Preparation problems
3-2. For the vectors shown in Fig. 3-19 (p. 61),
evaluate the vectors (a)[A + B],
(b) [A - B], (c) [A + C],
and (d) [A + B + C].
Solution:
We begin by writing each vector in its component form. This we
accomplish by remembering:
Giving us the vectors:
(a) Having written down each vector in component form,
it is a simple task to add them, as we must simply add their components.
So we get
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|
®
A
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+ |
®
B
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= [Ax + Bx] |
^
i
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+ [Ay + By] |
^
j
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|
|
Plugging in our values, we get
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[(8.19) + (-3.44)] |
^
i
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+ [(5.74) + (-4.91)] |
^
j
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| |
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(b) Here we need to subtract the components of
B from those of A. Being careful with the signs,
we get
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[Ax - Bx] |
^
i
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+ [Ay - By] |
^
j
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| |
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[(8.19) - (-3.44)] |
^
i
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+ [(5.74) - (-4.91)] |
^
j
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(c) Again we are adding the components of the vectors.
Plugging in our values, we get
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[Ax + Cx] |
^
i
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+ [Ay + Cy] |
^
j
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[(8.19) + (-3.38)] |
^
i
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+ [(5.74) + (7.25)] |
^
j
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(d) Finally, we are adding the components of all three
vectors. This gives us
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[Ax + Bx + Cx] |
^
i
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+ [Ay + By + Cy] |
^
j
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| |
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[(8.19) + (-3.44) + (-3.38)] |
^
i
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+ [(5.74) + (-4.91) + (7.25)] |
^
j
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3-9. A direct flight from Orlando, Florida, to Atlanta,
Georgia, covers 660 km and heads at 29° west of north. Your
flight, however, stops at Charleston, South Carolina, on the way to
Atlanta. Charleston is 510 km from Orlando, in a direction 9.3°
east of north. What are the magnitude and direction of the
Charleston-to-Atlanta leg of your flight?
Solution:
If we look at the direct flight to Atlanta and the Orlando-to-Charleston
leg as two vectors (A and C,
respectively), we see that the Charleston-to-Atlanta leg (call it
F) is found by
So we start out by writing each known vector in its component
form. To do this we must determine where we will call 0°, which I
choose as north. So the angle that A makes with this is
qa = -29°, while the vector C makes an
angle qc = 9.3°. Further, we must remember that the
j-direction is north and the
i-direction is east. Then the components of the vectors
become
Giving us the vectors:
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-319.97 km |
^
i
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+ 577.24 km |
^
j
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82.42 km |
^
i
|
+ 503.30 km |
^
j
|
|
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| |
|
Having written down each vector in component form, it is a
simple task to subtract them, as we must simply subtract their
components. Doing this we get
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|
|
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[Ax - Cx] |
^
i
|
+ [Ay - Cy] |
^
j
|
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|
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[(-319.97 km) - (577.24 km)] |
^
i
|
+ [(82.42 km) - (503.30 km)] |
^
j
|
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| |
|
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-402.39 km |
^
i
|
+ 73.94 km |
^
j
|
|
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| |
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Now we must convert this into a magnitude and direction. To
do this, we recall that the magnitude is given by
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| ___________
Ö (Fx)2 + (Fy)2
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æ
Ö
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(-402.39 km)2 + (73.94 km)2
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Finally, we can find the angle made with north by looking at the
resultant vector, and applying a little trigonometry. This tells us
that
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tan-1 |
æ
ç
è
|
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-402.39 km
73.94 km
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ö
÷
ø
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or, qf = 79.6° west of north.
3-23. Let A = 15 i - 40 j
and B = 31 j + 18 k. Find a vector
C such that A + B + C = 0.
Solution:
Since we know
then we immediately see that
Since each vector is already in its component form, we can very quickly
find the vector C:
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|
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- [Ax + Bx] |
^
i
|
- [Ay + By] |
^
j
|
- [Az + Bz] |
^
k
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|
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- [(15) + (0)] |
^
i
|
- [(-40) + (31)] |
^
j
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- [(0) + (18)] |
^
k
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3-28. A biologist studying the motion of bacteria notes
a bacterium at position r1 = 2.2 i + 3.7 j - 1.2 k mm (Note 1 mm = 10-6 m).
