Calculating the mass of an object pivoting on a point

1. The problem statement, all variables and given/known data

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?

A-----------------------
......Δ

2. Relevant equations

τ = d*f

3. 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.

 

thanks for your reply. im not sure if i understand now or not....

from your reply i set up the equation (first: 8kg*9.8N=78.4N)

removing all the units for clarity sake

this below is essentially Fd + Md = Md (where F is the 8kg, and M is the mass of the beam, and 'd' is distance)
(78.4)(0.5) + M(0.5) = M(1.7)
when i solve for M i get 32.666 repeating Newtons, and dividing by 9.8 i get 3.333333 repeating kilograms

but i still dont understand where im at. but i do notice that we have 3.3kg on both sides. do i just add the two sides together 3.33 + 3.33 to get 6.66 (rounded up to 6.7kg which is the answer in the book?)

if so is that how you set up the question for a solution?

 

i dont understand what you are saying

oh ok wait maybe i do

so there the force of 8kg 0.5m..... and since the center of gravity is at 1.1m can we say that all of the balancing force is at 1.1m away from A? (which is 0.6m to the right of the pivot point)

so the equation would be like.....
-> Fd = Fd
-> (78.4N)(0.5m) = (FN)(0.6m)
-> 39.2 = 0.6F
-> F=65.333repeating Newtons

-> F/9.8=kilograms
kilograms=6.66repeating (which is the answer when taken to two sig figs)

AHA i get it now... that is to say i might... if i interpretted that correctly and my answer isnt a cooncidence

 

i'd appreciate if someone could confirm that my last attempt is the logically correct way. i just bolded the portion of my last post that i am most concerned with.