The Addition and Resolution of Force Vectors using the Force Table

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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.
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The Addition and Resolution of Force Vectors using the Force Table

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