lecture 7, Pg 5 A special contact force: Friction What does it do? It opposes motion (velocity, actual or that which would occur if friction were absent!) How do we characterize this in terms we have learned? Friction results in a force in a direction opposite to the direction of motion (actual or, if static, then “inferred”)! j N FAPPLIED fFRICTION ma i mg g Physics 207: Lecture 7, Pg 6 Page 3 Physics 207 – Lecture 7 Friction... Friction is caused by the “microscopic” interactions between the two surfaces: Physics 207: Lecture 7, Pg 7 Friction... Force of friction acts to oppose motion: Parallel to a surface Perpendicular to a Normal force. j N F ma fF i mg g Physics 207: Lecture 7, Pg 8 Page 4 Physics 207 – Lecture 7 Static Friction with a bicycle wheel You are pedaling hard and the bicycle is speeding up. What is the direction of the frictional force? You are breaking and the bicycle is slowing down What is the direction of the frictional force? Physics 207: Lecture 7, Pg 9 Important notes Many contact forces are conditional and, more importantly, they are not necessarily constant We have a general notion of forces is from everyday life. In physics the definition must be precise. A force is an action which causes a body to accelerate. (Newton’s Second Law) On a microscopic level, all forces are non-contact Physics 207: Lecture 7, Pg 10 Page 5 Physics 207 – Lecture 7 Pushing and Pulling Forces A rope is an example of something that can pull You arm is an example of an object that can push or push Physics 207: Lecture 7, Pg 11 Examples of Contact Forces: A spring can push Physics 207: Lecture 7, Pg 12 Page 6 Physics 207 – Lecture 7 A spring can pull Physics 207: Lecture 7, Pg 13 Ropes provide tension (a pull) In physics we often use a “massless” rope with opposing tensions of equal magnitude Physics 207: Lecture 7, Pg 14 Page 7 Physics 207 – Lecture 7 Forces at different angles Case1: Downward angled force with friction Case 2: Upwards angled force with friction Cases 3,4: Up against the wall Questions: Does it slide? What happens to the normal force? What happens to the frictional force? Cases 3, 4 Case 2 Case 1 F N N ff F F N ff ff mg mg mg Physics 207: Lecture 7, Pg 15 Free Body Diagram A heavy sign is hung between two poles by a rope at each corner extending to the poles. Eat at Bucky’s A hanging sign is an example of static equilibrium (depends on observer) What are the forces on the sign and how are they related if the sign is stationary (or moving with constant velocity) in an inertial reference frame ? Physics 207: Lecture 7, Pg 16 Page 8 Physics 207 – Lecture 7 Free Body Diagram Step one: Define the system T2 T1 θ2 θ1 Eat at Bucky’s T2 T1 θ1 mg θ2 mg Step two: Sketch in force vectors nd Step three: Apply Newton’s 2 Law (Resolve vectors into appropriate components) Physics 207: Lecture 7, Pg 17 Free Body Diagram T1 T2 θ2 θ1 Eat at Bucky’s mg Vertical : y-direction Horizontal : x-direction 0 = -mg + T1 sinθ1 + T2 sinθ2 0 = -T1 cosθ1 + T2 cosθ2 Physics 207: Lecture 7, Pg 18 Page 9 Physics 207 – Lecture 7 Exercise, Newton’s 2nd Law P is the upward force being exerted on the crate by the person C is the contact or normal force on the crate by the floor, and W is the weight (force of the earth on the crate). !" #$%&!'(' )$& $$! & $'$ #$' ' &#*$+ $ & $ ($#'! ' &#, !" *!'*$''* , & & & $ #%&$- ./&$0 #$ *( ' ('&1$ 2 3 ! ' !$"%&1$4 A. B. C. D. P+C<W P+C>W P=C P+C=W Physics 207: Lecture 7, Pg 19 Mass We have an idea of what mass is from everyday life. In physics: Mass (in Phys 207) is a quantity that specifies how much inertia an object has (i.e. a scalar that relates force to acceleration) (Newton’s Second Law) Mass is an inherent property of an object. Mass and weight are different quantities; weight is usually the magnitude of a gravitational (non-contact) force. “Pound” (lb) is a definition of weight (i.e., a force), not a mass! Physics 207: Lecture 7, Pg 20 Page 10 Physics 207 – Lecture 7 Inertia and Mass The tendency of an object to resist any attempt to change its velocity is called Inertia Mass is that property of an object that specifies how much resistance an object exhibits to changes in its velocity (acceleration) r r a ∝ Fnet If mass is constant then If force constant r | a |∝ 1 m |a| Mass is an inherent property of an object m Mass is independent of the object’s