BFPM Practicum

BFPM Practicum

This experiment was meant to test our skills in forming free body diagrams as well as using SOH CAH TOA to solve for the missing vector for gravity.

a.     Picture of the diagram:

b. Free body diagram:

c. Making the free body diagram was quite simple, the tensions were both at an angle, causing me to split them up into components in the X and Y directions. All that was left was the force of gravity pointing downward. The work written out to solve the problem can be seen in the picture below. To start off, I knew that all the horizontal and vertical force vectors were congruent because the object was not in motion. That information tells me that the force of the 2 tension components in the Y direction, FT1y and FT2y, added together will equal the weight of the object because they are both half the total weight. We were given 2 preliminary angle measurements, 23 and 75 degrees, each half of a quadrant of the diagram meaning the other part of the angle must add to equal 90 degrees. This easily allowed me to find the measures for the other angle which were 65 and 15 degrees. Because I was also given the force of tension for both vectors I was able to use that value along with the angle to solve for the component in the Y direction. I decided to use cosine, which is adjacent over hypotenuse because I was given the hypotenuse, which was the force of tension, and I was trying to solve for the adjacent value to the angle measurement. I set up COS 65=FT1y/.9 and COS 15= FT2y/2.2. I multiplied the forces of friction by the cosine of the angle and got .380 Newtons for one of them and 2.125 Newtons for the other. To find the gravity I just added the numbers together.

d.  PREDICTED WEIGHT: .380 N + 2.125N= 2.5 Newtons

e. ACTUAL WEIGHT: 

f. PERCENT ERROR: