Vector Addition of Forces

Introduction:

This lab concentrates on the study of vector addition by graphing and by using the components. After all the calculations have been done a circular force table can be used to check the results.

Materials:
Circular force table, masses plus holders, string, protractor, and four pulleys.

Procedure:
1.) Gather the given magnitudes and angles and use the scale of 1 cm = 20 g. Then make a vector diagram showing the forces and find their resultant. Determine the magnitude and the direction of the resultant force using a ruler and protractor.  

2.) Make a second vector diagram and show the same three forces.Find the resulting vector but this time using components. On the diagram show both the resultant and the components plus the force that is required to cancel out the resultant.


3.) Mount the three pulleys on the edge of the force table at the given angles. Attach the strings to the center ring so that each run over the pulley and attach the mass holder as well. Hang the appropriate masses on the strings. Set up the fourth pulley at 180 degrees opposite from the angle calculated for the resultant of the first three vectors (record the mass and the angles). After adding the weights to the fourth pulley you should notice that the ring is at equilibrium if the resultants are correct.


4.) Check the results on the simulation. Add the vectors and obtain the resultant.







Vector Diagram

The diagram shows vectors A, B, C and D as the sum. (1 cm = 20 g)

Second Vector Diagram






































Vector Calculations:

Rx = (100Cos(0)) + (200Cos(71)) + (160Cos(144)) = 36.1

Ry = (100Sin(0))  + (200Sin(71))  + (160Sin(144))  = 283

ø = 82.7°              R = sqrt(283^2 + 36.1^2) = 285.3  

The simulation confirmed the results as correct. *Note: scale to 1=10.

 

Conclusion:
The lab sole purpose of the lab was to add the vectors and to find the counter balance so that the ring in the middle of the circular force table would be at equilibrium. The vector diagram shows the given vectors graphed on quadrant I with D as the resultant; the resultant D is the sum of the three given vectors.
In order to find the magnitude of the D component we had to find the Rx and Ry values, then we solved to find the magnitude of D by squaring the Rx and Ry values and taking the square root of both. To find the angle of the D component we just had to take the inverse tangent of Rx and Ry, giving us 82.7 degrees. Some of the errors that could’ve happed are that when trying to simulate the program with the given link, one could’ve miscalled the magnitude of the vectors resulting with a sum that is incorrect.