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Force Table with 3 Vectors at Equilibrium

Background

● Background Overview

There are two types of physical quantities: scalars and vectors. The number of attributes required to define a scalar and a vector distinguishes them. A scalar has just a magnitude telling us how large or small something is (or positive or negative); an example of a scalar quantity is temperature or speed (m/s)

A vector, meanwhile, is a quantity that is described by both a magnitude and direction. Examples of vectors are velocity (speed but with a direction; m/s at some angle) or force (a push or pull on an object with a specific magnitude at some direction).

A force vector example: a force may be 100 N in magnitude with direction 90° counterclockwise from the xx-axis. This force vector is written as 100 N @ 90°. We will use typefont of (with a note that bold isn’t always bold when viewing on some machines):

When several vectors act on an object, it is generally desirable to determine the sum of these vectors, called the resultant vector. Suppose force vectors F1\vec{F}_1 and F2\vec{F}_2 act on a body. The resultant R\vec{R} is defined by the vector sum of the two forces, thus

R=F1+F2.\vec{R} = \vec{F}_1 + \vec{F}_2.

If many forces act on the body, then we sum all the forces together

R=i=1NFi.\vec{R} = \sum_{i=1}^N \vec{F}_i.

The resultant force is a single force which can completely represent a number of individual forces acting. When the resultant force is zero, the object is said to be in equilibrium.

There are two methods of vector addition to consider:

● Graphical Method

Vectors F1\vec{F}_1 and F2\vec{F}_2 (Figure 1) are added graphically as follows: Beginning at a convenient point on a piece of graph paper, usually at the origin of a rectangular coordinate system draw one of the vectors as an arrow to scale and pointing in the proper direction. Place the second vector with its tail at the tip of the first, again drawn to scale and pointing in the proper direction. The resultant R\vec{R} is the vector drawn from the tail of the first vector to the tip of the second. The process is illustrated in Figure 2 demonstrating the addition operation does not depend on the order of addition. Thus, like scalar addition,

F1+F2=F2+F1=R.\vec{F}_1 + \vec{F}_2 = \vec{F}_2 + \vec{F}_1 = \vec{R}.
Two force vectors \vec{F}_1 and \vec{F}_2

Figure 1:Two force vectors F1\vec{F}_1 and F2\vec{F}_2

Adding 2 vectors \vec{F}_1 and \vec{F}_2, using the graphical (“tail-to-tip”) method.

Figure 2:Adding 2 vectors F1\vec{F}_1 and F2\vec{F}_2, using the graphical (“tail-to-tip”) method.

It is important that an appropriate scale be selected with which the vectors are drawn (e.g. 1 N = 10 cm). The magnitude of R\vec{R} is determined using a ruler, and the angle θ\theta is measured using a protractor. Since the negative of a vector is merely the vector pointing in the opposite direction, subtraction is addition with the negative vector pointing in the opposite direction. Errors can be significantly reduced by using a scale that makes the drawing as large as possible. Neatness counts!

● Method of Components

The method of components is a much more useful and quantitatively accurate method of vector addition. Each vector is resolved into components along the xx- and yy-axes. That is to say, the vector addition of the two components of the vector is the vector itself. Thus if two vectors are to be added, we add the components along each axis to form the components of the resultant.

Adding 2 vectors \vec{F}_1 and \vec{F}_2 using the method of components.

Figure 3:Adding 2 vectors F1\vec{F}_1 and F2\vec{F}_2 using the method of components.

