Vector Forces (last edited December 29, 2011)

Dr. Larry Bortner

 Learning Objectives

Demonstrate the principles of static equilibrium by adding force vectors of weights balanced in two dimensions.

Add two vectors graphically with a triangle, protractor, straight edge, and compass.

Draw arrows on an Excel plot.

 Purpose

To experimentally verify the vector laws of addition. To demonstrate that a system in equilibrium has no net force on it (i.e., the vector sum of the forces must be zero).

 Background

A two-dimensional vector is a mathematical quantity that needs two numbers to define it unambiguously. This is analogous to locating a point in a plane, where the tail of the vector is at the origin and the pointed head is at the point (see Figure 1).

Figure 1

Figure 1 2D displacement vector.

 Vector Representation

Typically for vectors the two numbers are a magnitude and a direction. For a displacement vector from the origin, this is the length (r in the example below) and the angle θ measured counter clockwise from the x-axis. This is equivalent to giving the polar coordinates of the point at the head of the vector. We’re probably all more familiar with giving the xy or Cartesian coordinates of a point and we can define a vector similarly. We give the x- and y-components. These are the lengths of a vector along the x-axis and one along the y-axis, the vector sum of which is the original vector. As pictured, the sum of these two right-angled vectors is the diagonal across the rectangular box suggested by the two components. Finding the magnitudes of these two vectors is called breaking a vector down into components.

 

For a displacement vector you can look at the vector representation as a set of directions to go from the origin to a particular point. The Cartesian description tells you to go a certain distance due east (or west), then turn and go another distance due north (or south). Polar coordinates tell you to go a certain distance in a particular direction. With both sets of direction you end up at the same location.

 

For a vector  at an angle θ measured in the counterclockwise direction from the positive x-axis that has x- and y-components x and y, the transformations between the two descriptions is

 

(1)                     

 Vector Rules

If you are adding a bunch of vectors, the analytical technique is to


1.     Break all the vectors down into components.

 

2.     Add all the x-components as straight numbers.

3.     Add all the y-components similarly.

4.     Convert back to the magnitude and direction description to state the final single vector.

 

Two vectors are equal if and only if they have the same magnitude and direction or, equivalently, if both the x-components are equal and the y-components are equal. Note that equal vectors do not have to occupy the same space. This means that as long as we keep the same orientation, we can put a vector anywhere we want it.

 Vector Forces

According to Newton's second law of motion, a body undergoes an acceleration that is directly proportional to the net force exerted on it. Since both these quantities are vectors, this means that the net force and acceleration are in the same direction and their magnitudes are directly proportional to each other. In static equilibrium, a body is not moving. Obviously, then, the acceleration is zero and from Newton's second law the net force must also be zero.

 

In this experiment, we apply forces to an object so that it is in an equilibrium condition, then measure the vector forces and sum them to see if they do indeed add to zero.

 

The apparatus used in this experiment is called a force table (Figure 2). The table consists of a circular top supported by a heavy tripod base. There is a small peg located at the center of the top and the perimeter of the table is graduated in degrees. Forces are applied to a small ring by means of strings connected over pulleys to weight hangers. By varying the total mass on each string as well as the direction at which each string acts, one can adjust the equilibrium position of the ring so that its center is the peg. This equilibrium configuration is the only one where the angles measured along the edge signify the correct direction of each string.

 

Figure 2

Figure 2 The force table.

    

There will be four strings and pulleys in the present system (see Figure 2; the fourth string is not used here but you will be using it). The vertical tension in each string is equal and opposite to the weight of the supported mass. Each pulley redirects this tension onto the ring. Therefore the net force acting on the ring can be written as

 

(2)                     

 

The vector sum has been set equal to zero because the acceleration of the ring is zero, as mentioned above. If all of the strings lie in a common plane, we can treat the forces in Eq. 2 as two dimensional vectors.

 

With the force table:


 

       The origin is at the center where the peg is.


 

       The positive x-axis is a ray starting at the peg and containing the θ = 0° mark.


 

       The positive y-axis is along a ray from the center through the θ = 90° mark.


 

 

Breaking the individual vectors into components, we rewrite Eq. 2 as:

 

(3)                     

 

where , θn is the counterclockwise angle between  and the x-axis, and and  are the unit vectors along the x- and y-directions, respectively. Note that for a vector to be zero, all of its components have to be zero. This leads us to the set of equations

 

(4)                     

 

These equations can be checked experimentally by setting up four arbitrary weights at arbitrary angles on the force table, as long as the ring is centered. The four angles are measured by the angular position of the pulleys. The magnitudes of the four vector forces acting on the ring are just the total weights hanging from the strings. Since gravity is constant to many significant figures in the lab, we’ll just use force units of gram weights, where the acceleration of gravity g=1:

 

(5)                     

 

where mn is the nth mass.