After 6.2 s, the bacterium is at
r2 = 4.6 i + 1.9 k mm. What is its average velocity? Express in unit
vector notation and calculate the magnitude.
Solution:
First we must recall from that average velocity is calculated
by
Since we know the components of the initial and final
positions, we can calculate the average velocity using
So all that remains is to find
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[(4.6 mm) - (2.2 mm)] |
^
i
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+ [(0.0 mm) - (3.7 mm)] |
^
j
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+ [(1.9 mm) - (-1.2 mm)] |
^
k
|
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| |
|
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(2.4 mm) |
^
i
|
+ (-3.7 mm) |
^
j
|
+ (3.1 mm) |
^
k
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. |
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|
Having found this, we simply plug into the equation above to get
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(2.4 mm) |
^
i
|
+ (-3.7 mm) |
^
j
|
+ (3.1 mm) |
^
k
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6.2 s
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0.387 |
^
i
|
- 0.597 |
^
j
|
+ 0.5 |
^
k
|
mm/s |
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|
Finally, we find the magnitude in the normal way. That is,
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| _________________
Ö (vx)2 + (vy)2 + (vz)2
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| _________________________
Ö (0.387)2 + (-0.597)2 + (0.5)2
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mm/s |
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Follow-up problems
2-83. You toss a hammer over the 3.7 m high wall of a
construction site, starting your throw at a height of 1.2 m above the
sidewalk (see Fig. 2-22 on page 44). (a) What is the minimum
speed at which you must throw the hammer for it to clear the wall?
(b) Assuming it's thrown with the speed given in part (a),
when will it hit the bottom of the excavation?
Solution:
We start out setting 0 m to be the bottom of the excavation. Then we
know the following information:
(a) In order to determine how fast you must throw the
hammer to get it over the wall, you need to focus only on the first part
of the flight. For this part of the flight, we can find the initial
velocity using the third equation from our constants and equation sheet,
giving us
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vx12 = vx02 + 2 ax (x1 - x0). |
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Solving for vx0, we get
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(0 m/s)2 - 2 (-9.8 m/s2) [(11.6 m) - (9.1 m)] |
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(b) We now need to find the time it takes for the
hammer to travel over the wall and down to the bottom of the excavation.
To do this, we can split the flight into two portions and add the time
of flight for each portion. The first portion will be the trip up to
the top of the wall. To find that time, which we can call t1, we can
use the equation
Solving for t1, we get
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(0 m/s) - (7 m/s)
-9.8 m/s2
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Now we can find t2, the time it takes for the hammer to fall from
the top of the wall. To do this we will use the first equation from our constants and
equation sheet, remembering that our initial conditions are at the top
of the wall:
|
x = x1 + vx1 t2 + |
1
2
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ax t22 . |
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Noting that vx1 = 0 m/s, we can solve for t2, getting
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2 [(0 m) - (11.6 m)]
-9.8 m/s2
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Putting these two answers together, we see the total time of flight is
2-91. The position of a particle as a function of time
is given by x = x0 sin(wt), where x0 and w are
constants. (a) Take derivatives to find expressions for the
velocity and acceleration. (Hint: consult the table of
derivatives in Appendix A) (b) What are the maximum values of
velocity and acceleration?
Solution:
(a) We know that the velocity is the derivative of the
position, or
|
v = |
dx
dt
|
= |
d
dt
|
[ x0 sin(wt) ]. |
|
Taking the derivative as shown in Appendix A, we get
We know that the velocity is the derivative of the position,
or
|
a = |
dv
dt
|
= |
d
dt
|
[ wx0 cos(wt) ]. |
|
Taking the derivative as shown in Appendix A, we get
(b) To find the maximum values of the velocity and
acceleration, we need to remember that the sine and cosine functions
have maximum and minimum values of 1 and -1, respectively. Once we
realize this, it is simple to substutute in the value necessary to get
the maximum velocity or acceleration.
For the maximum velocity, we then get
For the maximum acceleration, we get
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amax = - w2 x0 [-1] = w2 x0 . |
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Solutions translated from TEX by TTH, version 1.57.