In Figure 3, the magnitudes and directions are shown for the two force vectors F1\vec{F}_1 and F2\vec{F}_2. The magnitudes of the xx and yy components are calculated for this example as follows:

F1,x=F1cos(θ1)F2,x=F2cos(θ2)=1Ncos(0°)=1N=1Ncos(45°)=0.707NF1,y=F1sin(θ1)F2,y=F2sin(θ2)=1Nsin(0°)=0N=1Nsin(45°)=0.707N.\begin{aligned} F_{1,x} &= \vec{F}_1 \, \cos(\theta_{1}) & \qquad F_{2,x} &= \vec{F}_2 \, \cos(\theta_{2}) \\ &= 1\,\text{N} \cos(0°) = 1\,\text{N} & &= 1\,\text{N} \cos(45°) = 0.707\,\text{N} \\ F_{1,y} &= \vec{F}_1 \,\sin(\theta_1) & \qquad F_{2,y} &= \vec{F}_2 \,\sin(\theta_2) \\ &= 1\,\text{N} \sin(0°) = 0\,\text{N} & &= 1\,\text{N} \sin(45°) = 0.707\,\text{N}. \end{aligned}

Next, add the xx-components

Rx=F1,x+F2,x=1N+0.707N=1.707NR_x = F_{1,x} + F_{2,x} = 1\,\text{N} + 0.707\,\text{N} = 1.707\,\text{N}

Then, add the yy-components

Ry=F1,y+F2,y=0N+0.707N=0.707NR_y = F_{1,y} + F_{2,y} = 0\,\text{N} + 0.707\,\text{N} = 0.707\,\text{N}

The magnitude of the resultant is found using the relation

R2=Rx2+Ry2,R^2 = {R_x}^2 + {R_y}^2,

or

R=(1.707N)2+(0.707N)2=1.848N.R = \sqrt{\left(1.707\,\text{N}\right)^2 + \left(0.707\,\text{N}\right)^2} = 1.848\,\text{N}.

The angle θ\theta specifies the direction of the resultant and it can be calculated by noting that

tanθ=Ry/Rx\tan\theta = R_{y} / R_{x}

and

θ=arctan(Ry/Rx).\theta = \arctan (R_{y} / R_{x}).

For this example illustrated in Figure 4

tanθ=(0.707N)/(1.707N)=0.414,\tan \theta = (0.707\,\text{N}) / (1.707\,\text{N}) = 0.414,
θ=arctan(0.414)=22.5°.\theta = \arctan(0.414) = 22.5°.
Example of \vec{R}, the sum of 2 vectors as x- and y-components, as well as its magnitude and angle with the x-axis.

Figure 4:Example of R\vec{R}, the sum of 2 vectors as xx- and yy-components, as well as its magnitude and angle with the xx-axis.

In this laboratory, an object will be presented with two known forces acting on it. Equilibrium will be established by adding a third equilibrant force F3\vec{F}_3, such that the sum of the three forces is zero (i.e. at equilibrium). Thus, we must first find the resultant R\vec{R} of the two given forces F1\vec{F}_1 and F2\vec{F}_2. Then we can determine the equilibrant, F3=R\vec{F}_3 = -\vec{R}, equal in magnitude and opposite in direction to the resultant R\vec{R}. This is illustrated in Figure 5.

Illustration of the method to determine the force \vec{F}_{3} needed to balance two given forces \vec{F}_{1} and \vec{F}_{2}.

Figure 5:Illustration of the method to determine the force F3\vec{F}_{3} needed to balance two given forces F1\vec{F}_{1} and F2\vec{F}_{2}.

Here F3\vec{F}_3 is the equilibrant force necessary to equilibrate F1\vec{F}_1 and F2\vec{F}_2. Note that the resultant of all the vectors for today’s lab should add up to zero, i.e.

Ftotal=F1+F2+F3=0\vec{F}_{\text{total}} = \vec{F}_1+\vec{F}_2+\vec{F}_3 = 0

since the ring is in equilibrium.

Recall that R=F1+F2\vec{R}=\vec{F}_1+\vec{F}_2, therefore the equilibrant force F3\vec{F}_3 has the same magnitude as the resultant R\vec{R}, but acts in a direction opposite to R\vec{R}. So we now have

F3=R.\vec{F}_3 = -\vec{R}.

