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 1 st and 2 nd laws. Distinguish static and kinetic coefficients of friction Differentiate between Newton s 1 st, 2 nd and 3 rd Laws Assignment: HW4, (Chapters 6 & 7, due 2/18, 9 am, Wednesday) Read Chapter 7 1 st Exam Wednesday, Feb. 18 from 7:15-8:45 PM Chapters 1-7 Physics 207: Lecture 7, Pg 2 Page 1
Exercise, Newton s 2 nd Law A woman is straining to lift a large crate, without success. I t is too heavy. W e denote the f orces on the crate as f ollows: 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 Wis the weight (force of the earth on the crate). W h ich of following relationships between these forces is true, while the person is trying unsuccessfully to lift the crate? (Note: f orce up is positive & down is negative) A. P + C < W B. P + C > W C. P = C D. 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 Page 2
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 If mass is constant then a F net 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 5 Exercise Newton s 2 nd 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. increasing B. decreasing C. constant in time D. Not enough information to decide Physics 207: Lecture 7, Pg 6 Page 3
Exercise Newton s 2 nd 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 B. C C. D D. F E. G Physics 207: Lecture 7, Pg 7 Home Exercise Newton s 2 nd 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. 4 x as long B. 2 x as long C. 1/2 as long D. 1/4 as long Physics 207: Lecture 7, Pg 8 Page 4
Home Exercise Newton s 2 nd Law Solution Force Cart Air Track F = ma Since F 2 = 1/2 F 1 a 2 = 1/2 a 1 We know that under constant acceleration, v = a t So, a 2 t 2 = a 1 t 1 we want equal final velocities 1/2 a 1 / t 2 = a 1 / t 1 t2 = 2 t 1 (B) 2 x as long Physics 207: Lecture 7, Pg 9 Home Exercise Newton s 2 nd 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. The new distance the cart moves relative to D is: A. 8 x as far B. 4 x as far C. 2 x as far D. 1/4 x as far Physics 207: Lecture 7, Pg 10 Page 5
Home Exercise Solution Force Cart Air Track We know that under constant acceleration, x = a ( t) 2 /2 (when v 0 =0) Here t 2 =2 t 1, F 2 = F 1 a 2 = a 1 x x 1 1 2 a t a t 2 2 2 2 2 ( 2 t1 ) = = = 2 2 1 t 1 1 (B) 4 x as long 4 Physics 207: Lecture 7, Pg 11 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 12 Page 6
Moving forces around Massless strings: Translate forces and reverse their direction but do not change their magnitude (we really need Newton s 3 rd of action/reaction to justify) string T 1 -T 1 Massless, frictionless pulleys: Reorient force direction but do not change their magnitude T 1 -T 1 T 2 -T2 T 1 = -T 1 = T 2 = T 2 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/s 2 ). 10 N 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. 1.0 kg? What force does the fish scale now read? 1.0 kg Physics 207: Lecture 7, Pg 14 Page 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 15 Scale Problem T 3: T 2: 1: T W? 1.0 kg -T -mg -T -T Σ F y = 0 in all cases 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? Page 8
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 µ were 0. Magnitude: f is proportional to the applied forces such that f s µ s N µ s called the coefficient of static friction Physics 207: Lecture 7, Pg 18 Page 9
Friction: Static friction Static equilibrium: A block with a horizontal force F applied, Σ F x = 0 = -F + f s f s = F FBD Σ F y = 0 = - N + mg N = mg As F increases so does f s F mg N m 1 f s 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 µ were 0. Magnitude: f S is proportional to the magnitude of N f s = µ s N F N m f s mg Physics 207: Lecture 7, Pg 20 Page 10
Kinetic or Sliding friction (f k < f s ) Dynamic equilibrium, moving but acceleration is still zero FBD Σ F x = 0 = -F + f k f k = F Σ F y = 0 = - N + mg N = mg As F increases f k remains nearly constant (but now there acceleration is acceleration) v F mg N m 1 f k f k = µ k 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: f k is proportional to the magnitude of N f k = µ k N ( = µ K mg in the previous example) The constant µ k is called the coefficient of kinetic friction Logic dictates that µ S > µ K for any system Physics 207: Lecture 7, Pg 22 Page 11
Coefficients of Friction Material on Material steel / steel add grease to steel metal / ice brake lining / iron tire / dry pavement tire / wet pavement µ s = static friction 0.6 0.1 0.022 0.4 0.9 0.8 µ k = kinetic friction 0.4 0.05 0.02 0.3 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 µ S Static equilibrium: Set m 2 and add mass to m 1 to reach the breaking point. Requires two FBDs Mass 1 Σ F y = 0 = T m 1 g T m 1 m 1 g Mass 2 Σ F x = 0 = -T + f s Σ F y = 0 = N m 2 g f S N T = -T + µ S N m 2 m 2 g T = m 1 g = µ S m 2 g µ S = m 1 /m 2 Physics 207: Lecture 7, Pg 24 Page 12
A 2 nd experiment Two blocks are connected on the table as shown. The table has unknown static and kinetic friction coefficients. Design an experiment to find µ K. Dynamic equilibrium: Set m 2 and adjust m 1 to find place when a = 0 and v 0 Requires two FBDs Mass 1 Σ F y = 0 = T m 1 g T m 1 m 1 g Mass 2 Σ F x = 0 = -T + f f Σ F y = 0 = N m 2 g f k N T = -T + µ k N m 2 m 2 g T = m 1 g = µ k m 2 g µ k = m 1 /m 2 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. Design an experiment to find µ K. N T Non-equilibrium: Set m 2 and adjust m 1 to find regime where a 0 Requires two FBDs T m 1 m 1 g f k m 2 m 2 g Mass 1 Mass 2 Σ F y = m 1 a = T m 1 g Σ F x = m 2 a = -T + f k = -T + µ k N Σ F y = 0 = N m 2 g T = m 1 g + m 1 a = µ k m 2 g m 2 a µ k = (m 1 (g+a)+m 2 a)/m 2 g Physics 207: Lecture 7, Pg 26 Page 13
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 µ k for jelloium on steel? Physics 207: Lecture 7, Pg 27 Sample Problem Σ F x =ma = F - f f = F - µ k N = F - µ k mg Σ F y = 0 = N mg µ k = (F - ma) / mg & x = ½ a t 2 0.80 m = ½ a 4 s 2 a = 0.40 m/s 2 µ k = (1.00-0.20 0.40 ) / (0.20 10.) = 0.46 Physics 207: Lecture 7, Pg 28 Page 14
Inclined plane with Normal and Frictional Forces 1. At first the velocity is v up along the slide 2. Can we draw a velocity time plot? 3. What the acceleration versus time? v Friction Force Sliding Down Normal means perpendicular Normal Force Weight of block is mg Note: If frictional Force = Normal Force (coefficient of friction) F friction = µ F normal = µ mg sin θ then zero acceleration θ f k Sliding Up θ mg sin θ 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 Page 15