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Unformatted text preview: 1 Chapter 5 1. We apply Newtons second law (specifically, Eq. 52). (a) We find the x component of the force is F x = ma x = ma cos 20 = (1 . 00 kg)(2 . 00 m / s 2 ) cos 20 = 1 . 88 N . (b) The y component of the force is F y = ma y = ma sin 20 = (1 . 0 kg)(2 . 00 m / s 2 ) sin 20 = 0 . 684 N . (c) In unitvector notation, the force vector (in Newtons) is ~ F = F x + F y j = 1 . 88 + 0 . 684 j . 2. We apply Newtons second law (Eq. 51 or, equivalently, Eq. 52). The net force applied on the chopping block is ~ F net = ~ F 1 + ~ F 2 , where the vector addition is done using unitvector notation. The acceleration of the block is given by ~a = ~ F 1 + ~ F 2 /m . (a) In the first case ~ F 1 + ~ F 2 = (3 . 0 N) + (4 . 0 N) j + ( 3 . 0 N) + ( 4 . 0 N) j = 0 so ~a = 0. (b) In the second case, the acceleration ~a equals ~ F 1 + ~ F 2 m = (3 . 0 N) + (4 . 0 N) j + ( 3 . 0 N) + (4 . 0 N) j 2 . 0 kg = 4 . j m / s 2 . (c) In this final situation, ~a is ~ F 1 + ~ F 2 m = (3 . 0 N) + (4 . 0 N) j + (3 . 0 N) + ( 4 . 0 N) j 2 . 0 kg = 3 . 0 m / s 2 . 3. We are only concerned with horizontal forces in this problem (gravity plays no direct role). We take East as the + x direction and North as + y . This calculation is efficiently implemented on a vector capable calculator, using magnitudeangle notation (with SI units understood). ~a = ~ F m = (9 . 6 ) + (8 . 6 118 ) 3 . = (2 . 9 6 53 ) Therefore, the acceleration has a magnitude of 2 . 9 m/s 2 . 4. Since ~v = constant, we have ~a = 0, which implies ~ F net = ~ F 1 + ~ F 2 = m~a = 0 . Thus, the other force must be ~ F 2 = ~ F 1 = 2 + 6 j N . 5. Since the velocity of the particle does not change, it undergoes no acceleration and must therefore be subject to zero net force. Therefore, ~ F net = ~ F 1 + ~ F 2 + ~ F 3 = 0 . Thus, the third force ~ F 3 is given by ~ F 3 = ~ F 1 ~ F 2 = 2 + 3 j 2 k 5 + 8 j 2 k = 3  11 j + 4 k in Newtons. The specific value of the velocity is not used in the computation. 6. The net force applied on the chopping block is ~ F net = ~ F 1 + ~ F 2 + ~ F 3 , where the vector addition is done using unitvector notation. The acceleration of the block is given by ~a = ~ F 1 + ~ F 2 + ~ F 3 /m . (a) The forces exerted by the three astronauts can be expressed in unitvector notation as follows: ~ F 1 = 32(cos 30 + sin 30 j) = 27 . 7 + 16 j ~ F 2 = 55(cos 0 + sin 0 j) = 55 in Newtons, and ~ F 3 = 41 cos( 60 ) + sin( 60 ) j = 20 . 5  35 . 5 j in Newtons. The resultant acceleration of the asteroid of mass m = 120 kg is therefore ~a = (27 . 7 + 16 j) + (55 ) + (20 . 5  35 . 5 j) 120 = . 86  . 16 j m / s 2 . (b) The magnitude of the acceleration vector is  ~a  = q a 2 x + a 2 y = q . 86 2 + ( . 16) 2 = 0 . 88 m / s 2 ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Force

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