1

Ch3

3.1 Lazy with sig figs here…more worried about getting plus and minus signs correct.

�� � �26.0 ms �̂ 15.0 m� �̂ ���� � �26.0 m� �̂ � 15.0 ms �̂ ��� � 26.0 m� �̂ � 15.0 ms �̂ ��� � �15.0 ms �̂ 26.0 m� �̂

�� � �26.0 m� �̂ 15.0 ms �̂ �� � �26.0 m� �̂ 15.0 ms �̂ 3.2 Lazy with sig figs here.

3.3 I was lazy with sig figs here as it is unclear how many you can get from picture.

a) �� � �20 �� �̂ and ��� � 50 �� �̂ � 30 �� �̂ b) �� � 20 �� duesouth and ��� � 58.3 �� heading31.0°SofW

c) �� � �30 �� �̂ 40 �� �̂. See next step for

sketch…

d) Two examples are shown at right. The left

example shows �� ��� �� � ,��� while the

right example shows �� ��� �� � ,��� e) To perform component wise addition I like

to stack the equations as shown below. I leave off the units until the final step but I include 0’s to keep

things aligned well. �� � 0�̂ � 20� ̂��� � 50�̂ � 30� ̂�� � �30�̂ 40� ̂Then add the � ̂column and �̂ column separately to find ,��� � -./̂ � 0.1̂ � --. 234 56789:;-<. <°=>?@

22.6°

36.9°

A � BA�B � 13m C � BC���B � 5m

��

�����

,�����

�����,���

2

3.4 Watch out for the wording in this question. We know �� ��� �� � D�� but the problem statement does not give

you ��, ���, &��. If you read carefully the problem statement gives you ��, ���, &D��.

First I would rearrange �� ��� �� � D�� to give �� � D�� � G�� ���H. Then I would find D�� � 5.00�̂ 0�̂ �� � 10. 00�̂ � 17. 32�̂ ��� � �25. 98�̂ � 15. 00�̂ This time we must subtract the final two vectors from the first. You

should find �� � 20. 98�̂ 32. 32�̂ � 38. 53mheading32. 99°EofN

A sketch of �� only is also worthwhile to get a feel for it. That is

drawn at right.

The full graphical vector addition is shown below.

���

��� ���

D��

����

���� ���

D��

This shows �� � D�� � G�� ���H �� � D�� G���H G����H

This shows D�� � �� ��� ��

Watch out!

��� has same magnitude as �� but points opposite direction. ���� has same magnitude as ��� but points opposite direction.

� � B��B � M20. 98N 32. 32N

20. 98

32. 32

O � tanPQ 32. 3220. 98

3

3.4½ �� � �8.00 cos 30.0° �̂ 8.00 sin 30.0° � ̂�� � �6.928�̂ 4.000� ̂

��� � �10.0 sin 40.0° �̂ � 10.0 cos 40.0° � ̂��� � �6.428�̂ � 7.660�̂

The bottom figure at left shows �� ��� � ,���. �� � �6.928�̂ 4.000� ̂��� � �6.428�̂ � 7.660�̂ _________________________________________ ,��� � �0T. TU</̂ � T. <<.1 ̂

, � B,���B � MG0T. TU<H- GT. <<.H-

, � V0WX. TX 0T. 2. , � V0Y0. WX , � 0T. X2X

Notice the magnitude has four sig figs when following sig fig rules

while each component only has three!!!

Calling the angle to the horizontal Z

Z � [7:P0 \ T. <<.0T. TU<]

Z � [7:P0G.. -W2.H Z � 0U. T-° Strictly speaking, there are better ways to determine the digit of uncertainty for functions.

In general, this is discussed in lab courses.

,��� ^ 0T. XU56789:;0U. T°=>?_

���

,���

��

���

��� 40.0°

30.0°

Z

4

3.4¾ Watch out for the wording in this question. We know �� ��� �� � D��. If you read carefully the problem

statement gives you ��, ���, &D��. First I would rearrange �� ��� �� � D�� �� � D�� � G�� ���H

Then I would find D�� � �10. 0�̂ 0�̂ �� � 0�̂ 8.00� ̂��� � 3.000�̂ � 5.196� ̂

�� � G�10. 0�̂H � `G8.00�̂H G3.000�̂ � 5.196�̂Ha �� � G�10. 0�̂H � G3.000�̂ 2.804�̂H b��� � �0T. ../̂ � -. X.21 ̂Pay attention to the sig fig rules and keep the left most column

when adding terms…

b � Bb���B � MG0T. ..H- G-. X.2H-

b � V0<Y. . W. X<- b � V0W<. Y b � 0T. T.

