vectors - city university of new york
TRANSCRIPT
Brooklyn College 1
Vectors
Purpose
1. To study vectors and their addition graphically.
2. To study resolution of a two dimensional vector and the addition method through components.
Introduction
Vectors are quantities that have both direction and magnitude. The magnitude is the numerical value of the vector. A
quantity that has only magnitude (no direction) is called a scalar. A scalar could have any value including negative values,
but it does not have a direction. Examples of scalars are temperature, distance and mass. Examples of vectors are force,
velocity and acceleration. The direction of a force vector, is the direction in which the force is applied. The magnitude of
the force vector is the strength of the force. We draw vectors as ray with the direction as the arrow. When drawing a
vector we represent the magnitude of a vector by the length of that vector. So a strong force is represented by a longer
vector than a weak force.
As a notation, we will write a vector in bold βπ¨β, while its magnitude in normal font βπ΄β.
Part 1: Adding vectors
When we add vectors we have to take the direction of the vector into consideration. Two equal but opposite vectors
add up to zero. Since a vector is characterized by only magnitude (represented by length of the vector) and direction,
then we if we move the vector parallel to itself while keeping its length fixed, we still have the same vector (since in
moving the vector parallel to itself we preserved its direction and magnitude). A method to add vectors is to arrange
them in a way that the tip of the 1st vector is at the tail of the 2nd vector and the tip of the 2nd vector is at the tail of the
3rd vector and so on. Then the sum of the vectors, called Resultant vector, πΉ, is a vector that connects the tail of the 1st
vector to the tip of the last vector. See figure 1. πΉ = π¨1 + π¨2 + π¨3. If a
vector π¬ is equal and opposite to πΉ (π¬ = βπΉ), then π¬ + πΉ = 0. π¬ is called
the βequilibrant vectorβ. If the vectors represent forces, then π¬ is the force
that when added to the other force vectors will produce equilibrium.
Part 2: Resolving a vector
Consider the two vectors π¨π₯ (horizontal) and π¨π¦ (vertical) in Figure 2. If we add the vectors π¨π₯ and π¨π¦ using the
graphical addition method above, we get the Resultant vector. We will call it π¨ here. So, π¨ = π¨π₯ + π¨π¦. So in a plane,
any vector π¨ can be analyzed (resolved) to two components a horizontal component π¨π₯ and a vertical component π¨π¦.
Using trigonometry,
π΄π₯ = π΄πππ π and π΄π¦ = π΄π πππ ππππ . (1)
π‘πππ =π΄π¦
π΄π₯ πππ. (2)
Using Pythagoras theorem π΄ = π΄π₯2 + π΄π¦2 πππ. (3)
In order to add vectors, there is a method other than the graphical method mentioned above. This other method uses
vector resolution. First we resolve all the vectors to their x and y components then we add the x components (with their
signs) and we add the y components (with their signs) then find the magnitude of the resultant vector using eqn. 3, and
its direction using eqn. 2.
Figure 1: Adding vectors graphically
π¨1 π¨2
2 π¨3
π¨1
π¨2
π¨3
πΉ
Figure2: Resolving a vector π¨
π¨π₯
π¨π¦ π¨π¦
π¨π₯
ΞΈ π¨
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Running the experiment The data sheet is on page 5 & the graph paper is on page 6
Part 1: Adding vectors graphically and finding the equilibrium vector (Ask the instructor for your set of π¨π, π¨π, & π΄π)
If you do not have a protractor you may find these links useful (copy the link and paste it in your browser):
i) https://www.ginifab.com/feeds/angle_measurement/
ii) https://www.inchcalculator.com/wp-content/uploads/2016/04/protractor.pdf
1) Consider figure 3. Each square has a side equal 1 cm. Using the graphical method for adding vectors find the resultant
πΉ of the 3 vectors π¨1, π¨2, and π¨3. The tails of the vectors are at point (8, 8), & for their tips: ask your instructor. In this
graphical method you can, using a ruler, a protractor (or using the coordinates: for ex. π¨2 has a tip that is 5 units left & 5
units up from from its tail) & the graph paper given on page 6 here (follow the orientation given in the graph paper: π₯-
horizontal & π¦-vertical), to redraw π¨2 so that its tail is at the tip of π¨1, and π¨3 so that its tail is at the tip of π¨2, then
connect the tail of π¨1 to the tip of π¨3: this is the resultant vector, πΉ. See figures 4 and 5 for a method of redrawing the
vectors using the protractor with same angles (In fig. 4. the magnitudes are not to scale). With the ruler you draw the
vector with the same length (this represent same magnitude) and with the protractor you draw the vectors with the
same directions as demonstrated in figure 5. π1, π2, πππ π3 are examples to resemble π¨1, π¨2 πππ π¨3, respectively.
