A pivot is 0.50m from the end (called A) of a 2.2m uniform beam. A load of 8.0kg at end 'A' will cause the beam to be balanced on the pivot. What is the mass of the beam?
τ = d*f
The Attempt at a Solution
im close to understanding this but i just dont see how they deduce the mass of the beam.
from what i understand, there are a couple critical things to realize:
1) the clockwise torque = counterclockwise torque because its balanced (which means the torque on both sides are equal)
8kg*9.8=78.4N (this is the counterclockwise force 0.5m away from pivot, thus producing 39.2N*m torque
so the clockwise torque will be equivalent. the end of the beam to the right is 1.7m from the pivot and i dont know exactly how relevent that is.
2)additionally, i later realized that the center of gravity will be at 1.1m (halfway due to uniform beam) and im not sure but i suspect there is 9.8N pushing counterclockwise here.
i just dont see how i take all this information and deduce the mass of the beam. i would really appreciate some help, preferably a step by step solution using words rather than numbers.