webct4 - Boxes with strings (2) 1. More about the boxes!...

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Boxes with strings (2) 1. More about the boxes! Three boxes labeled A, B and C with masses m , m and M ( M > m ), respectively, are attached with ideal massless strings as shown in the figure below. The string goes through an ideal pulley. Neglect friction and air resistance. Initially, the boxes are at rest and box C is at a height h = 3 m from the floor. The system is released at t = 0, and 1.3 s after, box C hits the floor. 1. If m = 150 g, what is M , the mass of box C? All the masses must have the same acceleration if the string length is constant. From the given information, we can compute the acceleration of box C. 2 00 2 2 1 2 1 3( 1 . 3 ) 2 3.55 / y yy vt a t a am s =+ + = = The net force on the whole system (made of three masses, all moving together so all having the same acceleration) is Mg (tensions are internal forces and they cancel each other out). For the whole system, then, Newton’s second law reads: (2 ) 2 () 2 2 (1) −= = −−−−−−−−−−−> M gM m a Mg Ma ma Mg a m a ma M ga .
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Substituting acceleration in equation (1), 2 2(150)(3.55) 9.81 3.55 170 = = = ma M ga M Mg 2. 2. Find the tension on the string between box A and box B. The only force acting on box A is T AB , the tension on the string between boxes A and B. For box A, then, Newton's second law read: AB Tm a = Hence, 2 0.15 3.55 / 0.533 == × = AB a k g m s N
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2 books 1. Two books lie one on top of the other on a table. The top book (B 1 ) has a mass m 1 = 1.0 kg and the bottom book (B 2 ) has a mass m 2 = 2.0 kg. The friction between the table and B 2 can be neglected, but the coefficients of kinetic and static friction between B 1 and B 2 are µ k = 0.2 and s = 0.4, respectively. The bottom book (B2) is pushed with a force F = 10 N (see figure). Draw the free body diagrams for both books. (Third-law pairs are marked in color) What is the magnitude of the acceleration of B 1 ? Book B 1 will move with the same acceleration as B 2 and in the same direction… as long as the friction between both is large enough to make them move together. We’ll check that this is the case in part 9. This acceleration can be found by applying Newton’s second law to the whole system: W 1 N on 1 by 2 f S on 1 by 2 m 1 W 2 N on 2 by 1 f S on 2 by 1 m 2 N on 2 by table F
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() 12 2 10 N 3.3 m/s 3 kg Fm m a F a mm =+ == = + 2. What is the magnitude of the acceleration of B 2 ? The acceleration of both books is the same. . as long as they move together. So the answer is 3.3 m/s 2 . 3. What is the magnitude of the net force on B 1 ? According to Newton’s second law: 2 net 1 1 (1 kg)(3.3 m/s ) 3.3 N a = 4. What is the magnitude of the net force on B 2 ? According to Newton’s second law: 2 net 2 2 (2 kg)(3.3 m/s ) 6.6 N a = 5. What is the magnitude of the frcition force done by B 2 on B 1 ? Now we really need the free-body diagram drawn at the beginning of the problem. The only horizontal force on B 1 is the friction due to B 2 (what else could be exerting a horizontal force on B 1 ?). So it must be equal to the net force on B 1 and is equal to 3.3 N.
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webct4 - Boxes with strings (2) 1. More about the boxes!...

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