The Addition and Resolution of Force Vectors using the Force Table

Page 1 of 6Print these lab instructions before coming to lab.You will need to use all your lab time to do the solutions below.Lab #7The Addition and Resolution of Force Vectorsusing the Force TablePhysics 107, Spring 2014Note: Follow the Vector Addition#1 sample for the VectorAddition#2 and Vector Addition#3 Examples.Student:Partner 1:Partner 2:Partner 3:________________________________________________________________________________________________________________________Lab Date: ________________________Due Date: ____________________________

Page 2 of 6The Addition and Resolution of Force Vectors using the Force Table1.0) Introduction: In this lab, you will use the graphical “tip to tail method” to calculate the resultantvector coming from the addition of two or more vectors. The resultant vector will also be calculatedmathematically by resolving each force vector into its X and Y components and adding them to getthe resultant vector.2.0) Equipment:1) Welch force table2) force table pulleys3) force table center ring4) force table center pin5) roll pulley string (length as needed)6) weight hooks, (50 g hook base)7) 100 g weights8) 50 g weights9) assortment of additional 100, 50, 20, 10, and 5 gram weights10) sheets paper for diagrams11) circular or semi-circular protractor12) cm rule, and13) calculator, pencils, and loose leaf paper as needed3.0) Procedure:3.0.1) Leveling of force table - Check to see of your force table is level. Make any neededadjustments by means of leveling screws at the base of the table.All force vectors angles will be referenced to the right horizontal, or “0 0 Standard PositionAngle” as discussed in lecture. All angles will be measured in a counter-clockwise direction fromthe right horizontal and will be assigned a positive angle.3.0.2) Symbology - Any vector which has yet to be resolved into its x and y components will belabeled with the following notation for the magnitude and direction of the vector. A force vectorof 3920 Newtons pointing 30° above the left horizontal, would be represented as:F = 3920 N / 150°In word processing text, a force vector symbol is shown in boldface with the standard positionangle underlined. On hand-drawn diagrams or hand written analysis you can place a smallarrow above the vector symbol to emphasize that it is a vector.Example: FThe magnitude of the force exerted by any one hook and weight combination will be calculated as:f = ma y = mgwhere, m is the mass of the weights and hook in kg (kilograms),g is the uniform acceleration constant of near-earth, 9.8 m/s 2 , andf is the force in N, Newtons, or kg-m/s 2 .

Page 3 of 63.1) Vector Addition #1 – In the first addition we will be adding two vectors,and,F 1 = m 1·g / θ1° = (0.250 kg)·( 9.80 m/s 2 ) / 30° = 2.45 N / 30°F 2 = m 2·g / θ2° = (0.250 kg)·( 9.80 m/s 2 ) /120° = 2.45 N / 120° .3.1.1) Graphical Analysis #1: “Tail to Tip” graphical analysis – Draw F 1 on a Cartesian coordinatesystem drawn on a piece of paper using a rule and protractor using a scale of 0.4 N per cm ( 0.4Newtons per centimeter). Label it with it’s symbolic name, F 1 and include the magnitude anddirection as shown above. Now, at the tip of the arrowhead of the first vector, draw the secondvector to the same scale. Label F 2 in a similar fashion.Create the resultant vector by drawing a vector from the tail of the first vector to the arrowheadof the second vector. Label it F R and determine the F R vector by measuring its length andmultiplying it by the scale of 0.4N/cm. Measure its standard position angle (relative to righthorizontal) with a protractor.Label the upper right corner with your name and “Vector Analysis #1.”3.1.2) Analytical Analysis – In a relatively open area on your “Vector Analysis #1” paper, show thevector addition of these two force vectors by converting each vector into its x and ycomponents. Neatly create an organized table that includes each vector and it’s correspondingx and y components. Use row and/or column headings to clearly label all values. Add the x andy components of F1 and F2 to find the x and y components of your resultant vector, FR. Neatlycalculate the magnitude and direction of FR as we have previously done. Clearly label this resultappropriately.3.1.3) Experimental Analysis – Set a center pin in the receptacle on the center of the force table(see Fig. 3-1 on the next page). Place a ring over the pin so that it lays flat on the table. For F1,clamp a pulley at 30° relative to the right horizontal. Tie a long enough pulley string between the50 g weight hook and the center ring to drape the hook appropriately over the pulley as shownbelow. When you are ready to perform the experimental force analysis as indicated below, youwill place two 100 g weights on the weight hook yielding a total mass of approximately 250 g(0.250 kg) yielding a force of 2.45 Newtons. Use the actual total mass for your weightcalculation. You will ignore the mass and weight of the ring and string. Do not place the 100 gweights on the hook now, wait until instructed to do so in the next steps.

Page 4 of 6Figure 3-1 A force table, with it’s top shown vertically for demonstration. In the laboratory, the table willbe horizontal, and a more compact style of pulley clamp will be used.For F2, attach a pulley clamped at 120° relative to the right horizontal, and a string and weight hook tiedto the center table ring. Once again, as indicated below, when you are ready to perform theexperimental force analysis, you will place two 200 gram weights on the weight hook yielding a totalmass of 250 grams (0.250kg) yielding a force of approximately 2.45 N.The resultant of two or more force vectors is found by balancing the forces with another force attachedto the ring centered on the central pin. The balancing force is not the resultant vector FR. Rather, it isthe “equilibrant” force vector E, the force that balances the resultant of the other forces holding the ringin equilibrium.The equilibrant vector E is the vector force of equal magnitude, but the opposite direction (i.e.,displaced 180°) of the resultant force FR. Accordingly,or,E = -FRFR = -E = |E|/ θE° + 180 0(note the addition of 180 0 . Why is this needed?)Clamp a pulley at the angle 180° from the angle you believe you will find the resultant vector FR. Add astring and a hook and drape it over the pulley. At this point, add masses to the hooks for F1 and F2 asappropriate. Now add masses to the hook associated with force E, until the system appears to bebalanced around the pin on the table. The ring should be able to sit on the center of the table withoutthe support of the pin. When your system is in balance, record the mass of the equilibrant anddetermine the weight associated with that mass. This weight produces the equilibrant force vector E.Draw this vector E on your graphical solution. Using vector E, determine vector FR.Place your graphical, analytical, and experimental results in Table 3-1.Show your force table results to the lab instructor for verification and/or corrections.

