4.1: a) For the magnitude of the sum to be the sum of the magnitudes, the forces must be parallel, and the angle between them is zero. b) The forces form the sides of a right isosceles triangle, and the angle between them is 90 . Alternatively, the law of cosines may be used as   ,cos22 2 2 22  FFFF  from which cos 0   , and the forces are perpendicular. c) For the sum to have 0 magnitude, the forces must be antiparallel, and the angle between them is 180 . 4.2: In the new coordinates, the 120-N force acts at an angle of 53 from the x -axis, or 233 from the x -axis, and the 50-N force acts at an angle of 323 from the x - axis. a) The components of the net force are N32323cos)N50(233cos)N120(  x R .N124323sin)N50(233sin)N120()N250(  y R b) ,N128 22  yx RRR     104arctan 32 124 . The results have the same magnitude, and the angle has been changed by the amount )37(  that the coordinates have been rotated. 4.3: The horizontal component of the force is N1.745cos)N10(  to the right and the vertical component is N1.745sin)N10(  down. 4.4: a) ,cos  FF x  where  is the angle that the rope makes with the ramp (  30θ in this problem), so .N3.69 30cos N0.60 cos    x F F F  b) N.6.34tansin  θFθFF xy 4.5: Of the many ways to do this problem, two are presented here. Geometric: From the law of cosines, the magnitude of the resultant is .N49460cos)N300)(N270(2)N300()N270( 22 R The angle between the resultant and dog A’s rope (the angle opposite the side corresponding to the 250-N force in a vector diagram) is then   .7.31 N494 )N300(120sin arcsin           Components: Taking the x -direction to be along dog A’s rope, the components of the resultant are N42060cos)N300()N270(  x R ,N8.25960sin)N300(  y R so   .7.31arctan,N494)N8.259()N420( 420 8.259 22  θR 4.6: a) N10.8)9.126(cos)N00.6(120cos)N00.9( 21  xx FF N.00.3)9.126(sin)N00.6(120sin)N00.9( 21  yy FF b) N.8.64N)(3.00N)10.8( 2222  yx RRR 4.7: 2 s/m2.2kg)(60N)132(/  /mFa (to two places). 4.8: .N189)m/skg)(1.40135( 2  maF 4.9: kg.16.00)m/sN)/(3.000.48(/ 2  aFm 4.10: a) The acceleration is 2 s)(5.00 )m0.11(2 2 s/m88.0 22  t x a . The mass is then kg.9.90 2 m/s0.88 N0.80  a F m b) The speed at the end of the first 5.00 seconds is m/s4.4at , and the block on the frictionless surface will continue to move at this speed, so it will move another m0.22vt in the next 5.00 s. 4.11: a) During the first 2.00 s, the acceleration of the puck is 2 m/s563.1/ mF (keeping an extra figure). At s00.2t , the speed is m/s13.3at and the position is m13.32/2/ 2  vtat . b) The acceleration during this period is also 2 m/s563.1 , and the speed at 7.00 s is m/s6.26s)00.2)(m/s(1.563m/s13.3 2  . The position at s00.5t is m125s)2.00sm/s)(5.00(3.13m13.3 x , and at s00.7t is m,21.89s))(2.00m/s3(1/2)(1.56s)m/s)(2.00(3.13m12.5 22  or 21.9 m to three places. 4.12: a) .m/s31.4kg5.32/N140/ 2  mFa x b) With m215,0 2 2 1 0  atxv x . c) With m/s0.43/2,0 0  txtavv xxx . 4.13: a) 0 F  b), c), d) 4.14: a) With 0 0  x v , .m/s1050.2 )m1080.1(2 )m/s1000.3( 2 214 2 262      x v a x x b) s1020.1 8 s/m1050.2 s/m1000.3 214 6     x x a v t . Note that this time is also the distance divided by the average speed. c) N.1028.2)m/s1050.2)(kg1011.9( 1621431   maF 4.15: .N1094.2)m/s80.9)(m/s12)(N2400()/( 322  gawmaF 4.16: .m/s0.22)m/s80.9( 2.71 160 / 22         g w F gw F m F a 4.17: a) kg49.4)m/s80.9/()N0.44(/ 2  gwm b) The mass is the same, 4.49 kg, and the weight is .N13.8)m/s81.1)(kg49.4( 2  4.18: a) From Eq. (4.9), kg.327.0)s/m80.9/()N20.3(/ 2  gwm b) N.137)s/m80.9)(kg0.14( 2  mgw 4.19: N.825)s/m15)(kg55( 2  maF The net forward force on the sprinter is exerted by the blocks. (The sprinter exerts a backward force on the blocks.) 4.20: a) the earth (gravity) b) 4 N, the book c) no d) 4 N, the earth, the book, up e) 4 N, the hand, the book, down f) second g) third h) no i) no j) yes k) yes l) one (gravity) m) no 4.21: a) When air resistance is not neglected, the net force on the bottle is the weight of the bottle plus the force of air resistance. b) The bottle exerts an upward force on the earth, and a downward force on the air. 4.22: The reaction to the upward normal force on the passenger is the downward normal force, also of magnitude 620 N, that the passenger exerts on the floor. The reaction to the passenger’s weight is the gravitational force that the passenger exerts on the earth, upward and also of magnitude 650 N. .m/s452.0 2 m/s/9.80N650 N650N620 2    m F The passenger’s acceleration is 2 s/m452.0 , downward. 