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david morin solution manualStudents also viewed Modul 1, Praktikum Fisika Dasar Fisika-dasar Modul Tutorial 6 Fidas IA 2018-2019 Soal Ayunan Fisis final Difusi disolusi Quiz 4 - Quiz4 Other related documents TB Chapter 14 - Lecture notes 1 EL3110 06 13216024 Exam 18 July 2017, questions and answers Euroland Foods Case Solution 38468 100110 1 PB - Grade: 9 Cakul BAB VI Analisa Vektor 18102015 Preview text SOLUTIONS MANUAL Introduction to Classical Mechanics With Problems and Solutions David Morin Cambridge University Press TO THE INSTRUCTOR: I have tried to pay as much attention to detail in these exercise solutions as I did in the problem solutions in the text. But despite working through each solution numerous times during the various stages of completion, there are bound to be errors. So please let me know if anything looks amiss. Also, to keep this pdf file from escaping to the web, PLEASE distribute it to anyone, with the exception of your teaching assistants. And please make sure they also agree to this. Once this file gets free, no going back. In addition to any comments you have on these solutions, I welcome any comments on the book in general. I hope enjoying using it. David Morin (Version 2, April 2008) c David Morin 2008 2 CHAPTER 1. STRATEGIES FOR SOLVING PROBLEMS 1.13. machine (a) This gives a1 0. (Half of m2 balances each of m1 and m3.) (b) Ignore the m2 m3 terms, which gives a1 (Simply in freefall.) (c) Ignore the terms involving m1, which gives a1 3g. (m2 and m3 are in freefall. And for every meter they go down, a total of three meters of string appears above them, so m1 goes up three meters.) (d) Ignore the m1 m3 terms, which gives a1 g. (m2 goes down at g, and m1 and m3 go up at g.) (e) This gives a1 (Not obvious.) 1.14. Cone frustum The correct answer must reduce to the volume of a cylinder, h, when a b. Only the 2nd, 3rd, and 5th options satisfy this. The correct answer must also reduce to the volume of a cone, when a 0. Only the 1st, 3rd, and 4th options satisfy this.http://ehconsultores.com/userfiles/evenflo-breast-pump-manual-review.xml
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The correct answer must therefore be the 3rd one, ab b2 1.15. Landing at the corner The correct answer must go to infinity for. Only the 2nd and 3rd options satisfy this. The correct answer must also go to infinity for. Only the 1st and 2nd options satisfy this. The correct answer must therefore be the 2nd one. 1.16. Projectile with drag Using the Taylor series for, we have y(t) 1 g gt v0 sin 1 (1 t2 g gt 2 v0 sin t 1 2 1 2 (v0 sin gt v0 sin 2 2 (4) If sin then the third term is much smaller than the second, and we obtain the desired result. Letting T equal 1 gives a final position of 1.264. Letting T equal 10 and 100 gives final positions of and respectively. These approach 2. In the x 2 case, the only change in the entire program is in the 6th line, where we now have the square of x1: the given equation Letting T equal 1, 10, 100, and gives final positions of 1.099, 3.044, 5.302, 7.600, and 9.903, respectively. Looking at the successive differences between these values, we see that they approach roughly 2.3. This constant difference for inputs of powers of 10 implies a log dependence on the time. Chapter 2 Statics 2.20. Block under an overhang break up the forces into components parallel and perpendicular to the overhang. Let positive Ff point up along the overhang. Balancing the forces parallel and perpendicular to the overhang gives, respectively, Ff M g sin M g cos N M g sin M g cos and (5) N must be positive, so we immediately see that must be at least if there is any chance that the setup is static. The coefficient tells us that. Using Eq. (5), this inequality becomes M g(sin cos g(sin cos tan (6) We see that we must have 1 in order for there to exist any values of that satisfy this inequality. If then can be as small as, but it be any smaller. 