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# ch05 - Chapter 5 1 We are only concerned with horizontal...

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171 Chapter 5 1. 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 magnitude-angle notation (with SI units understood). a F m   9 0 0 8 0 118 30 2 9 53 . . . . b g b g b g Therefore, the acceleration has a magnitude of 2.9 m/s 2 . 2. We apply Newton’s second law (Eq. 5-1 or, equivalently, Eq. 5-2). The net force applied on the chopping block is F F F net 1 2 , where the vector addition is done using unit-vector notation. The acceleration of the block is given by a F F m 1 2 d i / . (a) In the first case 1 2 ? ? 3.0N i 4.0N j 3.0N i 4.0N j 0 F F so a 0 . (b) In the second case, the acceleration a equals 2 ? [(3 3) i (4 4) j] N ˆ 3.2 m/s j 2.5 kg a (c) In this final situation, a is 2 ? [(3 3) i (4 4) j] N ˆ 2.4 m/s i 2.5 kg a 3. We apply Newton’s second law (specifically, Eq. 5-2). (a) We find the x component of the force is 2 cos 20.0 1.00kg 2.00m/s cos 20.0 1.88N. x x F ma ma    

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CHAPTER 5 172 (b) The y component of the force is 2 sin 20.0 1.0kg 2.00m/s sin 20.0 0.684N. y y F ma ma     (c) In unit-vector notation, the force vector is ? ? i j (1.88 N)i (0.684 N)j . x y F F F 4. Since v = constant, we have a = 0, which implies F F F ma net 1 2 0 . Thus, the other force must be 2 1 ? ( 2 N) i (6 N) j. F F     5. The net force applied on the chopping block is F F F F net 1 2 3 , where the vector addition is done using unit-vector notation. The acceleration of the block is given by a F F F m 1 2 3 d i / . (a) The forces exerted by the three astronauts can be expressed in unit-vector notation as follows: 1 2 3 ? ? ˆ (32 N) cos 30 i sin 30 (27.7 N)i (16 N)j j ? ˆ (55 N) cos 0 i sin 0 (55 N)i j ? ? ˆ (41 N) cos 60 i sin 60 (20.5 N)i (35.5 N)j. j F F F     The resultant acceleration of the asteroid of mass m = 120 kg is therefore 2 2 ? ? ? 27.7i 16 j N 55i N 20.5i 35.5j N ? (0.86m/s )i (0.16m/s )j . 120 kg a (b) The magnitude of the acceleration vector is 2 2 2 2 2 2 2 (0.86 m/s ) 0.16 m/s 0.88 m/s . x y a a a (c) The vector a makes an angle with the + x axis, where
173 2 1 1 2 0.16 m/s tan tan 11 . 0.86 m/s y x a a   6. Since the tire remains stationary, by Newton’s second law, the net force must be zero: net 0.

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