 

The method of evaluating  by calculating Fx and Fy according to Eqs. 4 is the analytical or component method. It is the approach used most commonly in textbooks for adding vectors.

Note that subtracting a vector is the same thing as adding a vector of the same magnitude in the opposite direction.

 

 Graphical Vector Addition

There are two methods of adding vectors graphically. One way is the triangle method, where the second vector that is added to the first is drawn with its tail at the head of the first vector, as shown in Figure 3. The vector sum or resultant of the two vectors  and  is the vector drawn from the tail of  to the head of .

Figure 3a

Figure 3 Triangle method of graphical vector addition. (Head-to-tail)

 

In cases where the tails of the vectors share a common point a better technique is the parallelogram method as shown in Figure 4. As drawn these vectors are two adjacent sides of a parallelogram. Parallel lines of equal length can be added as shown to complete the parallelogram. The resultant is then the diagonal from the common tail point to the opposite corner. In practice you only need one of the other sides of the parallelogram. In essence, this is a variation of the triangle method; you are redrawing one of the vectors so that you can complete the correct triangle.

Figure 3 triangle.jpg

Figure 4 Parallelogram method of graphical vector addition. (Tail-to-tail)

 

As with numbers, vector addition is a binary process (you add two at a time) and it is commutative (the order doesn’t matter;  ). This graphical method can be used for more than two vectors; add any two, then add the third to the resultant, and continue until you run out of vectors.

 

To get accurate results in graphically adding vectors, you must draw the direction and magnitude of the vectors as precisely as possible. For this reason, the parallelogram method is be  tter than the triangle method because you want to depict the angles measured in the same coordinate system. The way to do this is in practice is to use a straightedge and drafting triangle as shown in Figure 5.

 

Figure 4

Figure 5 Using a triangle and straightedge to move a vector.

The drafting triangle is a right triangle, with two sides and a hypotenuse. Label the longer side edge #1 and the shorter side edge #2.

 

Take these steps to graphically add two vectors using a straight edge, triangle, and compass, assuming that the vectors are already drawn tail-to-tail:

1.     Align edge #1 with the vector to be moved, . The vertex opposite edge #1 should be pointed in the same general direction as the vector to be added to, .

2.     Put the straight edge along edge #2.

3.     Slide the triangle and the straight edge as one unit along the length of  until the tip of the arrow depicting  falls on the adjoining line of the triangle and the straight edge.

4.     Holding the straight edge in place, slide the triangle along the straight edge until the right angle outer vertex (where the lines containing edge #1 and edge #2 intersect) coincides with the head of the vector to be added to. (Second picture in Figure 5.)

5.     Draw a line along edge #1 at least as long as  .

6.     Fix the radius of a drawing compass to be the length of the originally drawn  .

7.     Mark this length along the line drawn in step 5, starting at the tip of .

 

Try this process with Figure 4.

Experimental Technique

Static friction in the pulley bearings can keep the pulleys from transferring the full amount of tension to the ring. The equilibrium position of the ring is made uncertain by the randomness with which static friction manifests itself in the system. Before passing final judgment on the equilibrium position of the ring, it is good experimental practice to lightly nudge one of the weights a few times. This gives the system a chance to assume its most appropriate configuration and thereby yield the most accurate results.

 Vectors and Uncertainty

Two numbers, the magnitude and the direction, are needed to measure a 2D force vector as in today's experiment. Both of these numbers will have uncertainties associated with them:

       the uncertainty in the magnitude, and

       the uncertainty in the direction  

 

These uncertainties are used to express the uncertainty in the vector , which is also a vector. But we want to express this vector uncertainty as a single number, that is, a scalar, not a vector. So we report the magnitude of the uncertainty vector, which is different from the uncertainty of the magnitude of the vector This is a calculated value, given by

 

(6)                     

Here we have introduced the shorthand convention of using a capital U to mean the magnitude of the uncertainty of a vector.

An experimental value of u{θ} is determined by finding the smallest change in angle of the pulley that disturbs the equilibrium position of the ring around the peg. Similarly, we get a value of u{f} by finding the smallest mass that can be added to the weight hanger that moves the ring.

 

Propagating the error of the net force in Eq. (2), we get

 

(7)                     

 

Because we expect friction to be the dominant source of error and because the amount of friction in any particular pulley is expected to vary linearly with the size of the supported load, it is reasonable to assume that the relative (percentage) errors of all four forces will be equal. Let’s call this ratio κ (kappa):

 

(8)                     

 

Combining Eqs. 7 and 8 gives the uncertainty of our resultant F:

 

(9)                     

 

Once this is known, we can determine if our experimental observations are consistent with the predictions of Newton's second law. That is, we compare the magnitude of the resultant with zero.

 Procedure

You need the following items:

*        force table


*        slotted mass sets (at least 2)

*        triple beam balance

*        drafting triangle

*        straightedge (ruler)

*        drawing compass

*        protractor

*        colored pencils

 

Your instructor assigns to each student group specific values in degrees for θ1 and θ2 when the workstations are assigned. Your objective is to center the ring by varying the four load masses and the two remaining pulley positions (angles).        

