ISM_CH05

# ISM_CH05 - Chapter 5 1 We are only concerned with...

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Unformatted text preview: 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 8 0 118 30 2 9 53 . . . . b gb g b g Therefore, the acceleration has a magnitude of 2.9 m/s 2 . 2. We apply Newton’s second law (specifically, Eq. 5-2). (a) We find the x component of the force is ( 29 ( 29 2 cos 20.0 1.00kg 2.00m/s cos 20.0 1.88N. x x F ma ma = = °= °= (b) The y component of the force is ( 29 ( 29 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 (in newtons) is F F F x y = + = + . . . i j i j 188 0 684 3. 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 ( 29 ( 29 ( 29 ( 29 1 2 ˆ ˆ ˆ ˆ 3.0N i 4.0N j 3.0N i 4.0N j F F + = + +-+ -= so a = 0. (b) In the second case, the acceleration a equals 179 CHAPTER 5 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 1 2 ˆ ˆ ˆ ˆ 3.0N i 4.0N j 3.0N i 4.0N j ˆ (4.0m/s )j. 2.0kg F F m + +-+ + = = (c) In this final situation, a is ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 1 2 ˆ ˆ ˆ ˆ 3.0N i 4.0N j 3.0N i 4.0N j ˆ (3.0m/s )i. 2.0 kg F F m + + + -+ = = 4. 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 (in newtons) exerted by the three astronauts can be expressed in unit-vector notation as follows: ( 29 ( 29 ( 29 ( 29 ( 29 1 2 3 ˆ ˆ ˆ ˆ 32 cos 30 i sin 30 27.7i+16 j j ˆ ˆ ˆ 55 cos 0 i sin 0 55i j ˆ ˆ ˆ ˆ 41 cos 60 i sin 60 20.5i 35.5 j. j F F F = ° + ° = = ° + ° = =-° +-° =- The resultant acceleration of the asteroid of mass m = 120 kg is therefore ( 29 ( 29 ( 29 2 2 ˆ ˆ ˆ ˆ ˆ 27.7i 16 j 55i 20.5i 35.5j ˆ ˆ (0.86m/s )i (0.16m/s )j . 120 a + + +-= =- (b) The magnitude of the acceleration vector is a a a x y = + = +-= 2 2 2 2 2 086 016 088 . . . . b g m / s (c) The vector a makes an angle θ with the + x axis, where = F H G I K J =-F H G I K J = -°--tan tan . . . 1 1 016 086 11 a a y x 5. We denote the two forces F F 1 2 and . According to Newton’s second law, F F ma F ma F 1 2 2 1 + =-= so , ....
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ISM_CH05 - Chapter 5 1 We are only concerned with...

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