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Vector Addition: Experimental Verification using Graphical and Component Methods, Lab Reports of Art

A lab experiment to verify the rules of vector addition through graphical (scale drawing) and component methods using a force table with pulleys and masses. Students are guided to find the equilibrant force (fe) and then determine the resultant force (fr) by finding the opposite direction of fe. The procedure includes steps to set up the apparatus, find the angle of fe, and record the masses and angles. The document also includes instructions to find the approximate error due to friction.

Typology: Lab Reports

Pre 2010

Uploaded on 08/08/2009

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VECTOR ADDITION
PURPOSE:
To experimentally verify the rules for
vector addition by graphical (scale drawing)
and by components.
APPARATUS:
Force Table with Pulleys
50g mass holders
Assortment of masses from 1g to
100g
METHOD:
Forces produced by masses attached
to strings over pulleys are vectors used in this
experiment. The directions of the vectors can be varied by positioning the pulleys at different
points around the rim of the circular force table. The magnitudes of the vectors are changed
by varying the masses on the strings. At the center of the force table, all of the strings are
attached to a plastic ring. (See Figure 1).
Fa
F
F
F
bR
EFigure 1
You are given two (or three) forces, which are called Fa and Fb (or Fa, Fb and Fc),
and your problem is to find their resultant FR. FR is the result of the combined actions of Fa
and Fb (or Fa, Fb and Fc). What you actually find on the force table is the equilibrant, FE of
the given forces: the force that balances them, so there is no tendency for the central ring to
start moving when FE is applied along with the given forces. The resultant, FR has the same
magnitude as FE, and its direction is opposite to that of FE. The direction of FR is found by
subtracting 180° from the angle found for FE.
PROCEDURE:
1. Given forces are Fa = 500 at 0° and Fb = 500 at 120°.
2. Place pulleys at the positions of the given forces and add the necessary masses to
produce Fa and Fb (include the mass of the holder). Remove any unused mass
holders from their strings.
3. Find the angle for FE by positioning another pulley roughly opposite the given forces
and pulling the string with your hand over the pulley. If you cannot center the plastic
ring on the central pin by pulling the string down over the pulley, the correct angle has
not been found. Move the pulley for FE in steps of about 5° along the rim toward the
direction where the ring touches the central pin, until you find the position where the
ring can be centered by pulling the string over the pulley. It might be necessary to
slide the strings slightly where they are tied to the ring to make the line of the string
pass through the center of the pin. You may also need to use finer angular
adjustments as you approach the proper position.
4. Put a mass holder on the string you have been pulling and add masses until the
plastic ring is centered, with no tendency to move when you tap the force table with
your hand.
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VECTOR ADDITION

PURPOSE:

To experimentally verify the rules for

vector addition by graphical (scale drawing)

and by components.

APPARATUS:

Force Table with Pulleys

50g mass holders

Assortment of masses from 1g to

100g

METHOD:

Forces produced by masses attached

to strings over pulleys are vectors used in this

experiment. The directions of the vectors can be varied by positioning the pulleys at different

points around the rim of the circular force table. The magnitudes of the vectors are changed

by varying the masses on the strings. At the center of the force table, all of the strings are

attached to a plastic ring. (See Figure 1).

F

a

F

F

F

b

R

E

Figure 1

You are given two (or three) forces, which are called F a and

F

b

(or F a,

F

b and

F

c),

and your problem is to find their resultant F R.

F

R is the result of the combined actions of

F

a

and F b (or

F

a,

F

b and

F

c). What you actually find on the force table is the equilibrant,

F

E of

the given forces: the force that balances them, so there is no tendency for the central ring to

start moving when F E is applied along with the given forces. The resultant,

F

R has the same

magnitude as F E, and its direction is opposite to that of

F

E.

The direction of F R is found by

subtracting 180

from the angle found for F E.

