* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Physics 207: Lecture 2 Notes
Survey
Document related concepts
Coriolis force wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Specific impulse wikipedia , lookup
Renormalization group wikipedia , lookup
Equations of motion wikipedia , lookup
Classical mechanics wikipedia , lookup
Centrifugal force wikipedia , lookup
Jerk (physics) wikipedia , lookup
Fictitious force wikipedia , lookup
Modified Newtonian dynamics wikipedia , lookup
Center of mass wikipedia , lookup
Relativistic mechanics wikipedia , lookup
Seismometer wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Classical central-force problem wikipedia , lookup
Transcript
Lecture 7 "Professor Goddard does not know the relation between action and reaction and the need to have something better than a vacuum against which to react. He seems to lack the basic knowledge ladled out daily in high schools." New York Times editorial, 1921, about Robert Goddard's revolutionary rocket work. "Correction: It is now definitely established that a rocket can function in a vacuum. The 'Times' regrets the error." New York Times editorial, July 1969. Physics 207: Lecture 7, Pg 1 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, due 2/18, 9 am, Wednesday) Read Chapter 7 1st Exam Wednesday, Feb. 18 from 7:15-8:45 PM Chapters 1-7 Physics 207: Lecture 7, Pg 2 Exercise, Newton’s 2nd Law A woman is straining to lift a large crate, without success. It is too heavy. We denote the forces on the crate as follows: 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). Which of following relationships between these forces is true, while the person is trying unsuccessfully to lift the crate? (Note: force up is positive & down is negative) A. B. C. D. P+C<W P+C>W P=C P+C=W Physics 207: Lecture 7, Pg 3 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 4 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) a Fnet If mass is constant then If force constant | 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 5 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 6 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 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 13 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 14 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 15 Scale Problem 3: T” T’ 1: 2: T ? W 1.0 kg -T ’ S Fy = 0 in all cases -T -T ? -mg 1: 0 = -2T + T ’ 2: 0 = T – mg T = mg 3: 0 = T” – W – T ’ (not useful here) Substituting 2 into 1 yields T ’ = 2mg = 20 N (We start with 10 N but end with 20 N) 1.0 kg Physics 207: Lecture 7, Pg 16 No Net Force, No acceleration…a demo exercise In this demonstration we have a ball tied to a string undergoing horizontal UCM (i.e. the ball has only radial acceleration) 1 Assuming you are looking from above, draw the orbit with the tangential velocity and the radial acceleration vectors sketched out. 2 Suddenly the string brakes. 3 Now sketch the trajectory with the velocity and acceleration vectors drawn again. Physics 207: Lecture 7, Pg 17 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 m were 0. Magnitude: f is proportional to the applied forces such that fs ≤ ms N ms called the “coefficient of static friction” Physics 207: Lecture 7, Pg 18 Friction: Static friction Static equilibrium: A block with a horizontal force F applied, S Fx = 0 = -F + fs fs = F FBD S Fy = 0 = - N + mg N = mg As F increases so does fs N F m fs 1 mg Physics 207: Lecture 7, Pg 19 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 m were 0. Magnitude: fS is proportional to the magnitude of N fs = ms N N F m mg Physics 207: Lecture 7, Pg 20 fs Kinetic or Sliding friction (fk < fs) Dynamic equilibrium, moving but acceleration is still zero S Fx = 0 = -F + fk fk = F S Fy = 0 = - N + mg N = mg As F increases fk remains nearly constant (but now there acceleration is acceleration) FBD v N F m fk 1 mg fk = mk N Physics 207: Lecture 7, Pg 21 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 = mk N ( = mK mg in the previous example) The constant mk is called the “coefficient of kinetic friction” Logic dictates that mS > mK for any system Physics 207: Lecture 7, Pg 22 Coefficients of Friction Material on Material ms = static friction mk = 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 23 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 mS N T Static equilibrium: Set m2 m2 and add mass to m1 to reach the breaking point. Requires two FBDs Mass 1 S Fy = 0 = T – m1g fS T m1 m2g m1g Mass 2 S Fx = 0 = -T + fs = -T + mS N S Fy = 0 = N – m2g T = m1g = mS m2g mS = m1/m2 Physics 207: Lecture 7, Pg 24 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 mK. 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 S Fy = 0 = T – m1g N m2 m2g m1g Mass 2 S Fx = 0 = -T + ff = -T + mk N S Fy = 0 = N – m2g T = m1g = mk m2g mk = m1/m2 Physics 207: Lecture 7, Pg 25 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 mK. T Non-equilibrium: Set m2 T and adjust m1 to find regime where a ≠ 0 m1 Requires two FBDs m1g Mass 1 S Fy = m1a = T – m1g fk m2 m2g Mass 2 S Fx = m2a = -T + fk S Fy = 0 = N – m2g = -T + mk N T = m1g + m1a = mk m2g – m2a mk = (m1(g+a)+m2a)/m2g Physics 207: Lecture 7, Pg 26 Inclined plane with “Normal” and Frictional Forces 1. At first the velocity is v up along the slide “Normal” means perpendicular 2. Can we draw a velocity time plot? Normal Force 3. What the acceleration versus time? v Friction Force Sliding Down q fk Sliding Up q mg sin q Weight of block is mg Note: If frictional Force = Normal Force (coefficient of friction) Ffriction = m Fnormal = m mg sin q then zero acceleration Physics 207: Lecture 7, Pg 29 Recap Assignment: HW4, (Chapters 6 & 7, due 2/18, 9 am, Wednesday) Read Chapter 7 1st Exam Wednesday, Feb. 18 from 7:15-8:45 PM Chapters 1-7 Physics 207: Lecture 7, Pg 30