Calling the angle to the horizontal Z

Z � [7:P0 \-. X.20T. ..]

Z � [7:P0G.. -0UWH Z � 0-. 0W° b��� ^ 0T. T56789:;0-. -°=>?@

Figure at right shows the graphical vector addition of b��� � ,��� � Gc��� d���H b��� � ,��� G�c���H G�d���H

Watch out!

��� has same magnitude as �� but points opposite direction. ���� has same magnitude as ��� but points opposite direction.

��

���,���30.0°

���� ����

b��� � ,��� G�c���H G�d���H

,���

5

3.5 Get the equations for each vector in the x and y directions.

For the second leg of the trip use �e � 80 fg� h and �i � 80 sin h.

For the third leg use �e � �� sin 20 and �i � �� cos 20.

Since �� ��� �� � 0 we find that �e �e �e � 0 and �i �i �i � 0.

One way to proceed from here is to rearrange these equations to �e � ��e � �e and �i � ��i � �i.

Then we find 80 cos h � ��e � �e and 80 sin h � ��i � �i.

By squaring each equation and adding them together one can eliminate the unknown angle! This leaves only one

unknown in the equation, in this case C, to solve for. In particular we find

j80 cos hkN j80 sin hkN � j��e � �ekN G��i � �iHN

80N � j�e �ekN G�i �iHN

80N � G80 j�� sin 20kHN G60 j�� cos 20kHN

From here it is a lot of algebra. Notice that you will get a quadratic formula for C. This means you could have 0, 1,

or 2 solutions (see below for how this might work out). In this case I believe the math should give you two solutions

for C (one short and one long) but I haven’t checked it yet... Each of these solutions gives a different choice of

angle for the second leg of the trip.

I thought about this geometrically as

well. The first leg is nothing unusual.

For the second leg, we know only how

far Boanventur walks. That means he

could end up anywhere on the circle that

has a radius equal to the distance he

walked. Finally, we know the last leg

must connect from somewhere on that

circle to the starting point AND be

along the heading stated. In this case

that gives rise to two possible answers.

If the size of the circle was just perfect

for the final heading, there would only

be one answer. This occurs when the

final vector lies tangent to the circle.

Finally, if the 2nd leg distance is too

short, the 3rd leg never intersects the

circle. Remember, we were told the

heading and final position of the 3rd leg

so we are not free to slide it around…only to change the length. In this case the problem makes no sense…or the

weed was really good.

Start

First leg = ��

Other possible 2nd

leg = �N�����

One possible 2nd

leg = �Q�����

3rd leg = �� connects from �Q����� or �N����� to

start

6

3.6 For practice I will assume three sig figs on all numbers and track the sig figs.

a) This first case is solved quickly with a picture and equation 3.2. Consider the

figure at right. One finds �� ∙ ��� � j10.0kj20.0k cos 150° � �173. 2.

b) The case is more easily solved using equation 3.1. I found �� ∙ ���� � �e�e �i�i �m�m �� ∙ ���� � j1kj�3k j�2kj2k j3kj1k �� ∙ ���� � �11. 0

c) First get the vector �� in Cartesian form (i-hats and j-hats) �� � �10.0 sin 30.0° �̂ 10.0 cos 30.0° � ̂�� � �5.000�̂ 8.660�̂ Don' t forget I asked for cn…not c���!!!

�o � ��� � �5.000�̂ 8.660�̂10.0 � �0.5000�̂ 0.8660�̂ d) To get �o � p�p we first need the magnitude � � V�eN �iN �mN � Vj1kN j�2kN j3kN � √14 ^ 3.742 �o � 0.2673�̂ � 0.5345�̂ 0.8018rs

e) I used my sketch to find the angle between �� & ��� was 150°. That said, to practice the process

htu � cosPQ \�� ∙ ����� ] � cosPQ v �173. 2j10.0kj20.0kw � 150. 0° Note: if you didn’t know the angle between the two vectors, the numerator inside the parentheses is found

using �� ∙ ��� � �e�e �i�i.

f) I used hpx � cosPQ `p�∙x���pxa � cosPQ ` PQQ√Qy√Qya � 141. 8°. Note: I got the numerator using the result of part b. The denominator terms were found using the process

similar to part d. Rather than memorizing this equation, hopefully you remember the procedure and feel

comfortable with the equations.

g) You could follow the procedure in the notes or be clever and use �o ∙ rs � cos hm. I found hm � 36. 70°.