PROCEDURE:

  1. Given forces are F a = 500 at 0

and F b = 500 at 120

  1. Place pulleys at the positions of the given forces and add the necessary masses to

produce F a and

F

b

(include the mass of the holder). Remove any unused mass

holders from their strings.

  1. Find the angle for F E by positioning another pulley roughly opposite the given forces

and pulling the string with your hand over the pulley. If you cannot center the plastic

ring on the central pin by pulling the string down over the pulley, the correct angle has

not been found. Move the pulley for F E in steps of about 5

along the rim toward the

direction where the ring touches the central pin, until you find the position where the

ring can be centered by pulling the string over the pulley. It might be necessary to

slide the strings slightly where they are tied to the ring to make the line of the string

pass through the center of the pin. You may also need to use finer angular

adjustments as you approach the proper position.

  1. Put a mass holder on the string you have been pulling and add masses until the

plastic ring is centered, with no tendency to move when you tap the force table with

your hand.

  1. Record all of the angles and masses with their uncertainties in accord with your

teacher's instructions.

  1. Repeat steps 2 through 5 for F a = 300 at 30

and F b = 400 at 80

  1. Repeat steps 2 through 5 for F a = 200 at 0

, F

b = 100 at 70

and F c = 100 at 160

  1. Find the approximate error due to friction by putting 500g on one string and 500g on a

second string directly opposite. Find how much mass must be added to one side for

the ring to move noticeably

(a) When the mass is added gently.

(b) When the table is tapped as they are added.

ANALYSIS:

  1. Find the experimental value of F R for each of the three different setups.

Remember

that the magnitude of F R is the same as that for

F

E and that the angle for

F

R is that of

F

E minus 180

. Show any calculations and record these values on your calculations

page.

  1. Find F R for each setup using the graphical method (scale drawing).

Choose a

reasonable and convenient length scale to represent the vector magnitudes, then

using a protractor and ruler lay out the given vectors head to tail. F R runs from the tail

of the first to the head of the last vector. Use a scale sufficiently large that the

completed drawing nearly fills half of an 8

× 11 page. Label all magnitudes and

angles on the drawing and list the scale used.

  1. Find F R

by the method of components (using trigonometry) and show your work on

the calculations page. See your text if you don't remember how to do this.

  1. Find the percent difference comparing the magnitudes only of the experimental results

to the component method results and also comparing the magnitudes of the graphical

results to the component method results.

  1. Calculate the percent error expected due to friction from your data in step 8-b.

Your results table should look something like this:

|< Include angles >|

Setup FR

Exp.

FR

Graph.

FR

Comp.

% Diff.

Experimental

% Diff.

Graphical

Friction

% error

Vector Addition

Now we will add the same three vectors above mathematically. Once again, they

are: A =20m@45deg., B =25m@300deg., and C =15m@210deg.

First draw a sketch of the vectors:

+ x

+ x

+ y + y

+ x

+ y

Then find the x components:

A

x =A⋅cosθ A =20m⋅cos(45°)=14.1m

B

x =B⋅cosθ B =25m⋅cos(300°)=12.5m

C

x

=C⋅cosθ C

=15m⋅cos(210°)=-13.0m

R

x

= A

x+

B

x+

C

x

= 13.6m

Find the y components:

A

Y =A⋅sinθ A =20m⋅sin(45°)=14.1m

B

Y

=B⋅sinθ B

=25m⋅sin(300°)=-21.7m

C

Y

=C⋅sinθ C

=15m⋅sin(210°)=-7.5m

R

Y

= A

Y+

B

Y+

C

Y = -15.1m

Now that you have the components of the resultant vector, use the Pythagorean

theorem to find the magnitude of the resultant and the inverse tangent function

to find the direction angle:

R =

2 2 2 2

R R ( 13. 6 m ) ( 15. 1 m )

X Y

+ = + − =20.2m

θ R

− −

m

m

R

R

X

Y

tan tan

1 1

-48° is the same as a positive 312°. So our resultant vector is R =20.2m@312°.

This shows that the graphical method used above was fairly accurate but not

exact.