Physics Lab 1.7 Vector Frog Racing
Overview
In vector frog races, each contestant uses a piece of
string that is threaded through a plywood frog to move
the frog in such a way that it travels along the floor
towards a finishing line. The person who is best able to
regularly lift the frog without allowing it to fall over
backwards and then let it fall forwards (onto its face)
as far as possible and as fast as possible will win the
race.
Vector frogs use
components of forces and friction to control their
movement. The purpose of this lab is to investigate the
use of force vectors and to identify the factors that
affect the rate at which vector frogs can be propelled
along a floor.
Vector Frogs
A vector frog consists of a piece of plywood (or board)
cut to the shape of a frog and with a hole drilled
through it slightly above its center of gravity. To make
such a frog, make an enlarged copy of the diagram on Page
2. (The height of the frog should be between 12’’
and 14") Use this to make a template to mark the
outline of the frog on a piece of plywood. Cut out the
shape, mark it and paint it in such a way that it
resembles a frog. Drill a ¼" hole through the frog
in the position indicated on the diagram. Make a number
of these frogs as required.
Tie a piece of string (roughly
5 meters - or 15 ft) long to a wall or post behind the
frog, thread the string through the hole in the frog. The
distance from the floor to the point at which the line is
anchored should be roughly the same as the distance from
the base of the frog to the hole through its center.
When vector frogs are
raced, their strings are anchored roughly 1 meter (3ft)
apart.
A finishing line is marked
on the floor in front of the competitors. This line is
parallel to, and about 4 meters from, the wall or post
that anchors the lines.
By pulling the string, the
frog is lifted to a position that is nearly vertical.
When the string is released, the frog flops forward. When
the frog is again lifted, its base moves forward along
the floor until it reaches its highest position again and
flops forward when the string is released. By pulling the
string and releasing the tension in the string in a
constant pattern, the person racing the frog gets the
frog to move forward consistently. The person whose frog
is the first to cross the line wins the race.
There is a certain amount
of skill involved in this game and previous practice does
provide an advantage. The correct amount of tension needs
to be applied to the string at the right time. The
tension needs to be released at the correct time and the
cycle repeated consistently.
Experiment
1.7.1 Tension on either side of the frog
Overview
Vector frogs provide a
good illustration of how vectors work. The tension in
this string on each side of the frog can be divided into
a vertical component and a horizontal component. The
vertical component lifts the frog. The horizontal
component is cancelled by the horizontal component of the
tension on the other side of the frog. The vertical
component of the tension on one side of the frog is
probably close to half of what is needed to lift the frog.
Purpose
The purpose of this
experiment is to estimate the angle between the string
and the horizontal by measuring the weight of the frog
and the tension in the string needed to lift the frog.
Measurements
The tension force in the
string needed to raise the frog depends on the weight of
the frog and the angle between the string and the
horizontal. The smaller the angle, the greater the force.
In this lab, we’ll measure the forces involved using
a spring balance and estimate the angle between the
string and the horizontal.
Procedure
- Attach a small wire
hook to the center of the frog.
- Use a spring balance
to estimate the mass of the frog.
- If the balance is
graduated in grams, it will provide a direct
reading of the frog’s mass (at the earth’s
surface.)
- The force in Newtons,
needed to lift the frog will be its mass in
kilograms multiplied by the gravitational force
vector. (9.81 Newtons per kilogram)
- To measure the
tension in the string, attach the spring balance’s
hook to the string. (The measurement obtained
from the spring balance will be slightly
inaccurate because it is calibrated for vertical
use.)
- The tension in the
string (in Newtons) will be equal to the reading
in kilograms (or grams divided by 1000)
multiplied by 9.81.
- Measure the actual
angle between the string and the floor and
compare this with the angle estimated from the
force vectors. (One method: Measure a distance of
2 meters along the floor below the string and
mark this distance. Measure the distance in
meters from the floor to the string directly
above this point. Call this distance A (m). The
tan of the angle will be A/2.The angle will thus
be tan-1 of A/2.)
Calculations
Using the measurements obtained in the video lab, the
force needed to lift one of the frogs was equivalent to
the weight of 360 grams. This was 0.36kg x 9.81 N/kg = 3.14
N.
The tension needed to lift
the frog at the chosen angle was equivalent to 1000 Grams
or 1 kilogram.
The tension on one side of
the frog only provided part (roughly 50%) of the force
needed to lift the frog.
The force needed to lift
the frog was 3.14 Newtons. Half of this force was
therefore 1.77 N
If the angle between the
vertical and the string was q and the tension in the
string was equivalent to the weight of 1kilogram, the
tension on one side was equivalent to 9.81 Newtons.
The vertical component of
this tension was 1.77 N.
For a right angle
triangle, Sin q = side opposite divided by the hypotenuse.
If we draw the vectors as
shown below, the side opposite is equivalent to 1.77 N
and the hypotenuse is equivalent to 9.81 N.
The sin of the angle q was
thus 1.77/9.81 which = 0.18
From this we can determine
(using sin-1 of 0.18) that the angle was 10.39deg.
Questions
for Discussion
- What tends to happen
when the frog is lifted too quickly?
- Is there any
advantage to standing up as opposed to sitting
down while racing a frog? Why?
- The base of the frog
moves forward when it is lifted as a result of an
increase in tension in the string. Why does the
string not slip through the hole in the frog when
the tension in the string is increased?
Answers
- The frog tends to
flip past the vertical and will fall backwards
when the tension in the string is released.
- There may be an
advantage. Less tension will be needed to lift
the frog and it may be easier to prevent the frog
from flipping over backwards. However if the
string is kept low, this will limit the maximum
height to which the frog may rise during each
cycle. Keeping the string low – while
sitting down – may be a better way to ensure
that the frog keeps moving forward.
- The friction between
the string and the frog depends on the force
between the two surfaces. When the string is
released, the frog can flop forwards because the
friction between the string and the plywood is
very small and the friction between the base of
the frog and the floor is relatively large. When
there is tension in the string, the force between
the plywood and the string increases. The process
of lifting the frog also reduces the friction
between the base of the frog and the floor.
Increased friction between the frog and the
string fixes the position of the upper portion of
the frog. Reduced friction between the base of
the frog and the floor allows the base to slide
forward.
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