Table 1 Graphically table solution Analytical Table solution % Error of magnitud
ID: 1532500 • Letter: T
Question
Table 1
Graphically table solution
Analytical Table solution
% Error of magnitude for Experimental compared to Analytical = ____________ %
% Error of direction for Experimental compared to Analytical = ____________ %
% Error of magnitude Graphical compared to Analytical = ____________ %
% Error of direction Graphical compared to Analytical = ____________ %
Question
1.A Find the resultant of these two applied forces by scaled graphical construction using the parallelogram method. Using a ruler and protractor, construct vectors whose scaled length and direction represent F1 and F2. For example, a convenient scale might be 1.00 cm = 0.100 N. Be careful to note that all directions are given relative to the force table, and this must be taken into account in the graphical construction to ensure the proper angle of one vector to another. Read the magnitude and direction of the resultant from your graphical solution and record them in the appropriate section of Graphical Table.
1.B Using trigonometry, calculate the components of F1 and F2 and record them in the Analytical solution portion of Calculations Table 1. Add the components algebraically and determine the magnitude of the resultant by the Pythagorean Theorem. Determine the angle of the resultant from the arc tan of the components. Record these results in Analytical Table.
1.C Calculate the percent error of the magnitude of the experimental value of FR compared to the analytical solution of FR. Also calculate the percent error of the magnitude of the graphical solution for FR compared to the Analytical solution for FR. For each of these comparisons, also calculate the magnitude of the difference in the angle.
Force Mass(kg) Force(N) Direction (degrees) F1 0.050 0.504 20.0 degrees F2 0.100 0.980 90.0 degrees Equilibrant -1.225 60.0 Degrees Resultant 1.225 248 DegreesExplanation / Answer
A) the resultant of these two applied forces = 1.225 N
B) Net Force in horizontal direction = 0.504 * cos20
= 0.473 N
Net Force in vertical direction = 0.504 * sin20 + 0.98
= 1.152 N
Thus, equivalent force = sqrt(0.4732 + 1.1522)
= 1.245 N
C) the percent error = (1.245 - 1.225)/1.245
= 1.6 %
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