 


 

1.     Position two pulleys at the assigned angles. These pulley positions must remain fixed during the experiment.

2.     Choose initial values for θ3 and θ4, with θ2 < θ3 < θ4. Follow the allowed ranges on the screen. (That is, θ3 and θ4 cannot be within ±15° of exactly opposite the first two angles.) Position the two remaining pulleys at these angles. Place the ring over the central peg and run the support strings over the pulleys.

3.     Add to each hanger some initial additional mass (between 100 and 300 g).

4.     Adjust the four masses and θ3 and θ4 to get the ring in its proper equilibrium configuration. Keep the added mass to any one hanger greater than 100 g and keep the angles within the allowed ranges.

5.     Choose one of the four masses. Determine the smallest mass that appreciably changes the equilibrium over the peg. Record this mass increment as u{fn}, where n is the number of the mass you chose.

6.     Remove the mass increment. Now, for the same mass, find the smallest change in angular position of the pulley that disturbs the equilibrium. Record this u{θn} in degrees.

7.     Return the system to its proper equilibrium position. Nudge the weights a few times to overcome static friction to make sure the system returns to its proper configuration. Fine tune the angle and masses as necessary.

8.     When finished, record the angles in degrees.

9.     Remove each hanger with its associated weights and measure the total mass (the hanger and everything on it) that had hung from each string, using the balance. Record the values of each of the four masses in grams.

Why not just add everything up and record that number? It would be a lot simpler. But in terms of experimental technique and establishing protocols, we always want to make an absolute distinction between observation and analysis, between measurement and calculation, no matter how trivial that calculation. We need separate records of all measurements and all calculations so we that can go back and check if needed.

 

 Analysis

Again, use gram-weights (g) as the force unit in your calculations, not newtons or dynes. Note that all of the following is a part of your Data Analysis and is not raw Data. (It does not go on your data sheet.)

 

 


1.     On a blank sheet of paper (not your data sheet) plot a vector diagram showing the x- and y-axes in one color and the force vectors f1, f2, f3, and f4 in separate colors. (All vectors start at the origin.)

a.     Use an appropriate scale factor like 1 inch = 100 g.

b.     Use decimal inches on the ruler.

c.     Leave enough space around the diagram to use the parallelogram method to determine the net force vector.

2.     Use the parallelogram method to find a graphical indication of F.

a. There are two vector additions you need to perform here, the first two, then the remaining two. If you were to attempt to add the results of these two operations, it would be difficult; you would have two large vectors that are close to 180° apart, making a very thin parallelogram

b.     Draw and label the resultant of each of these operations (each having a different color). The color of the moved side of the parallelogram should be the same as the original side.

c.      You must finish this and hand it in to the TA by the end of class.

d.  Draw and label the resultant of each of these operations (each having a different color). The color of the moved side of the parallelogram should be the same as the original side. In your <em>Experimental Results and Conclusions</em> discussion, you must comment on how close these two results are equal and opposite.

3.     There is no template for this experiment. Starting from the Excel Template> generic lab spreadsheet, input your data in shaded cells, with adequate headings and clear indications of what units each quantity are in.

4.     Use the component method to sum the vectors , , , and , obtaining the resultant vector .

5.     Use Eqs. 6, 8, and 9 to calculate the magnitude of the uncertainty in the net force, U{F}.

6.     Is F consistent with Newton's second law to within experimental uncertainty? 

 Vector diagram in Excel

7.     In the spreadsheet, plot the calculated components of the forces as single points.

g.     There are 4 points. Highlight your x- and y-components (They should be in adjacent columns.), Click on Insert> Charts> Scatter> Scatter with only Markers.

h.     Click on Chart Tools> Design> Chart Layouts> Layout 1.

i.      Move and resize the graph. Enter appropriate chart and axes titles.

j.      Click on Chart Tools> Layout> Insert> Shapes and choose the single-headed arrow (Figure 6).

 

layout arrow

Figure 6 Navigating to a Drawing Toolbar arrow to place on a graph.

 

k.     Draw an arrow from the origin to one of the four points.

l.      Repeat this for the other 3 points.

m.   With the chart still selected, click on Insert> Text Box then click near the head of one of the arrows. Type in a descriptor such as f1. (All of the formatting is the same in a text box, so you can’t have subscripts.)

n.     Repeat this for the other 3 arrows or copy a text box (Ctrl click and drag the border of a box.) and edit the contents.

 Questions

1.     If all the hangers have the same mass, is it possible to neglect their weights because their contributions “cancel out” when the vector forces representing the string tensions are added together?  Explain

2.     Propagate the error of the expression a=gh/L, where g is a constant and h and L are measured quantities having uncertainties u{h} and u{L}, respectively.