The Addition and Resolution of Force Vectors using the Force Table

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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.
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