Solucionario Fisica

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 [pic]. Alternatively, the law of cosines may be used as
[pic]
from which [pic], and the forces are perpendicular. c) For the sum to have 0 magnitude, the forces must beantiparallel, and the angle between them is [pic].


4.2: In the new coordinates, the 120-N force acts at an angle of [pic] from the [pic]-axis, or [pic] from the [pic]-axis, and the 50-N force acts at an angle of [pic] from the [pic]-axis.

a) The components of the net force are
[pic]
[pic]
b) [pic] [pic]. The results have the same magnitude, and the angle has beenchanged by the amount [pic] that the coordinates have been rotated.

4.3: The horizontal component of the force is [pic] to the right and the vertical component is [pic] down.

4.4: a) [pic] where [pic] is the angle that the rope makes with the ramp ([pic] in this problem), so [pic]
b) [pic]


4.5: Of the many ways to do this problem, two are presented here.
Geometric: From thelaw of cosines, the magnitude of the resultant is
[pic]
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
[pic]
Components: Taking the [pic]-direction to be along dog A’s rope, the components of the resultant are
[pic]
[pic]
so [pic]


4.6: a) [pic]
[pic]
b) [pic]

4.7: [pic](to two places).


4.8: [pic]


4.9: [pic]


4.10: a) The acceleration is [pic]. The mass is then [pic]
b) The speed at the end of the first 5.00 seconds is [pic], and the block on the frictionless surface will continue to move at this speed, so it will move another [pic] in the next 5.00 s.


4.11: a) During the first 2.00 s, the acceleration of the puck is [pic] (keeping an extrafigure). At [pic], the speed is [pic] and the position is [pic]. b) The acceleration during this period is also [pic], and the speed at 7.00 s is [pic]. The position at [pic] is [pic], and at [pic] is
[pic]
or 21.9 m to three places.


4.12: a) [pic]
b) With [pic].
c) With [pic].


4.13: a) [pic]
b), c), d)
[pic]



4.14: a) With [pic],
[pic]b) [pic]. Note that this time is also the distance divided by the average speed.
c) [pic]


4.15: [pic]


4.16: [pic]


4.17: a)[pic] b) The mass is the same, 4.49 kg, and the weight is [pic]


4.18: a) From Eq. (4.9), [pic]
b) [pic]


4.19: [pic] 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 forceon 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. [pic] The passenger’s acceleration is[pic], downward.


4.23:[pic]

4.24: (a)Each crate can be considered a single particle:
[pic]
[pic] (the force on [pic] due to [pic]) and [pic] (the force on [pic] due to [pic]) 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...