Solutions
Chapter 1
Problems:
1-12
Bacteria vary somewhat in size, but a diameter of 2.0μm is not
unusual. What would be the volume (in cubic centimeters) and
surface area (in square millimeters) of such a bacterium, assuming
that it is spherical?
Solution:
From Appendix A, the volume V of a sphere is given in terms of its
radius as
V 
4 3
r
3
while its surface area A is given as
A  4r 2
The necessary equalities for this problem are:
r
d
 1m
2
1m  10 6 m
1cm  10  2 m
3
1mm  10 3 m
6
4 3 4
3  10 m   1cm 
  2   4.23 1012 cm3
V  r   (1m) 
3
3
 1m   10 m 
6
2
2  10 m 

A  4r  4 (1m) 
 1m 
3
2
2
 1mm 
5
2
 3   1.33  10 mm
 10 m 
1-23:
A brass washer has an outside diameter of 4.50cm with a hole of
diameter 1.25cm and is 1.50mm thick. The density of brass is
8600kg/m3. If you put this washer on a laboratory balance, what it will
weight in grams?
Solution:
The mass can be calculated as the
product of the density and volume. The
volume of the washer is the volume Vd of
a solid disk of radius rd minus the volume
Vh of the disk-shaped hole of radius rh .
In general, the volume of a disk of radius
r and thickness t is πr2t. We also need
to apply the unit conversions
1 m3  106 cm3 and
1 g/cm3  103 kg/m3 .




V  Vd  Vh   rd2  rh2 t   2.25cm  0.625cm 0.150cm  2.20cm3
2
2
The density of the washer material is:
3
3
3
3
3
ρ= 8600 kg/m 1 g/cm  / 10 kg/m   8.60 g/cm .
The mass of the washer is:
mass   density  volume   8.60 g/cm3  2.20 cm3   18.9 g.
1-35:
A runner jogs around a circular track 150ft in diameter. (a) Clearly
sketch this runner’s displacement vector one he has completed oneforth of a lap; one-half of a lap; one lap. (b) Find the magnitude and
direction of the runner’s displacement vector for each case in part (a).
Solution:
The displacement vector d is directed from the initial position of the
object to the final position. In each case, its magnitude d is the length
of the line that connects points 1 and 2.
(a) The initial position (point 1), the final position (point 2), the
displacement vector
each case.
2
d
and its direction φ is shown in Figure. for
d
d
1
Quarter lap
2
•
1
Half lap
1, 2
Full lap
(b) Quarter lap: The magnitude is
d  r 2  r 2  2r  2  75 ft   106 ft.
From Figure
tan 
r
 1.0
r
Half lap: The magnitude is
or
  45
d  2r  150 ft. The direction is φ =1800
Full lap: The final position equals the initial position. The
displacement is therefore equal to zero and the direction is undefined.
Reflect: In each case, the magnitude of the displacement is less than
the distance traveled.
1-47:
Two vectors of equal magnitude act in a vertical plane perpendicular
to each other. If their resultant is 75 N directly downward, (a) sketch
these two vectors and the resultant, and (b) use components to find
the magnitude of each of the two vectors and the angle each makes
with the vertical.
Solution:
Use coordinates for which the  y axis is downward. Let the two

and B and let the angles they make on either side

vectors be A
  
of the  y axis be φA and φB . Since the resultant, R  A  B is
in the  y direction, Rx  0. The two vectors and their components
are shown in Figure(a).
Ax
Bx
x
A
φA
A
Ay
φB
450
R
B
By
450
(a)
(b)
y
(a) Rx  Ax  Bx   Asin  A  B sin B
Since A  B and Rx  0, we can conclude that
0
and  A  B . Similarly, since  A   B  90 ,
sin  A  sin B
 A   B  450
  
The vector addition diagram for R  A  B is given in Figure b.
(b) Ay  By  Ry and Ry  75 N. Acos A  B cosB  75N
AB
0
and  A  B  45
A  53 N.
so
2 A cos 450  75 N
and
1-63
A patient with a dislocated shoulder is put into a traction apparatus as


shown in Figure. The pulls A and B have equal magnitudes and
must combine to produce an outward traction force of 5.60 N on the
patient’s arm. How large should these pulls be?
Solution:
Use coordinates having a horizontal
x
axis and an upward  y
axis. Then Rx  5.60 N.
Acos 320  B sin 320  Rx
Ax  Bx  Rx and
Since A  B,
and
A
2 Acos 320  Rx
Rx
0
2 cos 32
 3.30 N