# Calculate weight needed to balance a beam

I have found many sources to determine the weight or force needed for a balance moment, but not one that included the weight of the beam as a variable. I assume this is a very simple equation. Please help an old guy out. Thanks.

The beam is uniform and I know very little about physics. Believe it or not, this is for an art project about the theoretical weight of color.

I can get the 50 * 3 = 7x (21.428), but this does't include the beam weight.

• What do you know about moments and weights of objects? What have you tried to do when answering this question? Note that we are not here to do homework for people. – JMac Apr 4 '17 at 13:16
• Is the beam uniform? – Utkarsh futous Apr 4 '17 at 13:18
• @WRICHIKBASU I'll give him a chance to edit it first. He posted the question 7 minutes ago and this is his first post to physics SE. I generally try and nudge people towards better questions instead of immediately voting to close. It seems to cause unnecessary drama when you close right away before they even have a chance to understand. After like half an hour, or a response suggesting they wont edit it, then I flag. If it's people who should know better I usually flag and only bother to comment half the time. – JMac Apr 4 '17 at 13:24
• Hint : Find COM of both sides and balance torque – Utkarsh futous Apr 4 '17 at 13:27
• @JMac I have my answer ready. He shows an attempt, and I answer it. I take back the -ve vote – BuddingPhysicist Apr 4 '17 at 13:31

Use the Principle of Moments.

The weight of the bar will act at 5m point.

So, by the above Principle, at equilibrium,

Sum of clockwise moments = Sum of anticlockwise moments

$Or, \quad 7×x+10×2=3×50$

$Or, \quad 7x=150-20=130$

$Or, \quad x = 130/7 \approx 18.5714285714 \, lbs$

In the equation, 10×2 is because the fulcrum is 2m from the 5m point where the weight of the bar acts. All arm lengths are taken from the fulcrum.