150°

��

��� z

{

7

3.7 For practice I will assume three sig figs on all numbers and track the sig figs.

a) The figure is the same mentioned in part a of the previous problem! The magnitude of the

cross-product is given by equation 3.10. One finds �� | ��� � j10.0kj20.0k sin 150° � 100.

Using the right hand rule explained in 3.11 the direction of �� | ��� is into the page. According

to the figure and the wheel of pain, out of the page is rs . Therefore �� | ��� � �100rs .

b) Using equation 3.9 for cross-products gives �� | ��� � G�4.00�̂ 3.00rsH | G17. 32�̂ � 10. 0�̂H �� | ��� � �69. 28�̂ | �̂ 40. 0�̂ | �̂ 51. 96rs | �̂ � 30. 0rs | � ̂�� | ��� � �69. 28G�rsH 0 51. 96j�̂k � 30. 0j��k̂ �� | ��� � 30. 0�̂ 51. 96�̂ 69. 28rs

In the third line above I used the wheel of pain trick and dropped �̂ | �.̂ With experience this can

sometimes be a real time saver. Notice it is important to keep everything in the proper right-to-left order or

you will introduce minus sign errors when doing cross-products. If you wanted to you could express this as

a magnitude times a unit vector like this �� | ��� � 91. 65G0.3273�̂ 0.5669�̂ 0.7559rsH

Notice that in 3D it is challenging to visualize cross-products and crucial to understand the math.

c) When we perform order of operations on �� | G�� ∙ ���H we would first take the dot product. The output of

the dot product is a scalar. One cannot proceed to cross vector �� with a scalar! In �� ∙ G�� | ���H the output

of the parentheses is a vector which then works as an input for the dot product.

150°

��

��� z

{

8

3.8 I got lazy with sig figs. Your final answers might differ in 3rd digit since I rounded to 3 early on.

a) From the wording of the problem statement we find �� ��� �� � 0 or b��� � �c��� � d���. WATCH OUT!

Most common error is to screw up the signs on ��. If you look in the picture you will see �� points up and to

the left. According to the coordinate system shown in the figure we thus know

�� � 0�̂ � 10 sin 20° �̂ 10 cos 20° rs �� � 0�̂ � 3.42�̂ 9.40rs Therefore�c��� is given by �c��� � ./̂ T. 2-1̂ � Y. 2.}n

��� � 20 sin 30° �̂ 20 cos 30° �̂ 0rs ��� � 10.0�̂ 17.32�̂ 0rs Therefore �d��� is given by �d��� � �0.. ./̂ � 0W. T-1̂ .}n

Doing the vector addition b��� � �c��� � d��� �c��� � ./̂ T. 2-1̂ � Y. 2.}n �d��� � �0.. ./̂ � 0W. T-1̂ .}n b��� � �0.. ./̂ � 0T. Y1̂ � Y. 2.}n

It is worth comparing this result to the initial figure. The third displacement is supposed to bring the moth

back to the origin. From the figure we can tell from both the � ̂and rs terms must be negative and probably

the �̂, too. Because the figure is not to scale I choose not to trust it completely for my �̂ estimate. Our result

agrees qualitatively with this observation.

b) The magnitude is � � Vj�10.0kN j�13.9kN j�9.40kN ^ 19.5.

c) The angle to the negative ~-axis is given by �o ∙ G�rsH � cos hm. One finds hm � cosPQ `PjP�.ykQ�.� a ^ 61.2°. Watch out! That extra minus sign inside the inverse cosine function comes from the wording of the

problem (to the negative axis…not the positive axis).