We conclude that the force necessary to equilibrate two or more forces is equal and opposite to the resultant of the two (or more) forces. This is additionally illustrated in an example of the force table apparatus we will use in Figure 6.

Apparatus

(Left) Illustration of the force table. (Right) An example of how to determine the force \vec{F}_{3} needed to balance two given forces \vec{F}_{1} and \vec{F}_{2}.

Figure 6:(Left) Illustration of the force table. (Right) An example of how to determine the force F3\vec{F}_{3} needed to balance two given forces F1\vec{F}_{1} and F2\vec{F}_{2}.

The apparatus for this experiment consist of a force table, weight holders, and weights (see Figure 6). The force table consists of a circular tabletop mounted on a vertical rod held in a tripod support with leveling screws. The rim of the circular top has a 360° scale engraved on it along which it is possible to clamp a number of pulleys. At the center of the table is a small ring held in place by means of a removable pin. The ends of three cords are tied to the ring with each cord leading over a pulley and ending with a weight holder tied to its other end. When the forces along the cords acting upon the small ring are balanced, or in static equilibrium, the ring remains stationary. For this lab, the force pulling on the ring is the tension of the string, which is merely translated from horizontal on the table to vertical.

Each hanging mass has an associated weight (force) due to being pulled down by gravity. This weight is balanced by the upward tension force of the vertical part of the cord which is merely translated from vertical to horizontal tension by the pulley. Thus, the force pulling the ring in the direction of the cord is

Fweight=mg=Ftension\vec{F}_\text{weight} = m \cdot \vec{g} = \vec{F}_\text{tension}

For accurate measurements of the angles involved, each cord must be aimed directly at the peg at the center of the ring requiring that the equilibrium condition must be established when the ring is exactly centered around the pin on the table. Use consistently the same counterclockwise protractor.

Experimental Procedure

● Preview & Examples

For today’s lab, two different cases will be assigned involving two given vectors and the determination of the equilibrant vector. A third case involves the determination of the mass of two unlabeled masses by balancing the system from a single known mass.

○ EXAMPLE --- Finding the Equilibrant (First two cases)

For each of the first two cases, you will have two given masses at given angles (direction). Each mass, consisting of a 50 g hanger plus the necessary additional mass, will be hung from a cord routed over a pulley at the assigned angular positions and finally tied to the ring. A third cord, hanger, and pulley assembly is put in place for the third, unknown force vector. Each force is the weight of the hanging mass at the pulley angle. Determine the unknown forces for each of the two cases.

If, for example, you were given the following case to work with

Mass 1:200g@0Mass 2:250g@135\begin{array}{rcl} \text{Mass 1:} & 200\,\text{g} & @\,0^\circ \\ \text{Mass 2:} & 250\,\text{g} & @\,135^\circ \end{array}

you would place a pulley at 0° and add 150 g to the 50 g weight hanger for a total 200 g. As mentioned in (15), the tension on the cord is the same on both sides of the pulley so that the downward pull of gravity on the hanger and masses is equal to the tension in the cord leading to the ring. Thus,

F1=mg=0.20kg×9.8m/s2=1.96N @ 0°.\vec{F_1} = m \vec{g} = 0.20\,\text{kg} \times 9.8\,\text{m/s}^2 = 1.96\,\text{N} \text{ @ } 0°.

Similarly, add 200 g to a 50 g weight hanger and run its cord over a pulley mounted at 135°.

F2=mg=0.25kg×9.8m/s2=2.45N @ 135°.\vec{F_2} = m \vec{g} = 0.25\,\text{kg} \times 9.8\,\text{m/s}^2 = 2.45\,\text{N} \text{ @ } 135°.

Having established the given magnitudes and directions for each of the given forces, F1\vec{F}_{1} and F2\vec{F}_{2}, you would adjust both the amount of mass hanging on cord 3 and its angular position so that the ring is stationary at the center of the table. In order for angular measurements to be accurate, each cord must be on a line that crosses through the center of the table. Sighting along each cord towards the center pin can help you to easily and accurately check this.

