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Chapter05

# Chapter05 - 1 We apply Newton's second law(specifically Eq...

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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. xx Fm am a = = °= (b) The y component of the force is 2 sin 20.0 1.0kg 2.00m/s sin 20.0 0.684N. yy a == ° = ° = (c) In unit-vector notation, the force vector is ˆˆ ˆ ˆ i j (1.88 N)i (0.684 N)j . xy FF F =+= + G

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2. We apply Newton’s second law (Eq. 5-1 or, equivalently, Eq. 5-2). The net force applied on the chopping block is G G G FF F net =+ 12 , where the vector addition is done using unit-vector notation. The acceleration of the block is given by G G G aFFm di /. (a) In the first case () ( ) ( ) ˆˆ ˆ ˆ 3.0N i 4.0N j 3.0N i 4.0N j 0 ªº ª º += + +− + = ¬¼ ¬ ¼ GG so G a = 0 . (b) In the second case, the acceleration G a equals ( ) 2 3.0N i 4.0N j 3.0N i 4.0N j ˆ (4.0m/s )j. 2.0kg m ++ + + == (c) In this final situation, G a is ( ) 2 ˆ ˆ 3.0N i 4.0N j 3.0N i 4.0N j ˆ (3.0m/s )i. 2.0 kg m + +
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). G G a F m == ∠°+ ° =∠ ° 9 0 0 8 0 118 30 29 53 .. . . bg b g b g Therefore, the acceleration has a magnitude of 2.9 m/s 2 .

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4. We note that ma = (–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 = F 1 F 2 =(–34 N) i ^ (12 N) j ^ .
5. We denote the two forces G G FF 12 and . According to Newton’s second law, G G G G G G FFm a Fm aF 2 1 += =s o ,. (a) In unit vector notation G F 1 20 0 = . # Ni b g and () ( ) ( ) ( ) 22 2 2 ˆˆ ˆ 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 =− ° ° = − G Therefore, ( ) 2 ˆ 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 + G (b) The magnitude of G F 2 is 2 2 2 | | ( 32.0 N) ( 20.8 N) 38.2 N. xy F =+ = + = G (c) The angle that G 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 °− °=− ° .

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G G G G FF F m a net =+= = 12 0. Thus, the other force must be 21 ˆˆ (2 N ) i (6 N ) j . =− = − + GG 6. Since G v = constant, we have G a = 0, which implies
7. The net force applied on the chopping block is G G G G FF F F net =+ + 123 , where the vector addition is done using unit-vector notation. The acceleration of the block is given by G G G G aFFFm =++ di /. (a) The forces exerted by the three astronauts can be expressed in unit-vector notation as follows: () 1 2 3 ˆˆ ˆ 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 ˆ (41 N) cos 60 i sin 60 (20.5 N)i (35.5 N)j. j F F F + ° = + + ° = =− ° + ° = G G G The resultant acceleration of the asteroid of mass m = 120 kg is therefore ( ) ( ) 22 ˆ ˆ ˆ 27.7i 16j N 55i N 20.5i 35.5j N (0.86m/s )i (0.16m/s )j . 120kg a ++ + == G (b) The magnitude of the acceleration vector is 2 2 2 2 2 (0.86 m/s ) 0.16 m/s 0.88 m/s .

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Chapter05 - 1 We apply Newton's second law(specifically Eq...

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