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# Chapter-05 - Chapter 5 1 We apply Newtons second...

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175 Chapter 5 1. 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 = = °= °= (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 = + = + r 2. We apply Newton’s second law (Eq. 5-1 or, equivalently, Eq. 5-2). The net force applied on the chopping block is r r r F F F net = + 1 2 , where the vector addition is done using unit-vector notation. The acceleration of the block is given by r r r a F F m = + 1 2 d i / . (a) In the first case ( 29 ( 29 ( 29 ( 29 1 2 ? ? 3.0N i 4.0N j 3.0N i 4.0N j 0 F F = + + - + - = r r so r a = 0 . (b) In the second case, the acceleration r a equals ( 29 ( 29 ( ( 29 ( 29 ( 2 1 2 ? ? 3.0N i 4.0N j 3.0N i 4.0N j ˆ (4.0m/s )j. 2.0kg F F m + + - + + = = r r (c) In this final situation, r a is ( 29 ( 29 ( ( 29 ( 29 ( 2 1 2 ? ? 3.0N i 4.0N j 3.0N i 4.0N j ˆ (3.0m/s )i. 2.0 kg F F m + + + - + = = r r

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CHAPTER 5 176 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 magnitude-angle notation (with SI units understood). r r 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 . 4. We note that m a = (–16 N) i ^ + (12 N) j ^ . With the other forces as specified in the problem, then Newton’s second law gives the third force as F 3 = m a F 1 F 2 =(–34 N) i ^ - (12 N) j ^ . 5. We denote the two forces r r F F 1 2 and . According to Newton’s second law, r r r r r r F F ma F ma F 1 2 2 1 + = - = so , . (a) In unit vector notation r F 1 20 0 = . \$ N i b g and ( ( ( ( 2 2 2 2 ? ? ˆ 12.0 sin 30.0 m/s i 12.0 cos 30.0 m/s 6.00 m/s i 10.4m/s j. j a = - ° - ° = - - r Therefore, ( 29 ( ( 29 ( ( 29 ( 29 ( 29 2 2 2 ? ? 2.00kg 6.00 m/s i 2.00 kg 10.4 m/s j 20.0 N i ? 32.0 N i 20.8 N j. F = - + - - = - - r (b) The magnitude of r F 2 is 2 2 2 2 2 2 2 | | ( 32.0 N) ( 20.8 N) 38.2 N. x y F F F = + = - + - = r (c) The angle that r F 2 makes with the positive x axis is found from tan θ = ( F 2 y / F 2 x ) = [(–20.8 N)/(–32.0 N)] = 0.656. Consequently, the angle is either 33.0° or 33.0° + 180° = 213°. Since both the x and y components are negative, the correct result is 213°. An alternative answer is 213 360 147 °- ° = - ° . 6. Since r v = constant, we have r a = 0, which implies
177 r r r r F F F ma net = + = = 1 2 0 . Thus, the other force must be 2 1 ? ( 2 N) i (6 N) j. F F = - = - + r r 7. The net force applied on the chopping block is r r r r 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 r r r r 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: ( ( 29 ( 29 ( 29 ( 29 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 = ° + ° = + = ° + ° = = - ° + - ° = - r r r The resultant acceleration of the asteroid of mass m = 120 kg is therefore ( ( ( 2 2 ? ? ?

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Chapter-05 - Chapter 5 1 We apply Newtons second...

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