Physics I
Homework III
Chapter 3: 8, 13, 20, 24, 32, 36, 48
CJ
3.8. Visualize:
Solve: Vector E points to the left and up, so the components Ex and Ey are negative and positive, respectively,
according to the Tactics Box 3.1.
(a)
(b)
Ex  E cos
Ex   E sin
and
and
Ey  E sin .
Ey  E cos .
Assess: Note that the role of sine and cosine are reversed because we are using a different angle.  and  are
complementary angles.
3.13. Visualize:
Solve: (a) Using the formulas for the magnitude and direction of a vector, we have:
A  (4)2  (6)2  7.21
6
4
  tan1  tan1 1.5  56.3
 ry 
1  80 m 
  tan  50 m   58.0
r
 x
40
  tan1  tan1 2  63.4
20
1 2
  tan
 tan1 0.33  18.4
6
  tan1 
(b)
r  (50 m)2  (80 m)2  94.3 m
(c)
v  (20 m/s)2  (40 m/s)2  44.7 m/s
(d)
a  (2 m/s2 )2  (6 m/s2 )2  6.3 m/s2
Assess: Note that the angle  made with the x-axis is smaller than 45 whenever | Ey |  | Ex | ,   45 for | Ey |  | Ex |, and
  45 for | Ey |  | Ex | . But, of course,  in part (d) is with the y-axis, where the opposite of this rule applies.
3.20. Solve: A different coordinate system can only mean a different orientation of the grid and a different origin of
the grid.
(a) False, because the size of a vector is fixed.
(b) False, because the direction of a vector in space is independent of any coordinate system.
(c) True, because the orientation of the vector relative to the axes can be different.
3.24. Visualize:
Solve : (a)  E  tan1  11   45
 F  tan1  21   63.4
Thus  180    E  F  71.6
(b) From the figure, E  2 and F  5. Using
G 2  E 2  F 2  2EF cos  ( 2 )2  ( 5)2  2( 2 )( 5) cos(180  71.6)
 G  3.00.
sin sin(180  71.6)
Furthermore, using

   45
2.975
5
Since  E  45 , the angle made by the vector G with the x-axis is G  (   E )  45  45  90.
(c) We have
Ex  1.0, and Ey  1.0
Fx  1.0, and Fy  2.0
Gx  0.0, and Gy  3.0
G 
 0.0    3.0 
2
2
 3.0, and   tan 1
|Gy |
 3.0 
 tan 1 
  90
|Gx |
 0.0 
That is, the vector G makes an angle of 90 with the x-axis.
Assess: The graphical solution and the vector solution give the same answer within the given significance of figures.
3.32. Solve: We have r  (5iˆ  4 ˆj )t 2 m. This means that r does not change the ratio of its components as t increases,
that is, the direction of r is constant. The magnitude of r is given by r  (5t 2 )2  (4t 2 )2 m  (6.40t 2 ) m.
(a) The particle’s distance from the origin at t = 0 s, t = 2 s, and t = 5 s is 0 m, 25.6 m, and 160 m.
dr
dt 2
 (5iˆ  4 ˆj )
m/s  (5iˆ  4 ˆj )2t m/s  (10iˆ  8 ˆj )t m/s
(b) The particle’s velocity is v 
dt
dt
(c) The magnitude of the particle’s velocity is given by v  (10 t )2  (8t )2  12.8t m/s. The particle’s speed at
t = 0 s, t = 2 s, and t = 5 s is 0 m/s, 25.6 m/s , and 64.0 m/s .
3.36. Visualize:
(a) The figure shows Sparky’s individual displacements and his net displacement.
Solve: (b) Dnet  D1  D2  D3 , where individual displacements are
D1  (50cos45iˆ  50sin45 ˆj ) m  (35.4iˆ  35.4 ˆj ) m
D  70iˆ m
2
D3  20 ˆj m
Thus Sparky’s displacement is Dnet  (34.6iˆ  15.4 ˆj ) m.
(c) As a magnitude and angle,
Dnet  ( Dnet )2x  ( Dnet )2y  37.9 m,
 ( Dnet )y 
  24.0.
 |( Dnet )x | 
 net  tan1 
Sparky’s net displacement is 37.9 m in a direction 24.0 north of west.
3.48. Visualize:
Note that the tilted coordinate system is such that x-axis is down the slope.
Solve: Expressing all three forces in terms of unit vectors, we have F1  (3 N)iˆ, F2  (6 N) ˆj, and F3 
(5 N)sin iˆ  (5 N)cos ˆj.
(a) The component of Fnet parallel to the floor is (Fnet ) x  (F1  F2  F3 ) x  (3 N)iˆ  (5 N)sin30iˆ  (0.5 N) iˆ.
(b) The component of F perpendicular to the floor is (F )  (F  F  F )  (6 N) ˆj  (5 N)cos30 ˆj 
net
net y
1
2
3 y
(1.67 N) ˆj.
(c) The magnitude of Fnet is Fnet  (Fnet ) x  (Fnet ) y  (0.5 N)2  (1.67 N)2  1.74 N. The angle Fnet makes is
  tan1
(Fnet )y
|(Fnet )x |
 1.67 N 
 tan1 
  73.3
 0.5 N 
Fnet is 73.3 above the –x-axis in quadrant II.