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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
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To experimentally verify the rules for
vector addition by graphical (scale drawing)
and by components.
Force Table with Pulleys
50g mass holders
Assortment of masses from 1g to
100g
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).
a
b
R
You are given two (or three) forces, which are called F a and
b
(or F a,
b and
c),
and your problem is to find their resultant F R.
R is the result of the combined actions of
a
and F b (or
a,
b and
c). What you actually find on the force table is the equilibrant,
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,
R has the same
magnitude as F E, and its direction is opposite to that of
The direction of F R is found by
subtracting 180
from the angle found for F E.
and F b = 500 at 120
produce F a and
b
(include the mass of the holder). Remove any unused mass
holders from their strings.
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.
plastic ring is centered, with no tendency to move when you tap the force table with
your hand.
teacher's instructions.
and F b = 400 at 80
b = 100 at 70
and F c = 100 at 160
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.
Remember
that the magnitude of F R is the same as that for
E and that the angle for
R is that of
E minus 180
. Show any calculations and record these values on your calculations
page.
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
angles on the drawing and list the scale used.
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.
to the component method results and also comparing the magnitudes of the graphical
results to the component method results.
Your results table should look something like this:
|< Include angles >|
Setup FR
Exp.
Graph.
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:
Then find the x components:
x =A⋅cosθ A =20m⋅cos(45°)=14.1m
x =B⋅cosθ B =25m⋅cos(300°)=12.5m
x
=C⋅cosθ C
=15m⋅cos(210°)=-13.0m
x
x+
x+
x
= 13.6m
Find the y components:
Y =A⋅sinθ A =20m⋅sin(45°)=14.1m
Y
=B⋅sinθ B
=25m⋅sin(300°)=-21.7m
Y
=C⋅sinθ C
=15m⋅sin(210°)=-7.5m
Y
Y+
Y+
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:
2 2 2 2
X Y
θ R
− −
X
Y
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.