Vectors: The Force Table
Introduction
Any quantity that has both a direction and a magnitude associated with it can be represented
as a vector. If a hiker walks five miles due north, her displacement can be represented as a vector.
If an airplane is flying east at 100 miles/hour, its velocity can be represented as a vector. In physics,
as in life, we experience many quantities that can be represented by vectors (force, acceleration,
momentum, etc.). Thinking of these quantities as vectors provides a convenient way to describe
how they interact. For example, how does a cross wind affect the motion of a plane flying at 100
miles/hour? How do two forces add to produce an acceleration? These questions can be addressed
through the addition or subtraction of vectors.
Right triangle analysis
In analyzing vectors in 2053 and 2054, the method generally used is adding vector
components. Before we present that discussion, along with some in lab exercises, recall from
trigonometry the analysis of a right-triangle. In the figure below, we show a general right triangle
with angles and , as well as sides x, y, and r.
General terms:
1.
X = side adjacent to angle
2.
Y = side opposite to angle
3.
r = the hypotenuse
r
y
In terms of trigonometric functions:
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑦
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
𝑥
1. sin ∝= ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 = 𝑟 .
2. cos ∝= ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 = 𝑟 .
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
x
𝑦
3. tan ∝= 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 = 𝑥 .
If the angle is known, we can calculate the value of using: 90o - =
For a triangle with known sides x and y, we can calculate the hypotenuse r by utilizing the
Pythagorean Theorem:
𝑟2 = 𝑥2 + 𝑦2
𝑟 = √𝑥 2 + 𝑦 2 .
Many proofs of the above equations exist. The student is encouraged to look at a few for general
knowledge (http://mathworld.wolfram.com/PythagoreanTheorem.html).
EXCERSICES FOR RIGHT TRIANGLE ANALYSIS
Task 1: Consider the right triangle above and calculate the unknown for each problem in the set.
Make sure the calculations are neat. Drawings should be included as well. Keep your work as
you will combine it with the theoretical calculations below:
1. X = 20 units, y = 10 units, = _______________, r = _________________, =
________________.
2. y = 10 units, r = 15 units, = _______________, x = _________________, =
________________.
3. X = 20 units, r = 30 units, = _______________, y = _________________, =
________________.
4. x = 8 units, = 40o, y = ____________, r = ______________, = ________________
A special case would be if x = y, and thus = . This is called an isosceles right triangle. You
will encounter many of those as you move through general physics.
VECTOR COMPONENT ANALYSIS
In analyzing vectors in general physics, we often add vector components in combination with the
analysis you just learned above. Below is a drawing of a typical vector in 2-D space:
y-axis
A
r
Ay
Ax
x-axis
Note that in most instances, especially when analyzing forces acting on a mass particle, the
vector emanates from the origin (pointing outward in the plane). Ax is called the x-component or
the x-projection of the vector A. Ay is called the y-component or the y-projection of the vector A.
The hypotenuse, r, denotes the magnitude or length of the vector A. This is usually symbolized
as just A but without the bold font or vector sign above it, namely, ⃗𝐴⃗⃗ .
The above vector might represent one of the following physical quantities:
a. A displacement vector: A person walking on campus leaves from their starting point (the
origin) and walks at an angle north of east given the campus’ well-defined coordinate
system. The x-projection of their displacement vector would represent how far they
walked along the x-axis or due east. The magnitude of the vector (the hypotenuse of that
right triangle) would represent the total displacement (usually in meters) from their
starting point to their ending point.
b. A velocity vector: An aircraft was launched from the origin (the take-off point) and is
flying north of east at 100 m/s (the magnitude of the vector). The x-projection of that
velocity vector would represent how fast the air-craft is moving easterly. In determining
how long will the air-craft take to move from the origin to ground point B (an xcoordinate due east of the origin), the x-projection is the relevant quantity in this case.
Based on what you know from analysis of right triangles, apparently we can write the equations
of the vector A the following way:
Ax r cos
Ay r sin
A
tan
Ax2 Ay2
Ay
Ax
When you study forces in general physics, there are usually (but not always) several forces
(vectors) acting on a mass particle. So the question is: If two forces were acting on a mass, what
is the resultant of these? How will the mass respond? To answer this theoretically, we draw a
force (vector) diagram and then simply add up all like-components to find the net magnitude or
resultant. Look at the figure below:
y-axis
B
A
x-axis
To analyze the above scenario, you would now utilize the following equations:
Ax r cos
Bx r cos
Ay r sin
B y r sin
A
B B x2 B y2
tan
Ax2 Ay2
Ay
Ax
tan
By
Bx
.