9

3.9

a) I used the Pythagorean theorem to determine the distance from the origin to the point where the guy lines

touch the ground (28. 72m). The vector from D to A goes down (�rs) and out of the page ( �)̂ giving �� � 28. 72�̂ � 20. 0rs �o � 0.8206�̂ � 0.5714rs

b) From D to point B we expect the vector must go down (�rs), to the right ( �̂), and into the page (��)̂. ��� � �28. 72 sin 30° �̂ 28. 72 cos 30° �̂ � 20. 0rs �s � �0.410�̂ 0.711�̂ � 0.571rs

Stop worrying about sig figs now since…it’s my book & I’m lazy.

c) �o � �0.410�̂ � 0.711�̂ � 0.571rs . Same as ��� except go left instead of right.

d) The angle should be the same between any pair of wires by symmetry. The

angle is given by

h � cosPQ \�o ∙ �s1 ∙ 1] � cosPQG�0.0105H ^ 91° WOW! I can’t believe they turned out nearly perpendicular!

e) We are told �� ��� �rs �� 1000�̂ � 0 ��o ��s �rs 6000�o 1000�̂ � 0 �G0.821�̂ � 0.571rsH �G�0.410�̂ 0.711�̂ � 0.571rsH �rs � G2460�̂ 3266�̂ 3426rsH 1000�̂ � 0 j0.821� � 0.410�k�̂ j0.711�k�̂ j�0.571� � 0.571� �krs � 2460�̂ 4266�̂ 3426rs

Comparing �’̂s to �’̂s, �̂’s to �̂’s, and rs’s to rs’s gives three equations. 0.821� � 0.410� � 2460 0.711� � 4266 3426 0.571� 0.571� � �

Solving gives � � 9067N � 9070N. If you noticed I forgot to keep 4 sig figs on my unit vectors (to

avoid intermediate rounding), good eye! This final answer has approximately three sig figs…

30° B

D

28. 72 cos 30° 28. 72 sin 30°

10

3.10

a) ��Q��N � 2D� ̂b) ��Q��� � �2D� ̂c) We know the orange is centered above the gap in the lower set of oranges. We know this gap should occur

halfway towards the orange to the right and halfway to the orange behind. This tells us ��Q��y � �D�̂ D�̂ ~rs

Unfortunately we don’t know how far up the orange is. We do know, however, the distance between the

centers of the oranges is 2D. This tells us B��Q��yB � 2D � VDN DN ~N

Square both sides and solve for ~ to show ~ � √2D. Therefore the vector that describes the distance

between these two oranges is given by ��Q��y � �D�̂ D�̂ 2Drs

��Q��y � 2D \� 12 �̂ 12 �̂ √22 rs]

Notice I wrote the last line as a magnitude times a unit vector. This style is useful in other classes.

3.11 Let �e � 20 and �e � �30. We find the following equations

From equation 3.1 �� ∙ ��� � �1000 � �600 �i�i or �i � Py��u�

From equation 3.2 �� ∙ ��� � �1000 � �� cos 153.44° or �� � 1118. 0

From equation 3.4

� � M400 �iN

� � M900 �iN

Plugging in the last two equations into the second gives

M400 �iNM900 �iN � 1118. 0

Squaring both sides gives G400 �iNHG900 �iNH � 1249900

Plugging the result of the first equation for �i gives \400 160000�iN ] G900 �iNH � 1249900

Here I choose to multiply both sides by �iN to get rid of fractions. I bring the �iN inside the first paren’s to

clean things up.

�iN \400 160000�iN ] G900 �iNH � 1249900�iN

G400�iN 160000HG900 �iNH � 1249900�iN

Doing some algebra simplifies this down to a quadratic equation in �iN: �iy � 1824. 75�iN 360000 � 0

�iN � �G�1824. 75H ± MG�1824. 75HN � 4j1kj360000k2j1k

From there you find �i � ±40or ± 15. I made an Excel spreadsheet to check all four cases and found the

problem works when �i � ±40 giving �i � ∓10. To be clear, when �i is positive, �i is negative. The

magnitudes in each case are � � 22.36 and � � 50. Similarly when �i � ±15 giving �i � ∓26.67 giving � � 33.33 and � � 33.54. I sketched all four vectors roughly to scale to verify the angle between appeared to

be approximately 153.44° and all four cases looked good.