2.21. Pulling a block The Fy forces tell us that N F sin mg 0 N mg F sin And assuming that the block slips, the Fx forces tell us that F cos. Therefore, F cos F sin F.If 0, we have 0 and F 0. If we have and F mg. 2.22.http://gemicilojistik.com/userfiles/upload/evenflo-booster-seat-manual.xml Holding a cone Let F be the friction force at each finger. Then the Fy forces on the cone tell us that 2F cos 2N sin mg 0. But F. Therefore, cos 2N sin mg 0 N mg.When tan we have N So tan is the minimum allowable value of 2.23. Keeping a book up The result of Problem 2.4 is F cos assuming that sin cos is positive (that is, tan If it is negative, there is no solution for F. To find the maximum force, consider two cases: 5 6 CHAPTER 2. STATICS (a) Your force is directed upward 0): Then Ff points downward in the maximal F case. So Fy gives F sin Ff mg 0 Ff F sin mg. But Ff cos so we have mg F sin mg cos F, (9) sin cos assuming that sin cos is positive (that is, tan If it is negative, then F (sin cos mg is true for any F, so there is no upper bound in this case. (b) Your force is directed downward 0): Then Ff points upward in the maximal F case. So Fy gives (note that sin is negative here) F sin Ff mg 0 Ff sin mg. But Ff cos so we have mg, (10) sin mg cos F sin cos assuming that sin cos is positive (that is, tan If it is negative, then F (sin cos mg is never true, so there is no solution for F. This is the same result as in Problem 2.4, so it actually yield an upper bound on F. Putting all this together (along with the results from Problem 2.4): As a function of and for a generic value of less than 1, the values of F that keep the book up are signified the shaded region in Fig. 1. F mg mg mg F mg F Figure 1 2.24. Bridges (a) Looking at the Fx forces on the car, we see that the two inner diagonal beams must have equal tensions. Then Fy with these two beams tells us that the tensions are each 3. Then Fy on one of the the hinges gives the compression in the outer diagonal beams as 3. Then Fx with one of the upper hinges gives the tension in the top beam as 3. (b) Using Fy, we can start in the middle and work our along the diagonal beams to show that they all have equal forces of 3, alternating tension and compression. We can then work our way back in top beams (using Fx at the hinges) to the outer ones have 3 compression, and the middle one has 3 compression. Likewise, the outer bottom beams have 3 tension, and the inner bottom beams have 3 tension. 8 CHAPTER 2. STATICS fact, because the stick is massless), which contradicts the fact that the system is static. (2) If the stick is massive, there is now a torque from gravity (unless the stick is hanging vertically). This can cancel a nonzero torque from the hinge at the other end. 2.30. Ball on a wall Let T be the tension in the string. Then the friction force Ff from the wall must also be T, to provide zero net torque around the center. So if N is the normal force from the wall, then balancing the x forces quickly gives N T sin But Ff T sin sin Interestingly, this equals 1 for. 2.31. Cylinder and hanging mass If T is the tension in the string, then T mg. If F is the friction force from the plane, then balancing torques around the center of the cylinder gives F T, so F also equals mg. If N is the normal force from the plane, then balancing horizontal forces on the cylinder gives N sin F cos N tan Finally, balancing vertical forces on the cylinder gives N cos F sin M g T 0 mg cos (mg) sin mg M g tan sin M. (12) 1 sin If 0, then m 0. And if, then m These make sense. Alternatively, once we know that T mg, we can just use torque around the contact point on the plane, which require knowing F or N. The lever arm for the M g force is R sin and the lever arm for the T force is R(1 sin Balancing the torques around the contact point therefore gives (M g)R sin (mg)R(1 sin in agreement with the above result. 2.32. Ladder on a corner If Nc is the normal force from the corner, then balancing torques around the top end of the ladder gives Nc M cos Nc g cos And if Nw is the normal force from the wall, then balancing torques around the corner gives Nw sin M cos Nw g cos sin If Ff is the friction force at the corner, then balancing the horizontal forces gives Ff cos Nc sin, and so Ff g sin sin But we need Ff. Therefore, 2M g sin Mg 2M g cos 3 3 sin 3 sin 1.Alternatively, a quicker way to do the problem is to note that balancing torques around the pivot, along with forces along the line of the stick, tells us that your force must have components M g sin along the stick and M g cos perpendicular to it. In other words, your force must have magnitude M g and must make the same angle with the stick as the gravitational force does. 9 2.34. Stick and a cylinder (a) Balancing torques