Page 5 of 63.2) Vector Addition #2 – In the second addition you will be adding two vectors,and,F 1 = m 1·g / θ1° = (0.200 kg)·( 9.8m/s 2 ) / 20° = 1.96N / 20°F 2 = m 2·g / θ2° = (0.150 kg)·( 9.8m/s 2 ) /80° = 1.47N / 80° .for these values of F1 and F2, repeati) the graphical “tail-to-tip” analysis of section 3.1.1ii) the hand analysis of section 3.1.2, and,iii) adjust masses and use the experimental force table analysis of section 3.1.3.Enter your results in Table 3-1 as appropriate.3.3) Vector Addition #3 – In the third addition you will be adding three vectors,and,and,F 1 = m 1·g / θ1° = (0.100 kg)·( 9.8m/s 2 ) / 30° = 0.980N / 30° ,F 2 = m 2·g / θ2° = (0.200 kg)·( 9.8m/s 2 ) /90° = 1.96N / 90° , and,F 3 = m 3·g / θ3° = (0.300 kg)·( 9.8m/s 2 ) /330° = 2.94N / 330° .Find the resultant force vector FR, brought about by the addition of these three vectors using:i) the graphical “tip to tail” analysis of section 3.1.1ii) the hand analysis of section 3.1.2, and,iii) the experimental force table analysis of section 3.1.3.Enter your results in Table 3-1 as appropriate.3.4) Vector Resolution – You are given a force vector ofF = m 1·g / θ1° = (0.300 kg)·( 9.8m/s 2 ) / 60° = 2.94N / 60° .Note: This problem is not the same as the previous examples. You are given a single force and willcalculate the x and y components of the vector. You will then determine the equilibrant vectors to eachof the two components to see if it balances the single original force. You will resolve this vector into its xand y components, (i.e., F X and F Y ) with the following procedures:3.4.1) Graphical Resolution of a Vector - Draw F on a Cartesian coordinate system drawn on apiece of paper using a rule and protractor using a scale of 0.4 N per cm ( 0.4 Newtons percentimeter). Label it with its symbolic name, F as indicated above. Drop a vertical to the x axisfrom the tip of the vector arrowhead. The displacement along the x axis is the horizontalcomponent Fx; measure it with your rule and use your scaling factor t determine it value inNewtons . Draw a horizontal from the y axis to the arrowhead; the displacement along the yaxis is the vertical component Fy; once again determine it with a rule and your scaling factor.Label the upper right corner with your name and “Vector Resolution Example”3.4.2) Analytical Analysis – Compute Fx and Fy by mathematically calculating the x and ycomponents of vector F.3.4.3) Experimental Analysis – Clamp pulleys at 60°, 180°, and 270° on the force table. Place a totalof about 300 grams (0.300 kg) on the 60° weight hanger including the hanger; this is force F.Place weights on the 180° hanger and the 270° hanger until the system is in equilibrium. Theforce at 180° is the equilibrant of Fx; that is to say Fx has the same magnitude but points in theopposite direction, 0°. The force at 270° is the equilibrant of Fy; Fy has the same magnitudebut points in the opposite direction, 90°.Enter your Fx and Fy component magnitudes from all three methods in Table 3-1.Show your force table results to the lab instructor for verification and/or corrections.

Page 6 of 6VectorAddition #1Table 3-1 Results TableAddition of Vectors (Enter Addition Results using the correct notation.)Force(s) Graphical FR Analytical FRF1 = 2.45N / 30°F2 = 2.45N / 120°Mass (MR)needed toproduce FR(& E R )Experimental FR(Was E R = -F R ?)(Circle one)Record your E R ifsignificantly differentFR = FR = MR = E R = -F RorE R ≠ -F RE R = ________InstructorVerification(Initials)VectorAddition #2F1 = 1.96N / 20°F2 = 1.47N / 80°FR = FR = MR =E R = -F RorE R ≠ -F RE R = ________VectorAddition #3VectorResolutionF1 = 0.98N / 30°F2 = 1.96N / 90°F3 = 2.94N / 330°FR = FR = MR =E R = -F RorE R ≠ -F RE R = ________Resolution of a Vector into Components (Enter components Fx and Fy as magnitudes only.)Fx =Fx =Mx =E X = -Fx & E Y = -FyorF1 = 2.94N / 60°Fy =Fy =My =E X ≠ -Fx or E Y ≠ -FyE X = ________E Y = ________Are the resultant vectors (or components) the same for each solution method used? Account for differences appropriately.Staple your graphical & analytical (mathematical) solutions for each example problem to this lab sheet in theorder shown in the table above. Be sure your name is on each page, along with the appropriate Addition # orResolution label.9/9/2010 M. Loudis, MSC(Rev. H) 03/27/14 A.Thompson, MSC