To determine the equilibrant vector experimentally F3\vec{F}_{3}, you will record both the angular position (direction) and total mass (to determine magnitude of the balancing force from the total hanging weight) required to balance F1\vec{F}_{1} and F2\vec{F}_{2}. You will then compare this vector to the expected vector based on your theoretical calculations.

○ EXAMPLE --- Determining 2 Unlabeled Masses (Third case)

In the third case, you will experimentally determine two unlabeled masses by balancing the xx and yy components of the force of a single known mass. Place the pulley with the known mass M1=50gM_1 = 50\,\text{g} at θ1=0°\theta_1 = 0°. Experimentally determine the angles θ2\theta_2 and θ3\theta_3 for the unlabeled masses M2M_2 and M3M_3 that result in the ring being in equilibrium. For this case, assume angles θ2\theta_2 and θ3\theta_3 are actual values.

We can calculate the theoretical unlabeled masses using the method of components. You have to solve a pair of equations for the xx and yy components of force. The equation for equilibrium in the xx direction is

i=1n=3Fi,x=F1,x+F2,x+F3,x=(M1+M2cosθ2+M3cosθ3)g=0.\sum_{i=1}^{n=3} F_{i,x} = F_{1,x} + F_{2,x} + F_{3,x} = (M_1 + M_2 \cos\theta_2 + M_3 \cos\theta_3)g = 0.

Similarly, the equation for equilibrium in the yy direction is

i=1n=3Fi,y=(M2sinθ2+M3sinθ3)g=0.\sum_{i=1}^{n=3} F_{i,y} = (M_2 \sin\theta_2 + M_3 \sin\theta_3)g = 0.

In these equations, we are assuming we know M1M_1, θ1\theta_1, θ2\theta_2, θ3\theta_3, and that we want to determine M2M_2 and M3M_3. Of the various ways your could do this, one way is to multiply (19) by sinθ2\sin \theta_2 and (20) by cosθ2\cos \theta_2. The result is:

(M1sinθ2+M2cosθ2sinθ2+M3cosθ3sinθ2)=0(M_1\sin\theta_2 + M_2\cos\theta_2\sin\theta_2 + M_3\cos\theta_3\sin\theta_2) = 0
(M2sinθ2cosθ2+M3sinθ3cosθ2)=0(M_2\sin\theta_2\cos\theta_2 + M_3\sin\theta_3\cos\theta_2) = 0

You can then subtract the yy component (22) from the xx component (21)

(M1sinθ2+M2cosθ2sinθ2+M3cosθ3sinθ2)(M2sinθ2cosθ2+M3sinθ3cosθ2)=0(M_1\sin\theta_2 + M_2\cos\theta_2\sin\theta_2 + M_3\cos\theta_3\sin\theta_2) - (M_2\sin\theta_2\cos\theta_2 + M_3\sin\theta_3\cos\theta_2) = 0

to find an equation where the only unknown is M3M_3:

M1sinθ2+M3cosθ3sinθ2M3sinθ3cosθ2=0.M_1 \sin \theta_2 + M_3 \cos \theta_3 \sin \theta_2 - M_3 \sin \theta_3 \cos \theta_2 = 0.

After algebraically solving for M3M_3, you can use the known values to solve for the experimental M3M_3 and enter the value in your data table for later comparisons to the actual values.

After having algebraically solved for M3M_3 with (24), you could plug that equation back in to (22) to eliminate M3M_3 to find an equation where the only unknown is M2M_2:

M1sinθ3+M2cosθ2sinθ3M2sinθ2cosθ3=0.M_1 \sin \theta_3 + M_2 \cos \theta_2 \sin \theta_3 - M_2 \sin \theta_2 \cos \theta_3 = 0.

