Alabama Science in Motion
Vectors: Force Tables (Vector Addition)
Force Tables (Vector Addition)
Teacher Notes
Purpose:
Forces can be resolved into components called vectors. The purpose of these experiments is to
examine vector addition of forces in a progression of experiments.
Course of Study (2005):
ALCOS #1: Explain linear, uniform circular and projectile motions using one- and twodimensional vectors.
Pre-lab Activities: Available on the ASIM Physics Web-site under the supplemental materials
of the Force Tables heading.
1. Outside Vectors
2. Clinometer measuring of Vectors
3. Tug of War in three directions
Force Table Lab Time: 1 hour
Force Table Setup Time per Station: 5 minutes
Background/Theory:
Experimental Method
Two known forces are applied on the force table by hanging masses over pulleys positioned at
certain angles. A third unknown force is determined experimentally by finding the angle and
mass necessary to balance the other two forces. This third force is called the equilibrant (FE)
since it is the force which establishes equilibrium. The equilibrant is not the same as the
resultant (FR). The resultant is the addition of the two known forces. While the equilibrant is
equal in magnitude to the resultant, it is in the opposite direction because it
Y
F
balances the resultant. Therefore, the equilibrant is the negative of the
resultant.
Fy = Fsinө
ө
X
Fx= Fcosө
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Alabama Science in Motion
Vectors: Force Tables (Vector Addition)
Component Method
Sketching a free body diagram or FBD is critical to solving these problems. Two known forces
are added together by first breaking each vector’s polar coordinates (r,ө) into its rectangular or x
and y components using the simple right triangle trig equations, sine, cosine, and tangent. The
like components are added, x with x and y with y. The resulting rectangular sums are converted
back into polar coordinates. The equilibrant is the same magnitude but opposite in direction to
the resultant.
sin  = opp/hyp
cos =adj/hyp
tan =opp/adj
opp2 + adj2 = hyp2
Graphical Method
Adding vectors head to tail
Two known forces are added together by drawing them to scale using a
ruler and protractor. The second force (FB) is drawn with its tail to the
head of the first force (FA). The resultant (FR) is drawn from the tail of
FA to the head of FB. Then the magnitude of the resultant can be
measured directly from the diagram and converted to the proper force
using the chosen scale. The angle can also be measured using the
protractor. The equilibrant is the same magnitude but opposite in direction to the resultant.
ENGAGE:
Before the lab, engage students by guiding them in the following activities:

Engagement 1:
o Have students, working in teams, measure between two points that are perpendicular to
each other. Example: measure between two fixed points along one building (A to B) and
then at a 90 degree angle, measure over to a fixed point (B to C) along another building.
A
B

C
o
Have the students draw what they measured outside on a piece of paper to scale being
sure to orient their drawing to the compass directions corresponding to the actual
buildings. Be sure students use a large enough scale for their drawing to cover at least
half of a sheet of paper ex: 1cm = 1 meter outside. Have students figure out the third side
of the triangle (resultant or, hypotenuse) measuring their scale diagram and then by using
trigonometry.
o
Relate what they have done to solving for the resultant in their textbook
problems.
Engagement 2:
o Working in groups students should find a nearby tree and measure from a place they
mark on the ground to the base of the tree. Record this distance.
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Alabama Science in Motion
Vectors: Force Tables (Vector Addition)
o Standing at the place they marked on the ground, students should use a student made
clinometer to measure the height to the top of the chosen tree.
o Working with their group, students should draw their measurements to scale on a piece of
paper and graphically figure out the third side of the triangle.
o Students should verify the accuracy of their measurement by solving for the third side of
the triangle (resultant) using the component method (trigonometry).