on the cylinder around its center tells us that the friction forces from the plane and the stick are equal. Call them F. Balancing torques on the stick around the pivot tells us that the normal force from the cylinder on the stick is Let N be the normal force from the plane on the cylinder. Balancing horizontal forces on cylinder gives F F cos N sin and balancing vertical forces gives mg F sin N cos Solving these equations for N yields N (This can also be obtained more quickly balancing torques on the whole system around the pivot.) (b) The above equations give F sin cos The cylinder slip on the plane if F sin cos The cylinder slip under the stick if F 3 sin cos Both of these conditions must be satisfied, and the latter is more strict, so we have 3 sin cos If 0, then 0, as expected. If, then 3, which obvious. 2.35. Two sticks and a string (a) Balancing vertical forces on the whole system tells us that the normal forces at the bottoms of the sticks must sum to 2mg. Balancing torques on the whole system around the hinge then tells us that these normal forces must be equal, and hence both equal to mg. Finally, balancing torques on the right stick around the hinge tells us that the tension T in the string satisfies T cos sin sin T mg sin. 2 cos (14) (b) Look at the forces on the right stick. The mg forces (gravity and normal force) cancel. Therefore, the force from the hinge must cancel the tension. So the hinge force points up to the right (perpendicular to the stick) with magnitude mg sin cos 2.36. Two sticks and a wall Let Fx and Fy be the desired components. The masses of the bottom and top sticks are and cos respectively. So balancing torques on the whole system around the left end of the bottom stick gives Fx (L tan L cos L 2 Fx 2 1 cos.The leftover quadratic gives c 3 as the physical answer. 2.37. Stick on a circle (a) From Problem 2.18 (using the same notation), we have Ff N sin But Ff, so we must have sin cos 11 as we wanted to show. Since the result is true for N 1, it is therefore true for all N. For large N, this result behaves like d ln N. So for N we can make the blocks hang out infinitely far. But the result grows very slowly with N, like a log. If we want the overhang to be L, then we need d ln N L N, which grows very quickly with L. 12 CHAPTER 2. STATICS 14 CHAPTER 3. USING F M A 3.29. 3 Define all accelerations positive upwards.The F ma equations are T mg ma1, 2T 2mg 2ma2, T 3mg 3ma3. (22) Solving these (the first two quickly give a1 a2 ), along with a2 a3 gives a1 a2 and a3 3.30. 4 If T is the tension in the string connected to the right mass, then you can work your way down the pulleys to show that the tension in the string connected to the left mass is 2N T (where there are N pulleys, not including the rightmost one).This reasoning continues until the right mass has acceleration 2N a downward. The F ma equations for the left and right masses are then (with upward and downward taken to be positive, respectively) 2N T mg ma, mg T m(2N a). (23) N Multiplying the second equation 2 and adding the result to the first equation gives the acceleration of the left mass as a g(2N 1). The acceleration of the right mass is then 2N a 22N 2N 22N 1 1 1 g. (24) For N 0 we have a 0, as expected. For N we have a 0, but 2N a so the left mass hardly moves upward, while the right mass accelerates downward with an acceleration essentially equal to g. 3.31. 5 Draw a horizontal line between the two shaded pulleys. If the right pulley goes down d, then a length d of string appears above the line (because 2d appears above the top pulley, but d disappears right below it). This length d must disappear below the line. It gets divided evenly between the two pieces touching the bottom pulley, which therefore goes up So the downward acceleration of the top pulley is twice the upward acceleration of the bottom pulley. The F ma equations for the top and bottom pulleys are, respectively, mg T 2T ma1, 2T mg ma2. (25) Solving these, along with a1 2a2, gives a1 downward, and a2 upward. And T happens to equal 3.32. 