PHYSICS 112 SUMMER 2005 Homework for June 14, 2005
Due in recitation during the first five minutes; after that, worth only 50% as
much.
Name
Section
In parts (1a) through (1c), consider the two dimensionless vectors
t œ (3.10, 2.40) and F
t œ ( 3.60, 1.70).
E
t.
(1a) Evaluate E, the magnitude of E
E œ |Et| œ È(3.10)# (2.40)# œ È15.37 œ 3.92
t.
(1b) Evaluate F , the magnitude of F
F œ |Ft | œ È( 3.60)# (1.70)# œ È15.85 œ 3.98
(1c) Sketch and label these two vectors on an appropriately-labeled coordinate system
below, drawing them approximately to scale. Label the four quadrants in the usual way.
Name the quadrant in which each vector is found.
Et is in the first quadrant and Ft is in the second quadrant.
t in the BC plane has a magnitude of 6.40 N (6.40 newtons) and
(2a) Suppose the vector J
lies in the third quadrant, making an angle of 60° with the negative B axis. Make a sketch
of the coordinate system and draw the vector, showing the 60° angle. Then determine the
components of this vector and express the vector in the form (B componet, C component).
The vector is shown in the diagram.
Since it lies in the third quadrant, it must have
an B-component JB that is negative, and a Ccomponent JC that is negative. Jt is at an angle
) œ 240° (measured counterclockwise from the
positive B-axis).
Using trigonometry we see that
JB œ J cos (240°) œ (6.40 N)( 0.5000) œ 3.20 N
and
JC œ J sin (240°) œ (6.40 N)( 0.8660) œ 5.54 N
Thus Jt œ (JB , JC ) œ ( 3.20 N, 5.54 N).
(2b) Verify that your vector expression in part (2a) has the correct magnitude.
We expect to find J œ |Jt | œ ÉJB# JC# œ 6.40 N. Let's check it out.
J œ È(3.20 N)# (5.54 N)# œ È(10.24 N# ) (30.69 N# ) œ È40.93 N# œ 6.40 N.
(3) Consider these two vectors:
t has a length of 7.4 m and is directed 31° N of E
E
t has a length of 5.8 m and is directed 67° S of W
F
tF
t . Draw a sketch to
Determine the components, magnitude, and direction of E
accompany your solution.
The easiest way to do this is to determine the components of the two vectors. From the
sketch we expect Et to have positive components and Ft to have negative components, as shown
in the sketch. Calculating the components:
Et œ (7.4 m)(cos 31°), (7.4 m)(sin 31°) œ 6.3 mß 3.8 m
Ft œ (5.8 m)(cos 247°)ß (5.8 m)(sin 247°) œ Ð 2.3 mß 5.3 mÑ
Then Et Ft œ (8.6 mß 9.1 m)
whose magnitude is |Et Ft | œ È(8.6 m)# (9.1 m)# œ È156.77 m# œ 12.5 m which should
probably be written as 13 m, to 2 significant figures.
m
"
The direction of Et Ft is at an angle ) œ tan" ( 9.1
(1.06) œ 47°. This is
8.6 m ) œ tan
in the first quadrant, as seems correct when you check the sketch for the vectors Et and Ft .
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