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7. Newton's 2Nd Law in More Complicated Problems and Friction

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7. Newton's 2Nd Law in More Complicated Problems and Friction 7. Newton's 2nd Law in More Complicated Problems and Friction The Atwood's Machine is used below to help in understanding how Newton's 2nd law applies to a system of two connected masses. It is not a practical machine. Suppose two different masses M1 and M2 are attached to a rope which is placed over a pulley as indicated in the diagram below. There are TWO free- body diagrams since there are two masses in this problem. Pulley T2 Y Free-Body Diagram for M2 T1 M2 W2 M1 X W 1 Free-Body Diagram for M1 i) W1 is the force of gravity on mass M1 and W2 is the force of gravity on mass M2. W1 = M1 g and W2 = M2 g. ii) T1 is the tension force on M1 due to the attached rope and T2 is the tension force on M2 due to the attached rope. iii) Assume an ideal rope that is a mass-less rope that does not stretch so T1=T2. iv) The only motion is in the y-direction so we only need to consider Newton's 2nd law in the form Fy = M ay . Since M2 > M1 the pulley will rotate clockwise. v) Key Idea: Apply Newton's 2nd law to each mass separately. 2 7. Newton'sLaw ComplexProblems rev2.nb i) W1 is the force of gravity on mass M1 and W2 is the force of gravity on mass M2. W1 = M1 g and W2 = M2 g. ii) T1 is the tension force on M1 due to the attached rope and T2 is the tension force on M2 due to the attached rope. iii) Assume an ideal rope that is a mass-less rope that does not stretch so T1=T2. iv) The only motion is in the y-direction so we only need to consider Newton's 2nd law in the form Fy = M ay . Since M2 > M1 the pulley will rotate clockwise. v) Key Idea: Apply Newton's 2nd law to each mass separately. | 7. Newton'sLaw ComplexProblems rev2.nb 3 For Mass M1: F1=M1a1 where the total force on M1 in the y-direction is F1=T1-W1 and a1 is the acceleration of mass M1 in the upward or positive y- direction (a1 is a positive number). eqt. #1: T1-W1 = M1a1 (T1 is a positive number.) For Mass M2: F2=M2 a2 where the total force on M2 in the y-direction is F2=T2-W2. Since the acceleration of M2 is downward, the acceleration has a negative y-component (-a2) where a2 is a positive number. eqt. #2: T2-W2 = M2 H-a2) (T2 is a positive number.) Ideal Rope: The rope does not stretch and if M1 moves a distance then M2 moves an equal distance so a1 = a2 = a and Ideal Pulley: Mass-less and only serves change the direction of the tension force in the rope so T1 = T2 = T. Equations #1 and #2 simplify to T-W1 = M1a and T-W2 = M2 H-a) Subtract equation #2 from equation #1 and obtain W2 - W1 = HM1 + M2 L a Notice the tension T canceled out. Finally solving for the acceleration a you HM - M L g get a = W2-W1 = 2 1 . HM1 +M2 L HM1 +M2 L 1. Notice the effective inertia of the combined systems is the sum of the two masses HM1 + M2 L . 2. The net gravitational force on the two masses is due to the difference of the two masses HM2 - M1 L g . For Mass M1: F1=M1a1 where the total force on M1 in the y-direction is F1=T1-W1 and a1 is the acceleration of mass M1 in the upward or positive y- direction (a1 is a positive number). eqt. #1: T1-W1 = M1a1 (T1 is a positive number.) For Mass M2: F2=M2 a2 where the total force on M2 in the y-direction is F2=T2-W2. Since the acceleration of M2 is downward, the acceleration has a negative y-component (-a2) where a2 is a positive number. eqt. #2: T2-W2 = M2 H-a2) (T2 is a positive number.) Ideal Rope: The rope does not stretch and if M1 moves a distance then M2 moves an equal distance so a1 = a2 = a and Ideal Pulley: Mass-less and only serves change the direction of the tension force in the rope so T1 = T2 = T. Equations #1 and #2 simplify to T-W1 = M1a and T-W2 = M2 H-a) 4 Subtract 7. Newton'sLaw equation ComplexProblems #2 rev2.nbfrom equation #1 and obtain W2 - W1 = HM1 + M2 L a Notice the tension T canceled out. Finally solving for the acceleration a you HM - M L g get a = W2-W1 = 2 1 . HM1 +M2 L HM1 +M2 L 1. Notice the effective inertia of the combined systems is the sum of the two masses HM1 + M2 L . 2. The net gravitational force on the two masses is due to the difference of the two masses HM2 - M1 L g . A Numerical Example: Suppose M1 =3 kg and M2 = 7 kg. Determine the acceleration of the masses. M1 = 3.; M2 = 7.; g = 9.8; HM2 - M1L * g a = M1 + M2 3.92 2 So the acceleration is a=3.9 m/s and a1 = a2 = a. Atwood's Machine Experiment on the World Wide Web: Rutgers University Go to the web address http://paer.rutgers.edu/PT3/experiment.php?topicid=8&exptid=4 for a QuickTime video atwood.mov of the Atwood's machine which you may play through Safari or Explorer on your computer at home. Here M2 =200 gm and M1 =150 gm. Compute the acceleration using (note gm are converted into kg) M1 = 150. ê 1000; M2 = 200. ê 1000; g = 9.8; HM2 - M1L * g a = M1 + M2 1.4 2 a=1.4 m/s The masses are released from rest and observe how long it takes M2 to travel y=0.5 m (M2 starts out at 20 cm so observe the time it takes for M2 to get to 70 cm). H70 cm- 20 cmL = 0.5 m If you use the kinmatic equation y = 1 a t2 and solve for the time t it takes 100 cmêm 2 M2 to travel y = 0.5 m you get 7. Newton'sLaw ComplexProblems rev2.nb 5 2 a=1.4 m/s The masses are released from rest and observe how long it takes M2 to travel y=0.5 m (M2 starts out at 20 cm so observe the time it takes for M2 to get to 70 cm). H70 cm- 20 cmL = 0.5 m If you use the kinmatic equation y = 1 a t2 and solve for the time t it takes 100 cmêm 2 M2 to travel y = 0.5 m you get y = 0.5; 2 * y t = a 0.845154 So it should take M2 t = 0.845 sec to travel y = 0.5 m. The QuickTime has a frame rate of 30 fps where fps is frames per second. The inverse of this is second per frame Dt which works about to be 1 Dt = 30. 0.0333333 The number of frames Num it takes M2 to travel y = 0.5 m which is t = 0.845 sec is given by t Num = Dt 25.3546 which is about half way between 25 frames and 26 frames provided Newton's 2nd Law is valid for the Atwood's machine. This should be a bit of good evidence for you that Newton's 2nd law is true but of course many more values of M2 and M1 need to be tried before you really should be convinced. In fact, to be convincing you should work out Newton's 2nd law for all possible examples not just the Atwood's machine but also, for example a block sliding down an inclined plane. But of course it is impossible to work out all possible examples and so you can never really prove Newton's 2nd Law is universally true. | 6 7. Newton'sLaw ComplexProblems rev2.nb The Force Due to Friction : Friction is a force between two objects in relative motion. Friction is due to intermolecular forces between the atoms of the to solid objects. When we refer to friction, we mean the force of resistance between two solid object. "Air resistance" or "drag force" has some different properties and will be discussed elsewhere. Properties of the Friction Force : 1. acts in a direction opposite the direction of motion. 2. The magnitude of is given by = m N where m is called the coefficient of friction and N is the normal force discussed previously. 3. m is a number having no dimension in the range 0§mb1. m depends upon the materials involved. 4. Actually there are at least two different coefficients of friction: "static" friction ms and "kinetic" friction mk . The concepts are similar but satisfy different equations: s = ms N and k = mk N. s applies before motion begins and k after motion starts. 5. ms and mk depend on the materials involved. For example, for example for wood on wood ms=0.4 and mk = 0.2 For Teflon on Teflon ms=mk =0.04 m is smaller when the friction force is less. See table 4.2 for other examples. 6. ms is the static coefficient of friction that exists before motion takes place and mk is the kinetic coefficient of friction after relative motion between the block and the table top begins.
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