After algebraically solving for M2M_2, you can use the known values to solve for the experimental M2M_2 and enter the value in your data table for later comparisons to the actual values.

You would then measure M2,actualM_{2,actual} and M3,actualM_{3,actual} on the scale and enter the values in the table and comare.

● CASE 1 & 2 -- Finding the Equilibrant Vector (Balancing Force)

Table 1:Case 1 & 2 Given Values

Hanger (i.e. Vector)Mass (g)Angle (°)
Case 1
1150
215070°
3??
Case 2
110075°
2200115°
3??

For these first two cases, with given initial values in Table Table 1, experimentally determine the 3rd (equilibrant) vector and compare your results to your expected values.

  1. Create data tables for the first case. NOTE: The data layout for each of the first two cases is the same. Create for the first case and run the whole experiment, then you can copy/paste the same data table for the additional case(s).

    • Common data section with accepted value of gg (9.803 m/s²), mass of the hanger, list of slotted masses with their uncertainties (see table Table 2 later in the procedure), and any other common values. You will reference these values in the calculations.

    • An experimental data table (e.g. ● Case 1 & 2 Experimental Data) to record your experimental results with:

      • With three rows (1 for each of the 3 vectors).

      • Include columns for:

        • mim_i: hanging mass in kilograms (kg)

        • δmi\delta m_i: your estimate of the experimental uncertainty [± value] of the masses in kg

        • FiF_i: calculated magnitude of the force in Newtons (N)

        • θi\theta_i: direction of the force vector in degrees

        • δθi\delta \theta_i: your estimate of the experimental uncertainty of the angle in degrees

        • Fi,xF_{i,x}: calculated xx-component of each force vector in N

        • Fi,yF_{i,y}: calculated yy-component of each force vector in N

    • An analysis table (e.g. ● Case 1 & 2 Analysis - Resultant of Given Vectors & Experimental Total) to analyze the results of your measurements. This is effectively one row as there would be just a single value for each of the variables. The columns should include variables:

      • RxR_x: the xx component of the resultant vector R\vec{R} in N

      • RyR_y: the yy component of the resultant vector R\vec{R} in N

      • RR: magnitude of the resultant vector R\vec{R} in N

      • θR\theta_{R}: direction of the resultant vector R\vec{R} in degrees

      • Fx,total,experimentalF_{x\text{,total,experimental}} & Fy,total,experimentalF_{y\text{,total,experimental}}: total of all vector components in N

      • Ftotal,experimental\vec{F}_{\text{total,experimental}}: total overall magnitude of all vectors which we will also treat as δF3\delta \vec{F}_3 in N

    • A secondary analysis table (e.g. ● Case 1 & 2 Analysis - Theoretical Values) to analyze your expected values. This is also effectively one row as there would be just a single value for each of the variables. The columns should include variables:

      • F3,theoreticalMagnitudeF_{3,\text{theoreticalMagnitude}}: your expected or theoretical magnitude of the equilibrant force in N

      • m3,theoreticalm_{3,\text{theoretical}}: theoretical equilibrant mass in kg

      • θ3,theoretical\theta_{3,\text{theoretical}}: theoretical equilibrant direction in degrees

  2. Starting with the first case, add slotted masses to hangers 1 & 2 to set their masses equal to the given values in Table Table 1.

  3. Loosen the screws of the black pulleys to rotate them around the tabletop to their given angles and retighten.

  4. With hanger 3 (unknown vector), add or subtract slotted masses and scoot the pulley with hanger and masses around the force table until the system is at equilibrium.

  5. Double check the given vectors are still lined up with their given angles. If not, readjust as needed.

  6. Note your mim_i & θi\theta_i values.

  7. Also note both of your estimated uncertainties δmi\delta m_i & δθi\delta \theta_i.

    • For δθi\delta \theta_i, estimate this based on the force table’s increments and your confidence that the ring is still centered on the center pin. Example: move the pulley left or right until you’re no longer convinced the ring is centered; however far you’ve shifted it, that will be your δθi\delta \theta_i range.