surroundings Mass is independent of the method used to measure it Mass is a scalar quantity The SI unit of mass is kg Physics 207: Lecture 7, Pg 21 Exercise Newton’s 2nd Law An object is moving to the right, and experiencing a net force that is directed to the right. The magnitude of the force is decreasing with time (read this text carefully). The speed of the object is A. B. C. D. increasing decreasing constant in time Not enough information to decide Physics 207: Lecture 7, Pg 22 Page 11 Physics 207 – Lecture 7 Exercise Newton’s 2nd Law A 10 kg mass undergoes motion along a line with a velocities as given in the figure below. In regards to the stated letters for each region, in which is the magnitude of the force on the mass at its greatest? A. B. C. D. E. B C D F G Physics 207: Lecture 7, Pg 23 Home Exercise Newton’s 2nd Law A constant force is exerted on a cart that is initially at rest on an air table. The force acts for a short period of time and gives the cart a certain final speed s. Force Cart Air Track In a second trial, we apply a force only half as large. To reach the same final speed, how long must the same force be applied (recall acceleration is proportional to force if mass fixed)? A. B. C. D. 4 x as long 2 x as long 1/2 as long 1/4 as long Physics 207: Lecture 7, Pg 24 Page 12 Physics 207 – Lecture 7 Home Exercise Newton’s 2nd Law Solution Force Cart Air Track F = ma Since F2 = 1/2 F1 a2 = 1/2 a1 We know that under constant acceleration, v = a ∆t So, a2 ∆t2 = a1 ∆t1 we want equal final velocities 1/2 a1 / ∆t2 = a1 / ∆t1 ∆t = 2 ∆t 2 1 (B) 2 x as long Physics 207: Lecture 7, Pg 25 Home Exercise Newton’s 2nd Law A force of 2 Newtons acts on a cart that is initially at rest on an air track with no air and pushed for 1 second. Because there is friction (no air), the cart stops immediately after I finish pushing. It has traveled a distance, D. Force Cart Air Track Next, the force of 2 Newtons acts again but is applied for 2 seconds. A. B. The new distance the cart moves relative to D is: C. D. 8 x as far 4 x as far 2 x as far 1/4 x as far Physics 207: Lecture 7, Pg 26 Page 13 Physics 207 – Lecture 7 Home Exercise Solution Force Cart Air Track We know that under constant acceleration, ∆x = a (∆t)2 /2 (when v0=0) Here ∆t2=2∆t1, F2 = F1 ⇒ a2 = a1 1 a ∆ t 22 ( 2 ∆ t1 )2 ∆x2 2 = = =4 ∆ x1 1 a ∆ t 2 ∆ t12 1 2 (B) 4 x as long Physics 207: Lecture 7, Pg 27 Home Exercise: Physics in an Elevator Gravity and Normal Forces A woman in an elevator is accelerating upwards The normal force exerted by the elevator on the woman is, (A) greater than (B) the same as (C) less than the force due to gravity acting on the woman (REMEMBER: Draw a FREE BODY DIAGRAM) Physics 207: Lecture 7, Pg 28 Page 14 Physics 207 – Lecture 7 Moving forces around Massless strings: Translate forces and reverse their direction but do not change their magnitude (we really need Newton’s 3rd of action/reaction to justify) string T1 -T1 Massless, frictionless pulleys: Reorient force direction but do not change their magnitude T2 T1 -T1 | T1 | = | -T1 | = | T2 | = | T2 | -T2 Physics 207: Lecture 7, Pg 29 Scale Problem You are given a 1.0 kg mass and you hang it directly on a fish scale and it reads 10 N (g is 10 m/s2). 10 N 1.0 kg Now you use this mass in a second experiment in which the 1.0 kg mass hangs from a massless string passing over a massless, frictionless pulley and is anchored to the floor. The pulley is attached to the fish scale. What force does the fish scale now read? ? 1.0 kg Physics 207: Lecture 7, Pg 30 Page 15 Physics 207 – Lecture 7 Scale Problem Step 1: Identify the system(s). In this case it is probably best to treat each object as a distinct element and draw three force body diagrams. One around the scale One around the massless pulley (even though massless we can treat is as an “object”) One around the hanging mass Step 2: Draw the three FBGs. (Because this is a now a one-dimensional problem we need only consider forces in the y-direction.) ? 