Lab Report M-2

Community College of Philadelphia Philadelphia, Pennsylvania Experiment Number: M-2 Title: Force Vectors Written by: Physics 140, Section 003 Lab Instructor: Professor Indu Das Date: February 04, 2022

Experiment M-2 2 Purpose The purpose of this experiment is to determine the vector resultant of several concurrent, coplanar forces and to confirm that this resultant accurately predicts the combined effect of these forces. Procedure * Part I: Force Board 1. Mark the three spring balances F 1 , F 2 , and F 3 in counterclockwise order. Place each balance flat on the table, apply a small tension force, slowly reduce the force to zero, and take a zero reading. Mark the zero reading on the piece of masking tape attached to the balance and use that reading to correct all later readings before entering them on the data sheet. 2. Place the three balances on a piece of plain 8.5" x 11" paper on the force board, and attach the chains into the slots that meet the following 3 conditions: (1) The angle between F 1 and F 2 should be between 30  and 80  for trial I. (2) At least one force should be over three-quarters of the full-scale reading (over 15 N). (3) Each of the other two forces should be over one-fourth of the full-scale reading (over 5 N). 3. Adjust the paper under the balances until the center of the paper is beneath the ring and until the long dimension of the paper is parallel to F 1 . Lift the ring a very small distance from the table to slightly increase the tension force in each balance. Gently lower the ring and release it. 4. Near each balance and near the edge of the paper, write the reading of the balance and show the correction for the zero reading. Draw a dot on the paper directly below the center of the ring. Draw thin dim pencil lines along the long edges of the balances. Write the trial number on the sheet of paper and remove the paper from the force board. Draw thin dim pencil lines from the center dot parallel to each pair of lines for each balance. 5. Draw the three vectors in dark wider lines over the dim lines. 6. Determine graphically the magnitude and direction of the resultant R 12 of forces F 1 and F 2 by using the parallelogram method. Measure R 12 with a cm scale and apply the scale factor to calculate its magnitude in newtons. Use a protractor and measure the

Experiment M-2 4 Data Sheets * Part I: Force Board Trial F 1 F 2  12 Calculated Resultant Measured Equilibrant Percent Difference in Magnitude Percent Error in Angle R 12  1R F 3  13 N N degrees N degrees N degrees % % 1 11.5 8 64 16.6 25.6 17 154 2.38 -0.22 2 10 8 88 13 37.8 14 147 7.4 2.67 3 10 12 52 19.8 28.6 22.5 155 12.8 2 * Part II: Force Table Masses: (Include the 5 g mass of the hanger in addition to the mass on each hanger.) m 1 = 55 grams = 0.055 kg m 2 = 205 grams = 0.205 kg m 3 = 155 grams = 0.155 kg m 4 = N/A Forces: (Weights of the masses) F 1 = 0.539 N F 2 = 2 N F 3 = 1.52 N F 4 = 1.19 N Angles: (Measured counter-clockwise from +x)  1 = 0   2 = 75   3 = 210   4 = 102.6 

Experiment M-2 5 Calculated Resultant of F 1 , F 2 , and F 3 : R 123 = 1.20 N  R = 102.8  % difference in magnitude between F 4 and R 123 : 0.84 % % error in angle between F 4 and R 123 : -99.9%

Experiment M-2 7 Calculations  Part I: Force Board The formulas used in this part: R 12 = √ R x 2 + R y 2  1R = tan − 1 | R y R x | % Difference in Magnitude = | F 3 − R 12 | F 3 + R 12 x 200% % Error in Angle = ❑ 1 R + | ❑ 13 | − 180 180 x 100% * Trial 1: R x = F 1 + F 2 cos(64  ) = 11.5 + 8cos(64  ) = 15 (N) R y = F 2 sin(64  ) = 8sin(64  ) = 7.19 (N) R 12 = √ 15 2 + 7.19 2 = 16.6 (N)  1R = tan − 1 | 7.19 15 | = 25.6  % Difference in Magnitude = | 17 − 16.6 | 17 + 16.6 x 200% = 2.38% % Error in Angle = 25.6 + | 154 | − 180 180 x 100% = -0.22% * Trial 2: R x = F 1 + F 2 cos(88  ) = 10 + 8cos(88  ) = 10.3 (N) R y = F 2 sin(88  ) = 8sin(88  ) = 8 (N) R 12 = √ 10.3 2 + 8 2 = 13 (N)  1R = tan − 1 | 8 10.3 | = 37.8  % Difference in Magnitude = | 14 − 13 | 14 + 13 x 200% = 7.4% % Error in Angle = 37.8 + | 147 | − 180 180 x 100% = 2.67%

Experiment M-2 8 * Trial 3: R x = F 1 + F 2 cos(52  ) = 10 + 12cos(52  ) = 17.4 (N) R y = F 2 sin(52  ) = 12sin(52  ) = 9.5 (N) R 12 = √ 17.4 2 + 9.5 2 = 19.8 (N)  1R = tan − 1 | 9.5 17.4 | = 28.6  % Difference in Magnitude = | 22.5 − 19.8 | 22.5 + 19.8 x 200% = 12.8% % Error in Angle = 28.6 + | 155 | − 180 180 x 100% = 2%  Part II: Force Table * Forces: F = mg (g = 9.8 m/s 2 ) F 1 = 0.055 x 9.8 = 0.539 N F 2 = 0.205 x 9.8 = 2 N F 3 = 0.155 x 9.8 = 1.52 N F 4 = 1.19 N * Calculated Resultant of F 1 , F 2 , and F 3 : R x = F 1 + F 2 cos(75  ) + F 3 cos(210  ) = 0.539 + 2cos(75  ) + 1.52cos(210  ) = - 0.26 (N) R y = F 2 sin(75  ) + F 3 sin(210  ) = 2sin(75  ) + 1.52sin(210  ) = 1.17 (N) R = √ R x 2 + R y 2 = √ (− 0.26 ) 2 + 1.17 2 = 1.20 (N)  R = tan − 1 | R x R y | = tan − 1 | − 0.26 1.17 | = 12.8  R is in the quadrant II, so  R = 12.8  + 90  = 102.8 

Experiment M-2 10 Answer to questions 1. Has this experiment demonstrated that the vector resultant of forces predicts the combined effect of these forces? Explain. Yes, this experiment has because the vector resultant of forces is the sum of all the forces and it has the same effect as the original forces. 2. In Part II, what factors could have resulted in the deviation of calculations from experimental values? The factors that could have resulted in the deviation of calculations from experimental values are friction in the pulleys and non-radial cord. 3. In Part II, could all four pulleys have been placed in a single quadrant and still be in equilibrium? In two adjacent quadrants? Explain. No, all four pulleys could not be placed in a single quadrant/in two adjacent quadrants and still be in equilibrium. The reason is that forces would be concentrated on one side and cause force difference, so they could not be in equilibrium.

Experiment M-2 11 Analysis & Discussion of Results Force board and force table were used to determine the vector resultant of several concurrent. In part I, the resultant of two forces is approximately equal and opposite the third force. The percent difference in magnitude between the calculated resultant and the measured equilibrant of the three trials is acceptable, and the percent error in angle of the three trials is low, which is less than 3%. In part II, the calculated magnitude of the fourth force and its angle are approximately equal to measured result (with a deviation of 0.01 for magnitude and 0.2 for angle). The source of errors could be from incorrect manipulations during the experiment such as setting up the three bring balances and four pulley or drawing parallel lines/parallelogram. After completing this experiment, it is confirmed that this resultant accurately predicts the combined effect of these forces, so the purpose of this experiment was achieved.