Engagement 3:
o Use a rope with a second piece of rope tied in the middle of it.
o Let two student volunteers (preferably two of the strongest individuals in the class) pull
on either end of the rope. Ask them to pull fairly hard.
o Have a third student volunteer (preferably a less brawny individual) pick up the second
rope that is tied in the middle and pull on it. What happens to the two original rope
pullers? They can easily be moved by this third person.
o Explain the vector relationship between these three forces.
EXPLORE:
Students work in pairs or three at the most, taking turns attempting to find the ideal resultant force needed
for each scenario presented for the force tables. Students need to sketch the forces necessary to reach
equilibrium and then answer the analysis questions.
Sample Data:
Sketch for your Educated Guess: Student sketches will differ.
FR
FB
FA
FE
Case I
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Alabama Science in Motion
Vectors: Force Tables (Vector Addition)
FR
FC
FB
FA
FE
Case II
DATA TABLE
Case I:
Case II:
FA due to 50g @ 0˚; FB due to 100g @ 90o
FA due to 75g @ 0˚; FB due to 120g @ 30˚;
FC due to 50g @ 110˚
Resultant
Resultant
Equilibrant
Equilibrant
Best Guess
110g @ 62˚
110g @ 242˚
195g @ 35˚
195g @ 215˚
Experimental
113g @ 63˚
111g @ 243˚
195g @ 37˚
195g @ 217˚
Component
112g @ 63˚
112g @ 243˚
194g @ 33.5˚
194g @ 213.5˚
Graphical
111g @ 63˚
111g @ 243˚
195g @ 33.5˚
195g @ 213.5˚
NOTE: Please be reminded that this is the force due to the mass at the given angles. Therefore
the following calculations are left in terms of mass for consistency. Also, masses in negative x
direction are regarded as negative masses.
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Alabama Science in Motion
Vectors: Force Tables (Vector Addition)
X
50g
0g
50g
Sample Component Calculations:
Fa
Fb
∑F
Case I:
y
0g
100g
86.6g
Fax: Fa(cosθa) = (50g)(cos(0)) = 50g
Fay: Fa(sinθa) = (50g)(sin(0)) = 0g
FR
FBy
Fbx: -Fb(cosθb) = -(100g)(cos(90)) = 0g
Fby: Fb(sinθb) = (100g)(sin(90)) = 100g
FR : ( Fx ) 2  ( Fy ) 2 
FAx
FR : (50 g ) 2  (100 g ) 2  111.8 g =112g
FE = FR  112g
FE
FR = θR = 63.4˚
θE = θR + 180˚ = 243.4˚
Fa
Fb
Fc
∑F
Case II:
x
75g
103.9g
-17g
161.9g
y
0g
60g
46.98g
106.98g
Fax: m(cosθ) = (75g)(cos(0)) = 75g
Fay: m(sinθ) = (75g)(sin(0)) = 0g
FR
Fbx: m (cosθ) = (120g)(cos(30)) = 103.9g
Fby: m(sinθ) = (120g)(sin(30)) = 60g
FC
FB
FAx
FE
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Fcx: -m(cosθ) = -(50g)(cos(70)) = -17g
Fcy: m(sinθ) = (50g)(sin(70)) = 46.98g
Fx = 75+103.9-17.1=161.8g
Fy = 0+60+46.98=106.98g
FR : (161.8 g ) 2  (106.98 g ) 2  194 g
FE = FR  194g
θR = tan-1(106.98/161.8) = tan-1(0.66 )
θR = 33.5˚
θE = θR + 180˚Page
= 213.5˚
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Alabama Science in Motion
Vectors: Force Tables (Vector Addition)
Sample Graphical Solutions:
FR
FC
FR
FB
FB
FA
FA
FE
FE
Student Response Sheet
EXPLAIN/ EVALUATE:
As you wrap up ask yourself: Have your students had the opportunity to understand the purpose
of this lab?
Although answers will vary based on experiences, students should note
1) The equilibrant force is equal to and opposite in direction to the resultant force.
2) Forces can be fairly accurately guessed with a basic knowledge of vectors. Once an accurate
approximation is made, a more precise answer can be gotten by manipulating the force and
angle of the equilibrant.
3) The equilibrant of more than two forces can be accurately determined using a force table.
Questions from the procedure:
Part 1: For both Case I and II:

Is there another combination of mass, angle that could return balance to these forces? If
so, try it. Explain why or why not.
There are no additional combinations that could return balance to these forces. There are
specific vectors in each situation that solely balance the system.
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Alabama Science in Motion
Vectors: Force Tables (Vector Addition)
Part 2: The Component Method (attach your solution to both Case 1 and II)
Follow up Questions: For both CASE I (two known forces) and CASE I, (three known forces)
1. How well did your “Educated Guess” match your Experimental values for the Equilibrant?
Student responses will vary.
2. How do the equilibrant values for magnitude and direction from the two theoretical methods
(component and graphical) compare to the experimental magnitude and direction found using
the force table? What are the percent errors for each?
Case I
Component: m:
| 112  112 |
*100%  0%
112
θ:
| 243.4  243 |
*100%  0.2%
243
Graphical:
m:
| 112  111 |
* 100%  0.9%
112
θ:
| 243.4  243 |
*100%  0%
243.4
Component: m:
| 194  195 |
* 100%  0.5%
194
θ:
| 213.5  217 |
* 100%  1.6%
213.5
Graphical:
| 195  195 |
*100%  0%
195
θ:
| 213.5  217 |
* 100%  1.6%
213.5
Case II
m:
3. What are some of the possible sources of Experimental Error? Note: Human error is not an
acceptable answer.
Some sources of error include the accuracy of the ruler or protractor and the accuracy of the
angle of the pulley.
4. The hangers are all considered to be 5 grams each. Assuming the hangers all have the same
mass, why were you asked to include the mass of the hanger in determining your vectors?
How would you use the equipment to test your answer?
The mass of the hangers contribute a vector force which has magnitude and direction. Those
forces must be added/subtracted as vectors. Ignoring them would be treating their impact as
scalars. That error would change the mass and angle needed to return the system to equilibrium.
If I place the empty hangers at the angles found in part one, the system is not in equilibrium.
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Alabama Science in Motion
Vectors: Force Tables (Vector Addition)
Alternately, if I put the hangers as indicated, but reduce the mass added at each angle by 5
grams, the system will not be in equilibrium.
5. Explain mathematically the relationship between force and mass.
Force and mass are directly related as shown by Newton’s Second Law: F=ma.
6. How would you explain to a younger student what the purpose of this lab is?
Student responses will vary.
EXTEND:
Extension #1 Refer to the ASIM web-site under the Force Table web-page and copy the
vector resolution extra problems for students to work using the force tables.
Extension #2
1. As a lab group create a paragraph describing simple vector addition of two vector forces. On
a second piece of paper, draw to scale your forces, the resultant, and equilibrant.
2. Solve your vectors by using the component method (trigonometry).
3. Exchange just your paragraphs with one other group. Do NOT show the other group the
matching solutions at this time.
4. Now try to sketch the vector problem described in your neighbor’s paragraph. Once your
group agrees on the sketch, solve the vector problem using the graphical and trigonometric
methods.
5. Answer the following questions:

What problems did you encounter with your neighbors lab?

Does your graphical sketch match the sketch of the original group for this problem? If
not, explain why it did not match. Student responses may vary, depending on the drawing
done.

Does your component answer match the original group’s answer? Explain why or why
not? The component answers should be the same.
A comparison of graphical sketches and component answers should be discussed as a class.
Problems encountered should also be discussed here.
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