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

Info iconThis preview shows pages 1–8. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 + +
Background image of page 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 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 .
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 ^ .
Background image of page 4
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 °− °=− ° .
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 6
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 .
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 8
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 05/11/2010 for the course PHYS 2101,2102 taught by Professor Giammanco during the Spring '10 term at LSU.

Page1 / 104

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

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online