    • For δmi\delta m_i, add up the total uncertainty of each of your slotted masses. For today, assume each hanger is exactly 50 g, but the list of uncertainties for the slotted masses is below. Example: if you had measured 123 g from just slotted masses, you could have used one of each 100, 20, 2, 1 g, which would end up giving an uncertainty of 123±(1+0.2+0.04+0.03)123 \pm (1 + 0.2 + 0.04 + 0.03) g or 123±1.27123 \pm 1.27 g:

      Table 2:Slotted Masses

      Mass (g)Uncertainty (g)
      2002.0
      1001.0
      500.5
      200.2
      100.1
      50.1
      20.04
      10.03
  8. Calculate the hangers’ respective forces FiF_i using (15).

  9. Determine the hangers’ respective xx and yy components Fi,xF_{i,x} and Fi,yF_{i,y}. See (4).

  10. In your analysis table, determine RxR_x and RyR_y, the xx and yy components of the resultant vector R\vec{R} of this case’s two given force vectors F1\vec{F}_1 and F2\vec{F}_2.

  11. Determine the magnitude RR of the resultant vector (see Pythagorean Theorem in (7)).

  12. Determine the direction θR\theta_R of the resultant vector. See (10). Use the ATAN2() Excel function to get the angle as measured counterclockwise from 0°.

  13. Determine Fx,total,experimentalF_{x\text{,total,experimental}} & Fy,total,experimentalF_{y\text{,total,experimental}}, the sum totals of the xx and yy experimental components from all of your vectors, i.e. R\vec{R} and F3\vec{F}_3 (see also (2), (5), (6), and (13)). Could also be stated as i=1n=3Fi,x\sum_{i=1}^{n=3} F_{i,x} and i=1n=3Fi,y\sum_{i=1}^{n=3} F_{i,y}.

  1. Determine Ftotal,experimentalF_{\text{total,experimental}} (also treated as δF3\delta F_3), the sum total experimental magnitude of all three of your vectors determined in similar fashion to (7), but with Fx,total,experimentalF_{x\text{,total,experimental}} & Fy,total,experimentalF_{y\text{,total,experimental}}.

  2. In your secondary analysis table, based on the resultant R\vec{R} of the two given vectors, determine your theoretical equilibrant vector F3,theoretical\vec{F}_{3,\text{theoretical}} that would balance the resultant, and its associated expected mass (see (14)):

    • magnitude F3,theoreticalMagnitudeF_{3,\text{theoreticalMagnitude}}

    • direction θ3,theoretical\theta_{3,\text{theoretical}}

    • mass m3,theoreticalm_{3,\text{theoretical}}

  3. COMPARE your experimental results of hanger/vector 3 to the theoretical values. Does F3±δF3F_3 \pm \delta F_3 overlap (and therefore agree) with your theoretical value F3,theoreticalMagnitudeF_{3,\text{theoreticalMagnitude}}? What about m3m_3 and θ3\theta_3? If not, are there significant issues that may be contributing to the discrepancy? Discuss with instructor if so. To be further discussed in Section Post-Lab Submission --- Interpretation of Results.

  4. Repeat for the second case (see Table Table 1) once you have completed the entire analysis procedure for Case 1 and are satisfied in your values and calculations.

● CASE 3 -- Determining 2 Unlabeled Masses

Table 3:Case 3 Initial Given Values

Hanger (i.e. Vector)Mass (g)Angle (°)
Case 3
1 (empty, M1M_1)50 (just hanger, no extra masses)
2 (Pikachu-black, M2M_2)??
3 (corgi-white, M3M_3)??

Experimentally determine the mass of just the Pikachu (black figurine) and the corgi (white figurine) and compare to actual values. The angles for hangers 2 & 3 will be treated as given values once experimentally determined. Reminder, the hangers are 50 g each.

