SCI 301 Applications in the Natural World
Introduction. If an unbalanced force acts on an object, then the object's
motion will change. The object may speed up, slow down, or change direction. If it is
rotating, then an unbalanced turning force (called a torque) will cause it to
rotate faster or slower. If the motion of the object is constant, that is, its motion
does not change, then the object is said to be in equilibrium. Equilibrium does
not mean that there are no forces acting on the object; it does mean that
all the forces that are acting cancel out each other.
An object which is at rest is said to be in static equilibrium. In this laboratory
we will investigate whether or not the forces on an object in static equilibrium really do
Theory. Forces have both a magnitude (size) and a direction; for this reason
forces are vectors. Vectors may be thought of as being arrows with lengths
which are proportional to their magnitudes. Vectors can be added by making a scale
drawing of the vectors (keeping the appropriate lengths and directions) such that the
"head" of each touches the "tail" of the next, as in the figure at right (click on the
figure for a larger version):
The vector sum is then the vector which connects the tail of the first vector to the head of
the last. In the figure above, the black vector is the sum of vectors A, B,
C, and D. If the vectors form a complete loop then the vector sum is zero.
We hypothesize that if a body is in static equilibrium, then the vector sum of the forces
acting on that body is zero, or SF = 0.
Procedure. The apparatus you will use to test this hypothesis is the force table,
which appears in the figure at right (taken from the
Middle Tennessee State University's Physics Department website. You will also need a
center pin, four pulleys, a metal ring with four light strings attached, four mass hangers, an assortment
of masses, and a carpenter's level. For the data analysis you will need several sheets of graph
paper, a ruler, protractor, sharp pencil, and eraser.
Using the carpenter's level, verify that the surface of your force table is horizontal. If it
is not, adjust the table legs until it is satisfactorily level. Insert the pin in the hole at
the center of the table and through the metal ring, The light strings should be able to reach
unobstructed from the center of the table to several centimeters beyond the table edge. Fasten
the pulleys on the table so that the angular scale is clearly visible and thread the strings
over them. Attach mass hangers to three of the threads. Note that the mass of each hanger is 50 grams. You will need to include this as part of the total mass in the procedure below.
1. First attach a total mass (including the hanger) of 210 g to one string and position
it at angle of 0° on the force table. Next attach a total mass of 180 g to a second string positioned at angle of 70°. These we will call the "fixed" vectors. Find the mass required and the position of a third string such that the
metal ring is in equilibrium; that is, so the ring will not touch the pin. This will be the "determined" vector, that brings the system into equilibrium.
Estimate the uncertainty in the location and magnitude of the third mass, and determine the "best" value of the mass and its angular position. You can do this by slowly adding or
removing 1-g masses from the hanger of the third mass until the metal ring is no longer in equilibrium. There will be a range of values that will maintain equilibrium. Assume that the best value is the "midpoint" value, the average of the highest and lowest values. Record the best value for the third mass and its "error" as "the average ±half the range". For example, let's say you found that values from 140 gm through 146 gm maintained equilibrium. You would report the third mass as 143 ±3 gm. (This standard type of error notation says the value is 143 gm with an error of ±3 gm.)
Next, place the best mass on the third string and slowly vary the angle. As you did for the mass, report the value of the angle as "the average ±half the range". When finished, draw a sketch of the force vectors, clearly labeling the mass and angle of each. For the "fixed" vectors, simply record the mass and angle. For the "determined" vector, include the errors for the mass and angle.
2. Repeat the above procedure, but now use all four strings. Place 160 g at
0°, 90 g at 45°, and 200 g at 240°. As in the first case, find for the fourth string the best mass and angle, with their respective errors, such that the system is in equilibrium. Record all masses and angles in a sketch.
Data and Results For each of the two cases you investigated, determine the vector
sum of the forces acting on the ring using graph paper, your ruler, protractor, and pencil.
Although the "gram" is a unit of mass, we can simplify the work by using units of "gram force".
Thus, a mass of 180 g exerts a force of 180 grams force.
Start with the three vector diagram and choose an appropriate length scale for your vectors. For example, you may choose a scale of 1 cm
= 10 grams force. You may choose any scale you want, but it must be the same for all vectors
in your diagrams, and each diagram should cover at least half of a sheet of graph paper. Using a ruler and protractor, construct the vector addition diagram for the three force vectors from the first part of the exercise.
Remember to keep the angles between the vectors exactly as you recorded in your sketch. You are "moving" the vectors, but you must maintain both their length and direction. Once you have drawn all three vectors, it should form a closed triangle, representing zero total force. Of course, it may not close the diagram perfectly. In the conclusion section, we will describe how to use the errors to quantify how close to zero it is.
Now repeat using the four vectors from the second part of the procedure.
Conclusions Did the forces add up to zero in both of your diagrams? That is, did the vectors form closed loops? While it may be unlikely that your diagram shows the total force is exactly zero, it may be zero within the error of your measurements. Here's how to determine this.
For each case, draw a dotted line from the head of the last fixed vector to the tail of the first vector. This represent the "ideal" vector that would exactly close the diagram. (It should be very close to the determined vector.) Measure the length of this ideal vector. Convert this length to the mass it repesents. Within error, is this mass value the same as that of the determined vector? (For example, if the value of the vector you determined was 143 ±3 gm and the value for the ideal vector was 145 gm, then you can conclude that, within error, the values are the same.)
Next, measure the angle between the determined vector and the ideal vector. Is the ideal vector within the error in the angle for your determined vector?
If both the mass and angle of the ideal vector are within the error range of the determined vector, then you may conclude that, within error, your results show that the total force is zero.