Adding and subtracting vectors graphically:
Even if the vectors are not at right angles, they can be added graphically by using the “tail-to-tip”
method. Consider the vector addition below:
+
+
In order to subtract vectors, we define the negative of a vector, which has the same magnitude
but points in the opposite direction. Refer to the figure below:
To subtract two vectors, we simply add the negative vector:
+
EXPERIMENTAL ANALYSIS OF VECTORS
One of the most important problems to solve in physics is called static equilibrium. One aspect
of this relevant in this lab is the net force (due to two or more forces) must be zero. You will
learn about specific forces in your lecture course and in later experiments. For now, let us
assume that all forces acting on a mass particle in a 2-D plane (whatever they are) act such that
the net result is zero:
𝑁
∑ 𝐹𝑖 = 0
𝑖=1
That is,
∑ 𝐹𝑥 = 0
𝑖
∑ 𝐹𝑦 = 0
𝑖
In this lab, you will calculate the “balancing force”. That is, suppose you are given two forces
which provide a new force. Question: What is a third force that balances the system such that the
net forces (sum of all forces) equals to zero? Essentially, to determine the balancing force,
simply find the resultant of the two forces given and then the force opposite of that would
balance the system.
Task 2 (Theoretical exercises): Below, you are asked to determine vector components
and in two cases the balancing force given two other forces. Perform these calculations
neatly (include rough drawings) and include them with Task1 above. Scan and save the
file as “vector.pdf” and upload to the link provided. Note: If you are asked to calculate
the balancing force, remember to calculate the resultant and the balancing force would
point OPPOSITE to the resultant.
1. (Calculate the components of the force) 𝐹 = 150 𝑁 at an angle of 30o below the positive
x-axis.
𝐹𝑥 = ____________, 𝐹𝑦 = ______________,
2. (Calculate the balancing force) 𝐹1𝑥 = −10𝑁, 𝐹1𝑦 = +5𝑁, 𝐹2𝑥 = 0𝑁, 𝐹2𝑦 = −4N
𝐹𝐵𝑥 = ____________, 𝐹𝐵𝑦 = ______________, 𝐹𝐵,𝑛𝑒𝑡 = √𝐹𝑥2 + 𝐹𝑦2 = _____________
3. (Calculate the balancing force) 𝐹1𝑥 = +25𝑁, 𝐹1𝑦 = −8𝑁, 𝐹2𝑥 = −2𝑁, 𝐹2𝑦 = +4N
𝐹𝐵𝑥 = ____________, 𝐹𝐵𝑦 = ______________, 𝐹𝐵,𝑛𝑒𝑡 = √𝐹𝑥2 + 𝐹𝑦2 = _____________
A. The Force Table: A force table is a device consisting of a solid table supported by a rigid
stand, and pulleys which are attached to the table. These clamped pulleys can be adjusted
to hang anywhere around the table. An illustration of the table is seen below. Hanging
over the pulleys is a very light string attached to a ring located at the center of the table
and a hanging mass draped below the table surface. These hanging masses will act as our
forces. Note that all strings used will be attached to a small ring at the center of the table.
If this ring is centered at the very center of the table, being that the center of the ring is
centered at the vertical peg, the system is balanced. Given two or more force vectors, you
will theoretically and experimentally determine what resultant force (vector) would
balance the system.
Experimental procedure:
1. Download the power-point called “force-table.pptx”.
2. For each slide, place the balancing force where it should go and include some degree
of accuracy when it comes to the size.
3. To place the balancing force, simply use the insert/shape/arrow and place the arrow
according to where it would balance the system and size it to about the size you
would need it to have. You saw how the arrows were placed in the interactive
activity. You are now merely doing this manually with power-point. Note: The arrow
size may not be exact, but size it as best as you can.
4. Study the example given in the first two slides. The first slide shows the unbalanced
system. The next slide shows how you would correct it. Note: After placing the
balancing force, center the dashed ring and copy/paste one of the dashed radial lines,
the line running from the center of the peg to where the forces are drawn, so that you
can have a radial line for your force as well.
5. Note: You do NOT have to create another slide for your solution. Just edit the
slide for which you are given the forces that you are trying to balance. The twoslide example I gave you is just an illustration of before and after.
6. After you have completed this short activity, save the file as “Force-table.pdf” and
upload it in the link provided.
7. Please remember: This is an activity to help you “think” about physical vectors. You
will not be exact. But you should try to make it as close as possible so you get a good
“feel” for forces and their associated magnitudes and directions.
8. Finally, on the last page there is a piece of “graph paper”. The paper can be scaled.
For example, 4 blocks can represent 1 N of force. I will give you three forces. Draw
these forces. You will need a protractor as these forces have angles with respect to the
positive x-axis. The angles given are with respect to the positive x-axis
(counterclockwise).
9. Determine the balancing force, magnitude and direction, using the equations above
for vector analysis, and draw the balancing force on the graph paper.
10. Make sure ALL FORCES are properly labeled and scaled to the best of your ability.
11. Scan the paper and save it as “force-diagram.pdf” and upload to the link provided.
GIVEN FORCES:
|𝐹1 | = 3 𝑁, 𝜃1 = 40°
|𝐹2 | = 2 𝑁, 𝜃2 = 100°
|𝐹3 | = 3 𝑁, 𝜃3 = 310°

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