  1. Create data table for this case with columns for (e.g. ● Case 3 Data):

    • Experimental masses mi,experimentalm_{i\text{,experimental}} in kg

    • Given/actual angles θi\theta_i in degrees

    • Actual values for each mass mi,actualm_{i\text{,actual}}

    • The % Difference between the experimental and actual mass values in kg

  2. Place the figurines on their respective hangers and set the empty hanger (m1m_1) to its initial angle found in Table Table 3.

  3. Unscrew the black pulleys to rotate the figurines around the tabletop until you find equilibrium in similar fashion to the first two cases.

  4. Once you’ve found equilibrium, record the given mass m1m_1 and angle θ1\theta_1 as shown in the Table Table 3 for the empty hanger.

  5. Record your experimentally determined angles θ2,Pikachu-black\theta_{2,\text{Pikachu-black}} & θ3,corgi-white\theta_{3,\text{corgi-white}}.

  6. Determine m2,Pikachu-black,experimentalm_{2,\text{Pikachu-black,experimental}} & m3,corgi-white,experimentalm_{3,\text{corgi-white,experimental}}:

    • First algebraically solve (25) and (24) for M2M_2 and M3M_3, respectively. Do this by hand before plugging the equations into your spreadsheet. There should be scratch paper and extra pens & pencils at the front of the room if needed.

    • Calculate your experimental masses for both figurines in your speadsheet using now your now solved equations. Reminders: hangers are 50 g, and there are trigonometric identities you could look up that could simplify the equations for your spreadsheet.

  7. Measure and record the actual masses of each figurine, m2,Pikachu-black, actualm_{2,\text{Pikachu-black, actual}} and m3,corgi-white, actualm_{3,\text{corgi-white, actual}}, with a triple-beam balance.

  8. COMPARE your experimental figurine masses to their actual values. Do they generally agree? Calculate the % difference of your experimental m2m_2 and m3m_3 to their actual measured values (see (26)). What may be contributing to a larger or smaller difference? To be further discussed in Section Post-Lab Submission --- Interpretation of Results.

Post-Lab Submission --- Interpretation of Results

● Finalized Spreadsheets

● Post-lab Writeup

% Difference=Experimental ValueActual ValueActual Value×100%.\text{\% Difference} = \frac{\text{Experimental Value} - \text{Actual Value}}{\text{Actual Value}} \times 100\%.

The Whiteboard

Example data tables are shown below to assist you in building your spreadsheet for this lab. Additionally the original whiteboard summary is at the end of this section.

● Case 1 & 2 Experimental Data

Hanger/Vectormm (SI units)δm\delta m (SI Units)F\vec{F} (SI Units)θ\theta (°)δθ\delta \theta (°)FxF_x (SI Units)FyF_y (SI Units)
1
2
3

● Case 1 & 2 Analysis - Resultant of Given Vectors & Experimental Total

RxR_x (SI units)RyR_y (SI Units)RR (SI Units)θR\theta_R (°)Fx,total,experimentalF_{x,\text{total,experimental}} (SI Units)Fy,total,experimentalF_{y,\text{total,experimental}} (SI Units)Ftotal,experimentalF_{\text{total,experimental}} treated as δF3\delta F_3 (SI Units)

● Case 1 & 2 Analysis - Theoretical Values

F3,theoreticalMagnitudeF_{3,\text{theoreticalMagnitude}} (SI Units)θ3,theoretical\theta_{3,\text{theoretical}} (°)m3,theoreticalm_{3,\text{theoretical}} (SI Units)

● Case 3 Data

Vectormi,experimentalm_{i\text{,experimental}} (SI units)θi\theta_i (°)mi,actualm_{i\text{,actual}} (SI units)% diff. masses
1 (known)
2 (Pikachu-black)
3 (Corgi-white)

● Original Whiteboard Info