4.23: .s/m104.7 )kg100.6( )s/m80.9)(kg45( 223 24 2 EE E     m mg m F a 4.24: (a) Each crate can be considered a single particle: AB F (the force on A m due to B m ) and BA F (the force on B m due to A m ) form an action-reaction pair. (b) Since there is no horizontal force opposing F, any value of F, no matter how small, will cause the crates to accelerate to the right. The weight of the two crates acts at a right angle to the horizontal, and is in any case balanced by the upward force of the surface on them. 4.25: The ball must accelerate eastward with the same acceleration as the train. There must be an eastward component of the tension to provide this acceleration, so the ball hangs at an angle relative to the vertical. The net force on the ball is not zero. 4.26: The box can be considered a single particle. For the truck: The box’s friction force on the truck bed and the truck bed’s friction force on the box form an action-reaction pair. There would also be some small air-resistance force action to the left, presumably negligible at this speed. 4.27: a) b) For the chair, 0 y a so yy maF   gives 037sin  Fmgn N142n 4.28: a) b)  sinmgT  N2790.26sin)s/m80.9)(kg0.65( 2  4.29: tricycle and Frank T is the force exerted by the rope and g f is the force the ground exerts on the tricycle. spot and the wagon T  is the force exerted by the rope. T and T  form a third-law action-reaction pair, .TT    4.30: a) The stopping time is .s1043.7 4 s/m350 )m130.0(2 )2/( 0   v x v x ave b) .N848)kg1080.1( s)10(7.43 )s/m350( 3 4-    maF (Using xva 2/ 2 0  gives the same result.) 4.31: Take the x -direction to be along 1 F  and the y -direction to be along R  . Then N1300 2  x F and N1300 2  y F , so N1838 2 F , at an angle of 135 from 1 F  . 4.32: Get g on X: 2 2 1 gty  2 )s2.2( 2 1 m0.10 g 2 s/m13.4g N41.0)s/m03.4)(kg100.0( 2 XX  mgw 4.33: a) The resultant must have no y-component, and so the child must push with a force with y-component N.6.1660sinN)100(30sin)N140(  For the child to exert the smallest possible force, that force will have no x-component, so the smallest possible force has magnitude 16.6 N and is at an angle of 270 , or 90 clockwise from the x -direction. b) .N840)s/m80.9)(kg6.85(.kg6.85 2 s/m0.2 30cosN14060cosN100 2     mgwm a F . 4.34: The ship would go a distance m,25.506 )N100.8(2 )s/m5.1)(kg106.3( 2)/(22 4 272 0 2 0 2 0     F mv mF v a v so the ship would hit the reef. The speed when the tanker hits the reef is also found from m/s,17.0 )kg106.3( )m500)(N100.8(2 m/s)5.1()/2( 7 4 22 0     mFxvv so the oil should be safe. 4.35: a) Motion after he leaves the floor: ).(2 0 2 0 2 yyavv yyy  0 y v at the maximum height, 2 0 m/s80.9,m2.1  y ayy , so m/s.85.4 0  y v b) .m/s2.16)s300.0/()s/m85.4(/ 2 av  tva c) avav mawF  )s/m2.16)(s/m80.9/N890(N890 22 avav  mawF N1036.2 3 av F 4.36: N.107.3 )m108.1(2 )s/m5.12( kg)850( 2 6 2 2 2 0     x v mmaF 4.37: a) (upward) net mgFF  b) When the upward force has its maximum magnitude max F (the breaking strength), the net upward force will be max F mg and the upward acceleration will be .s/m83.5s/m80.9 kg80.4 N0.75 22 maxmax    g m F m mgF a 4.38: a) N539 mgw b) Downward velocity is decreasing so a  is upward and the net force should be upward. mgF  air , so the net force is upward. c) Taking the upward direction as positive, the acceleration is .s/m47.1s/m80.9 kg0.55 N620 s/m80.9 222 airair    m F m mgF m F a [...]... )(0.025 s)  (2.40  10 5 m / s 3 )(0.025 s) 2  3.0  10 2 m / s c) The acceleration as a function of time is a (t )  1.80  104 m / s 2  (4.80  105 m / s 3 ) t , so (i) at t  0, a  1.8  10 4 m / s 2 , and (ii) a (0.025 s)  6.0  103 m / s 2 , and the forces are (i) ma  2.7  104 N and (ii) ma  9.0  103 N 4.42: a) The velocity of the spacecraft is downward When it is slowing down, the acceleration . Differentiating, the velocity as a function of time is so,)s/m 1040 .2()s/m1080.1()( 23524 tttv  23524 )s025.0)(s/m 1040 .2()s025.0)(s/m1080.1()s025.0( v s./m100.3.    m F The passenger’s acceleration is 2 s/m452.0 , downward. 4.23: .s/m 104. 7 )kg100.6( )s/m80.9)(kg45( 223 24 2 EE E     m mg m F a 4.24: (a)