2) Using the pencil, ruler and the protractor, find the magnitude and direction of πΉ. Then find π¬ = βπΉ.
Figure 3: Adding vectors graphically
π¨1 π¨2
π¨3
Ξ²
Ξ³
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3) Open the simulator https://phet.colorado.edu/sims/html/vector-addition/latest/vector-addition_en.html
Click Lab. There are 2 sets of vectors, blue and orange. We are going to use only the blue set of vectors. Drag a blue
vector (we will call it π¨1) to the graph sheet of the simulator (see fig. 3 above), so that its tail is at coordinates (x, y) = (8, 8)
& its tip is at point (16, 14). Click values. The magnitude of the vector should show as 10 (default). Drag another blue
vector (we will call it π¨2) to the graph sheet, place it such that its tail is at the same tail of vector π¨1; tail at point (8, 8),
and tip of vector π¨2 is at (3, 13). The magnitude of π¨2 should be 7.1. Drag a third blue vector (we will call it π¨3) to the
graph sheet. Place it such that its tail is coinciding with the tail of π¨1 and π¨2 at (8, 8), and its tip is at (4, 5). Its magnitude
should be 5. Assume each square of the graph has side length equal 1centimeter, 1 cm.
4) Click Sum (for the blue vectors) at the top right of the simulator. This shows the sum vector πΊ for the blue vectors,
which up till now are only the three blue vectors π¨π, π¨π and π¨π, we will call the resultant πΉ. Drag this Sum vector, πΊ so
that its tail is at (8, 8). Click angle . This shows the direction. If it is difficult to see the values, you can uncheck the
values option at the top right of the simulator and move the mouse to the vector that you want to select & click the
vector, the magnitude of the vector & all its values will show above the graph. You can also alternatively keep the values
and angles options checked & drag the sum vector a bit away to see the values clearly.
Compare to your calculated πΉ in step 2 above. Compute the calculated π¬ = βπΉ, (Notice that magnitudes of π¬ & πΉ are
the same, but the direction of π¬ is opposite to the direction of πΉ. So the angle of π¬, is ππΈ counter clock wise, (CCW) from
the positive x-axis is angle ππ of πΉ plus 180o, ππ + 180π and from below the positive x-axis (clock wise, CW) the angle of
π¬ is βππΈβ² = β[360π β (ππ + 180π)] ). Record these values for πΉ and π¬ in the table of part 1, step 4) in the data sheet,
we will use them for the next step 5. In this table of part 1, record both ππΈ(CCW) and βππΈβ² (CW), from the positive x-axis.
5) Drag a fourth blue vector to represent the vector π¬ that we calculated in step 4, and place it with its tail at (8, 8),
magnitude as calculated in step 4 (= magnitude of πΉ found and recorded in step 4) and angle clock wise from the
positive x-axis equal to βππΈβ² = β[360π β (ππ + 180π)] such that it is opposite to πΉ that was found in step 4. Note that
as you place the fourth vector π¬ and while you are adjusting its magnitude and direction (angle), the sum vector, πΊ
changes. Can you explain why? (Hint: what does the sum vector, πΊ represent?). After you adjust the magnitude and
direction (angle) of the fourth vector π¬, what is the value of the new sum vector, πΊ ? Explain.