6 The string is one continuous piece, so the tension is the same throughout it. The force on the (massless) left pulley is therefore T 2T. But this force must be zero because the pulley is massless. Hence, T 0, which means that nothing is holding the masses up, so both are in freefall. The physical reason for this result is that the left pulley is free to fall however much is needed to provide enough string for the freefall motion of the masses. If the left pulley falls a distance d, then a length d of string appears above it, but a length 2d disappears below it. So a length d has been which allows each of the masses to fall a distance (as you can verify). Since there is nothing keeping the left pulley from accelerating downward at 3g, there is therefore nothing keeping the two masses from freefalling at g. 15 3.33. Accelerating plane Consider first the case of maximum a. In this case, the friction force points down the plane. The F ma equations along the plane and perpendicular to it are Ff mg sin (ma) cos N mg cos (ma) sin (26) These equations give Ff and N. Demanding that Ff gives g(sin cos.The F ma equation along the plane changes to mg sin (ma) cos (28) while the equation for the perpendicular direction remains the same. Ff now gives g(sin cos. (29) cos sin Putting these bounds together gives g(sin cos g(sin cos.Consider first the case of maximum a. In this case N3 0, because the top cylinder is just about to rise up off the right cylinder. So we have: Fx max on top cylinder N2 cos ma, N2 sin mg 0. (31) Solving these equations for a gives a 3. Now consider the case of minimum a. In this case N1 0, because the top cylinder is just about to fall down between the bottom two cylinders. So we have: Fy may on top cylinder Fx max on right cylinder N3 cos ma, Fx max on top cylinder N2 cos N3 cos ma, N2 sin N3 sin mg 0. (32) equations for a gives a 3. Combining the results gives 3 a 3. Fy may on top cylinder 3.35. Leaving the sphere The normal force is obtained from the radial F ma equation, which gives mg cos N N mg cos. (33) The friction force is, so the tangential F ma equation is mg sin cos ) So is given sin cos. (34) (35) 17 (b) It turns out that the answer depends on the relation between and The acceleration on the way up points down the plane with magnitude au g sin cos The time on the way up is simply tu v0. The acceleration on the way down points down the plane with magnitude ad g sin cos To find the time on the way down, we need to find the length of the trip, which we will then use in ad t2d The length is obtained from the kinematic equation for the upward motion, v0 tu au t2u v02 ) (which can also be seen running time backwards for the upward journey). So ad t2d gives v02 ) ad t2d td v0 au ad. The total time with friction is therefore tu td v0 v0 g 1 1 au au ad ! 1 1 sin cos sin2 cos2. (36) The total time without friction is simply the preceding result with 0, which gives T0 2v0 sin Letting x cos sin tan (note that the tan condition from part (a) implies x 1), the trip takes longer with friction if T 0 1 1 2. 1 x2 (37) Isolating the square root and then squaring gives 1 1 x2 (1 2x)2 (1 x)2 (1 2x)2 (1 x) 0 2. x(2x2 1) x (38) So if x 2 then T0. Recalling the definition of x, we see that the trip takes longer with friction if (tan the slowness on the way down wins out over the decreased distance (this obvious, except in the limit where is close to tan in which case the block takes a very long time to come back down). If is smaller than this, then the trip is quicker with the decreased length wins out over the slowness on the way down (which is no means the above calculation is required). (c) For a given if we factor out v0 sin from the total time, we see from part (b) that we want to minimize the function f (x) x) 1 x2.If t1 is the timep to hit the surface, then gt21 h y gives t1 2(h and so v gt1 2g(h y). The vertical speed is zero right after the p bounce, so the time it takes to hit the ground is given gt22 y. Hence t2 The horizontal distance traveled is therefore p d vt2 2 y(h y). Taking the derivative, we see that this function of y is maximum at y The corresponding value of d is dmax h. Second solution: Assume that the greatest distance, d0, is obtained when y y0, and let the speed at y0 be v0. Consider the situation where the ball falls all the way down to y 0 and then bounces up at an angle such that when it reaches the height y0, it is traveling horizontally. When it reaches the height y0, the ball will have speed v0 ( conservation of energy, which will be introduced in Chapter 5), so it will travel a horizontal distance d0 from this point. 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