quiz5_sol_sp10

quiz5_sol_sp10 - Physics 101 Classical Physics Spring 2010...

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Unformatted text preview: Physics 101 Classical Physics Spring 2010 Quiz 5 Instructions: The answer sheets must be handed in as soon as time is called. You have until 9:20am. Answer the following two multiple choice questions. Each question is worth 5 points. 1. A new floor is built in the University library. The floor is supported by four steel beams, each 4 m long and with a square cross-section whose sides are each 10 cm. Unfortunately, the engineers neglected to account for the weight of the books, and when the books are added the floor sinks by 1 mm. What is the mass of the books on this new floor of the library? For steel, Young's modulus is 2.0 1011 N/m2 . (a) 5 105 kg (b) 2 105 kg (c) 1.25 109 kg (d) 1.25 105 kg (e) 2 108 kg This is tensile stress problem. We know that the equation for stress in terms of strain is F L =Y A L where F is the force, A is the area over which it is exerted, Y is Young's modulus, and L is the unstretched or uncompressed length of the object. L is the change in length, which we are given is 1 mm or 1 10-3 m. So for the force on the beams we have F = LAY L (1) For the area, we have the cross section of each beam ((0.1m)2 ) times 4 because there are four beams, so F = (1 10-3 m)(4 (0.1m)2 )(2 1011 N/m2 ) = 2 106 N 4m (2) Since F = mg we therefore have m = 2 106 N/10m/s2 = 2 105 kg. And so the answer is b. 1 2. A new planet is discovered orbiting another star. The planet is the same distance from the star as Earth is from the Sun, but it takes only 3 months for it to complete an entire orbit around the star. What is the mass of the star M , relative to the mass of the Sun,M ? (a) M = 4M , (b) M = 16M , (c) M = M , (d) M = 0.25M , (e) Not enough information given. The state of motion of any orbiting body is that it is traveling in a circle at constant speed, 2 so that it has a centripetal acceleration of vplanet /R. The force that provides the centripetal acceleration is gravity, so for the planet around the star we have 2 vplanet GM mplanet = mplanet R2 R (3) We notice that the planet's mass cancels on both sides. We also need to remember that v = 2R , where T is the period of the orbit. T While we are given the period, we don't have R handy, but fortunately the question only asked for the relative mass of the star to the Sun. For the Sun and the Earth, the same equation holds v2 GM mEarth = mEarth Earth (4) R2 R and we divide the first equations by the second we get v2 M = planet 2 M vEarth Now plugging in v = have 2R , T (5) and realizing that Tplanet /TEarth = 3 months/1 year = 1/4, we 1 12 M = = M 4 16 (6) and so the answer is b. 2 For the following problem, you must Answer all sections of the problem. show your work to receive full credit. 3. On your homework, you had the problem of a crane whose center-of-mass was in its base, and you had to find the largest mass it could support. The mass of the crane's `boom', the part that sticks up, was ignored. A real crane boom has a mass, and the boom is usually thicker at the bottom so that the center-of-mass is closer to the base than the middle of the boom. As shown in the figure below, we have for this problem a crane whose base has a length l = 8 m, and whose boom has a length L = 12 m. The mass of the base is 4000 kg, and the mass of the boom is 1000 kg. The center-of-mass of the boom is located a distance d = L/3 = 4 m along the boom, as shown in the figure. You may need to know that cos 60 = 1 , sin 60 = 23 or tan 60 = 3. 2 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 L 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 d 1111111111 0000000000 1111111111 0000000000 o 1111111111 0000000000 1111111111 0000000000 60 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 l (a) (20 pts) Draw the free-body diagram for the crane, for the case where it is carrying the largest mass possible before it tips over. The figure below shows the free-body diagram of the crane. Note that the normal force FN at the corner doesn't vanish when the crane tips: what does go away is the normal force we would have drawn at the center of mass of the crane's base. Once the crane is tipping, the only point of contact is the corner and so the normal force must be there (and nowhere else). 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 F=-m g 11111111111 00000000000 boom 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 FN 11111111111 00000000000 11111111111 00000000000 T=-m g mass F=-m g base 3 (b) (40 pts) What is the largest mass the crane can hold without tipping over? As usual, once we have the forces on the free-body diagram, we write down Newton's second law for both the forces and the torques: Fx = 0 Fy = 0 = 0 (7) (8) (9) (10) There are no forces in the x-direction, so we ignore that one. In the y-direction we have FN - mbase g - mboom g - mmass g = 0 (11) We notice that in this equation, both FN and mmass are unknown, so we will have to use the torque equation to solve the problem. If we take the torques about the corner of the base (i.e., use this as the axis of rotation) then FN will not be involved so probably the force equation above won't be that useful. Like the homework problem, it is easier to deal with the `moment arms' here in calculating the torques. For the base itself, the moment arm is just half the length of the base, l/2. For FN , it is zero. For the tension from the rope at the end of the crane, this is just like the homework and so the moment arm is the horizontal distance from the l end of the crane's base, which is L cos - 2 . For the center of mass of the boom, the moment arm is similar, only it is on the other side of the axis of rotation so we have l - d cos . So adding all the torques together we would have 2 l l l = mbase g + mboom g( - d cos ) - mmass g(L cos - ) = 0 2 2 2 (12) Note the relative signs of the torques, which are obtained using the right hand rule. Solving for mmass we get mmass l l mbase 2 + mboom ( 2 - d cos ) = =0 l (L cos - 2 ) 1 2 (13) We were given that cos 60 = mmass and all the masses and distances so we have (14) 1 4000kg 8m + 1000kg( 8m - 4m 2 ) 2 2 = 9000kg = ( 12m - 8m ) 2 2 4 (c) (30 pts) Many cranes are `extendable', meaning they can be made longer. Assume that lengthening the crane (making L longer) does not change the position of the center-of-mass, so that we still have d = 4 m. If the mass being held by the crane is 500 kg, what is the largest L can be so that the crane does not tip over? For this problem, everything is the same except that now the mass is known, but L is not. So we have for the sum-of-torques again l l l = mbase g + mboom g( - d cos ) - mmass g(L cos - ) = 0 2 2 2 and we solve for L, not mmass . L= l (mbase 2 (15) + mboom + mmass ) - mmass d cos mmass cos (16) plugging in all the values gives L = 80 m. 5 ...
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This document was uploaded on 05/20/2010.

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