1.0 kg Physics 207: Lecture 7, Pg 31 Static and Kinetic Friction Friction exists between objects and its behavior has been modeled. At Static Equilibrium: A block, mass m, with a horizontal force F applied, Direction: A force vector ⊥ to the normal force vector N and the vector is opposite to the direction of acceleration if µ were 0. Magnitude: f is proportional to the applied forces such that fs ≤ µs N µs called the “coefficient of static friction” Physics 207: Lecture 7, Pg 32 Page 16 Physics 207 – Lecture 7 Case study ... big F Dynamics: x-axis i : y-axis j : max = F − µKN may = 0 = N – mg or N = mg F − µKmg = m ax so fk v j N F i ma ax fk µK mg mg g Physics 207: Lecture 7, Pg 33 Case study ... little F Dynamics: x-axis i : y-axis j : so max = F − µKN may = 0 = N – mg or N = mg F − µKmg = m ax fk v j N F i ma ax fk µK mg mg g Physics 207: Lecture 7, Pg 34 Page 17 Physics 207 – Lecture 7 Friction: Static friction Static equilibrium: A block with a horizontal force F applied, Σ Fx = 0 = -F + fs fs = F FBD Σ Fy = 0 = - N + mg N = mg N As F increases so does fs F m fs 1 mg Physics 207: Lecture 7, Pg 35 Static friction, at maximum (just before slipping) Equilibrium: A block, mass m, with a horizontal force F applied, Direction: A force vector ⊥ to the normal force vector N and the vector is opposite to the direction of acceleration if µ were 0. Magnitude: fS is proportional to the magnitude of N fs = µs N N F m mg Physics 207: Lecture 7, Pg 36 Page 18 fs Physics 207 – Lecture 7 Kinetic or Sliding friction (fk < fs) Dynamic equilibrium, moving but acceleration is still zero Σ Fx = 0 = -F + fk FBD fk = F v Σ Fy = 0 = - N + mg N = mg N F As F increases fk remains nearly constant (but now there acceleration is acceleration) m fk 1 mg fk = µk N Physics 207: Lecture 7, Pg 37 Sliding Friction: Quantitatively Direction: A force vector ⊥ to the normal force vector N and the vector is opposite to the velocity. Magnitude: fk is proportional to the magnitude of N fk = µk N ( = µK mg g in the previous example) The constant µk is called the “coefficient of kinetic friction” Logic dictates that µS > µK for any system Physics 207: Lecture 7, Pg 38 Page 19 Physics 207 – Lecture 7 Coefficients of Friction Material on Material µs = static friction µk = kinetic friction steel / steel 0.6 0.4 add grease to steel 0.1 0.05 metal / ice 0.022 0.02 brake lining / iron 0.4 0.3 tire / dry pavement 0.9 0.8 tire / wet pavement 0.8 0.7 Physics 207: Lecture 7, Pg 39 An experiment Two blocks are connected on the table as shown. The table has unknown static and kinetic friction coefficients. Design an experiment to find µS N T Static equilibrium: Set m2 m2 and add mass to m1 to reach the breaking point. Requires two FBDs Mass 1 Σ Fy = 0 = T – m1g fS T m1 m2g m1g Mass 2 Σ Fx = 0 = -T + fs = -T + µS N Σ Fy = 0 = N – m2g T = m1g = µS m2g µS = m1/m2 Page 20 Physics 207: Lecture 7, Pg 40 Physics 207 – Lecture 7 A 2nd experiment Two blocks are connected on the table as shown. The table has unknown static and kinetic friction coefficients. Design an experiment to find µK. T Dynamic equilibrium: Set m2 and adjust m1 to find place when T a = 0 and v ≠ 0 fk m1 Requires two FBDs Mass 1 Σ Fy = 0 = T – m1g N m2 m2g m1g Mass 2 Σ Fx = 0 = -T + ff = -T + µk N Σ Fy = 0 = N – m2g T = m1g = µk m2g µk = m1/m2 Physics 207: Lecture 7, Pg 41 An experiment (with a ≠ 0) Two blocks are connected on the table as shown. The table has unknown static and kinetic friction coefficients. N Design an experiment to find µK. T Non-equilibrium: Set m2 T and adjust m1 to find regime where a ≠ 0 m1 Requires two FBDs m1g Mass 1 Σ Fy = m1a = fk m2 m2g Mass 2 T – m1g Σ Fx = m2a = -T + fk Σ Fy = 0 = N – m2g = -T + µk N T = m1g + m1a = µk m2g – m2a µk = (m1(g+a)+m2a)/m2g Physics 207: Lecture 7, Pg 42 Page 21 Physics 207 – Lecture 7 Sample Problem You have been hired to measure the coefficients of friction for the newly discovered substance jelloium. Today you will measure the coefficient of kinetic friction for jelloium sliding on steel. To do so, you pull a 200 g chunk of jelloium across a horizontal steel table with a constant string tension of 1.00 N. A motion detector records the motion and displays the graph shown. What is the value of µk for jelloium on steel? Physics 207: Lecture 7, Pg 43 Sample Problem Σ Fx =ma = F - ff = F - µk N = F - µk mg Σ Fy = 0 = N – mg µk = (F - ma) / mg & x = ½ a t 2 0.80 m = ½ a 4 s 2 a = 0.40 m/s2 µk = (1.00 - 0.20 · 0.40 ) / (0.20 ·10.) = 0.46 Physics 207: Lecture 7, Pg 44 Page 22 Physics 207 – Lecture 7 Recap Assignment: HW4, (due 2/17, 9 am, Wednesday) Read Chapter 7 1st Exam Wednesday, Feb. 17 from 7:15-8:45 PM Physics 207: Lecture 7, Pg 45 Page 23