Parallel Forces Physics Lab

Parallel Forces Objective: Find FA and FB on the apparatus which are parallel to both Fulcrum A and B. Calculations: Theoretical FB = 0 +FB 0. 5 – (0. 1kg x g x 0. 1m) – (0. 2kg x g x 0. 4m) – (0. 05kg x g x 0. 7m) – (0. 1kg x g x 0. 3m) = 0 -[{(0. 1kg x 0. 1m) + (0. 2kg x 0. 4m) + (0. 05kg x 0. 7m) + (0. 1kg x 0. 3m)}x 9. 8] + 0. 5FB = 0 0. 5FB = [(0. 1 x 0. 1) + (0. 2 x 0. 4) + (0. 05 x 0. 7) + (0. 1 x 0. 3)]x 9. 8 FB = FB = 3. 04 N Experimental FB FB = mpanB g – mfulcrumB g FB = (0. 385kg x 9. 8) – (. 0816kg x 9. 8) = 3. 77 – . 800 FB = 2. 97 N Theoretical FA -[(0. 1kg x g) + (0. 2kg x g) + (0. 5kg x g) + (0. 1kg x g)] + FA + FB = 0 -[(0. 1kg x g) + (0. 2kg x g) + (0. 05kg x g) + (0. 1kg x g)] + FA + 3. 04N = 0 -4. 41 + FA + 3. 04 = 0 FA – 1. 37 = 0 FA = 1. 37 N Experimental FA FA = (mpanA x g) – (mfulcrumA x g) FA = (0. 205kg x 9. 8) – (. 0693kg x 9. 8) = 2. 01 – . 679 FA = 1. 33 N Conclusion: Since the distance of FB is greater than that of FA, the torque of FB is larger. The line of FA, lies directing on the 0. 2 m axis, causing a torque of zero for FA. The theoretical and experimental values for both forces are very close, supporting the theory of torque and parallel forces.