# Parallel Forces Physics Lab

Only available on StudyMode
• Published : March 30, 2013

Text Preview
Parallel Forces
Objective: Find FA and FB on the apparatus which are parallel to both Fulcrum A and B.

Calculations:
Theoretical FB
Στ = 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) + (0.05kg 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.