Part 2: Resolving a vector
Figure 6: Resolving a vector
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1) For the vector shown in Figure 6, (we will call it vector π), using equations 1 in the introduction, find the x-component
ππ₯ and the y-component ππ¦.
2) In the same simulator of part 1 (do not remove the vectors we will use them again in part 3), click βExplore 2Dβ at the
bottom of the window. Drag a blue vector π and place it such that its tail is at the origin (as shown in figure 6). Click
values and click angle. Adjust the vector π so that its magnitude is 10.4 and its angle counter clock wise from the positive
x- axis is 106.7o, as shown in figure 6.
3) In components click the icon so as to show the x and y components. Record the values and compare with
your calculated values in step 1.
Part 3: Adding vectors using components
1) In the same simulator, click βLabβ at the bottom of the window. You should see your vectors π¨1, π¨2 , πππ π¨3 from part
1. You can remove the 4th vector π¬ (drag it back to the blue vector box at the right). Using the magnitude and angle of
each vector, calculate the x and y components: π΄1π₯ , π΄1π¦ , π΄2π₯ , π΄2π¦ , π΄3π₯ , π΄3π¦ , and π π₯& π π¦ (you may need to uncheck
the values option, in order to see clearly the angles). Note that in equations 1, ΞΈ is the angle measured counter clock
wise from the positive x-axis. So, for example, for the vector π¨3, in fig. 7, the angle ππ΄3 is 360π β 143.1π .
2) Click components (you can uncheck the angles option and select the values option to see the values of the
components clearly. You can also drag the vectors a bit apart to see the components of each vector clearly. See fig. 7).
Record the values of π΄1π₯ , π΄1π¦ , π΄2π₯ , π΄2π¦ , π΄3π₯ , π΄3π¦ , and π π₯& π π¦ and compare to your calculated values in step 1.
3) a) Add the x components (including sign) and b) add the y components (including sign). Record your calculated x and y
components.
4) Compare your calculated values in step 3 to the values displayed by the simulator for π π₯& π π¦ .
Question:
1) State the condition for equilibrium in two ways (i) as illustrated in part 1, (ii) as illustrated in part 3.
Figure 7: Adding vectors using components
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Data sheet
Name: Group: Date experiment performed:
Part 1: Adding vectors graphically
Step 2) magnitude of πΉ: Direction of πΉ, ππ : magnitude of π¬: Direction of π¬, ππΈ:
Step 4)
Measured πΉ Calculated π¬ = βπΉ (use measured πΉ of step 4)
magnitude Direction ππ CCW from +ve x-axis
magnitude Direction ππΈ CCW from +ve x-axis
Direction ππΈβ² CW from
+ve x-axis
Step 5) Why does the sum vector, πΊ change as you add and while adjusting the fourth vector, π¬?
With the fourth vector, π¬ added, the sum vector πΊ = Explanation:
Part 2: Resolving a vector
Step 1) ππ₯ = direction of ππ₯ : ππ¦ = direction of ππ¦ :
Step 3)
Measured ππ₯ Measured ππ¦
magnitude direction magnitude direction
Part 3: Adding vectors using components
Step 1) Calculated:
π΄1π₯ π΄1π¦ π΄2π₯ π΄2π¦ π΄3π₯ π΄3π¦ π π₯ π π¦
Step 2) Measured:
π΄1π₯ π΄1π¦ π΄2π₯ π΄2π¦ π΄3π₯ π΄3π¦ π π₯ π π¦
Step 3) a) Addition of x components, βπ΄π₯ :
b) Addition of y components, βπ΄π¦ :
Step 4) a) Compare βπ΄π₯ to measured π π₯ :
b) Compare βπ΄π¦ to measured π π¦ :
Answer to question 1:
(The graph paper is on the next page, page 6. Each square is 1cm x 1cm).