11

3.12 Lazy with sig figs on this one…Only one solution is possible! From the problem statement we have

From equation 3.2 �� ∙ ��� � �6 � 5� cos h

From equation 3.10 B�� | ���B � 8 � 5� sin h

Taking the ratio of the first two equations gives tan h � �1.333. Plugging into your

calculator gives -53.1. WATCH OUT! Your calculator only gives an answer in the first

or fourth quadrants. Another possible solution occurs in the 2nd quadrant by adding

180°. The two possible angles are -53.13° or 126.9°. Using the either of the equations

above to solve for B shows -53.13° gives a negative value for the magnitude of ��� which

is impossible! Therefore only 126.9° is a valid solution with � � 2. You should find �i � ±1.323. If �i � 1.323 then �i � �4.984 and �e � 0.395 which agrees with � � 5 form the problem statement.

Now we know �� � �4.984�̂ 0.395� ̂and ��� � 1.5�̂ 1.323� ̂which looks like the

figure shown at right. Note: using the right hand rule the cross-product is out of the page

as expected.

3.13

a) If �� is anti-aligned with ���, (or aligned with – �̂), it has max energy. Think: if you try to purposefully

misalign magnets they have a lot of stored energy and will flip over as soon as you release them!

b) If �� is aligned with ���, (or aligned with �̂), it has min energy.

c) ∆� � ���e � ���� � ���j– �̂k ∙ ��̂� � ���j�̂k ∙ ��̂� � 2�� � 10.0 | 10PN�J d) The torque is given by �� � �� | ��̂. The cross product is zero whenever �� is aligned with ±�.̂ e) If �� � �rs then � � 0 and �� � �rs | ��̂ � ����.̂ f) � � 0 any time �� � ���. Whenever �� lies in the xz-plane.

��

��� 41.41°

85.47° From �� to ��� is 126.9°

12

3.14

a) �� � 1.00m�̂ 6.00mrs

b) The force is �� � 2690N�̂ � 571. 8Nrs

c) �� � �� | �� � G1.00m�̂ 6.00mrsH | G2690N�̂ � 571. 8NrsH � G�16140�̂ 571. 8�̂ 2690rsHN ∙ m

3.15

a) ��Q � G0.750�̂ 0.64�̂ 7.88rsHm and ��N � G�0.750�̂ 0.64�̂ 7.88rsHm

b) ��Q � ��Q | �� � G0.750�̂ 0.64�̂ 7.88rsHm | G�3000NrsH � j�1920�̂ 2250�k̂N ∙ m ��N � ��N | �� � G�0.750�̂ 0.64�̂ 7.88rsHm | G�3000NrsH � j�1920�̂ � 2250�k̂N ∙ m

3.16

a) We know the vector points to the right and up. It will be of the form �� � �N �̂ ~rs . To figure out ~ in terms

of the givens, use the Pythagorean Thm.

v�2wN ~N � �N

~ � ��N � v�2wN

Therefore the final result is

�� � �2 �̂ ��N � v�2wN rs

b) The �̂ part flips sign but the rs part remains the same.

13

3.17

a) Consider the triangle at right. Notice

cos 30° � �2�

� � �2 cos 30° � �√3

While we’re at it, notice sin 30° � {�

{ � � sin 30° � �2√3

This shows us that the center of the triangle is at a third of the height…

b) To travel from the front left ball to the origin we must move �N to the right, �N√� into the page, and some other distance up. Because the string is angled

we cannot say we move distance � upwards. Notice a right triangle is

formed between the sides with lengths ~, �, and �√�.

~N v �√3wN � �N

~ � ��N � �N3

Make note of the coordinate system I chose. In this case �̂ is to the right

and into the page is ��.̂ The displacement vector is thus

�� � � �2√3 �̂ �2 �̂ ��N � �N3 rs

c) To get the angle from the vertical axis I choose to do �� ∙ rs � ‖��‖BrsB cos hm

Don’t forget, even though �� looks ugly, we already know ‖��‖ � �!!!

Also, the magnitude of any unit vector is 1 by definition (BrsB � 1).

Finally �̂ ∙ rs � 0 and �̂ ∙ rs � 0. Plug it all in to find �� ∙ rs � ‖��‖BrsB cos hm

��N � �N√3 � �j1k cos hm

cos hm � M�N � �N3� � �1 � �N3�N

hm � cosPQ ��1 � �N3�N 

You could also look at the picture and use Z¡ � 49:P0 v ¢√T£w

�

�2

30° � {

�

�2

�2√3

~

�√3

hm

~ {

z