# Find the mass M so that the net torque is zero

## Homework Statement

Hello, I have a problem I don't want how to approach, it is a little weird. It is a problem of equilibrium (torque)

The situation and my approach is in the picture I uploaded. A beam is supported by the blue structure. On the beam it is a person(100N) at a distance of 3m in
the overhang part. In order to avoid that the beam turns over, it is placed a counterweight of mass M in the middle of the supported part. The goal is to find
the needed mass of the counterweight.

Any help will be greatly appreciated.

T net=0
Fy net=0

## The Attempt at a Solution

My approach: Since 2m of the beam is supported by a structure, I think the structure will produce a distrubuted load of reaction, but that does not matter since we can
use use a single reaction in the middle of the distributed load ( at 1m from the end in this case, as I showed in the figure). The problem is the next:
If I apply torque in the red point ( the point under the counterweight), only the 100N will excert a torque, which is wrong because the system must be in equilibrium.
I would get that -400 N.m =0

If I apply torque in the blue point (at the right end of the blue support structure), all forces will excert torque. I set up my equation and I got -300-R-M=0
Then I find the second equation with the summation of forces, and I got R=100+M. For last I replaced this last equaqtion into the first to find M.
But again I got -400 N.m = 0.

I do not think I've forgotten some reaction force somewhere ... I am thinking right now that I don't have to take into account that reaction force in that way, because once the beam starts to turn over, the normal force (reaction force) will vanish. I meant, I will only have a reaction force in the "pivot ( blue point
in the drawing)". In that case the problem will be very easy to solve, but I am not sure of that approach. The second image I uploaded shows that approach.

Any help will be greatly appreciated.

#### Attachments

• _20180522_141104.JPG
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• Sin título.png
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Gold Member
unless the beam is attached to the blue structure it wont supply any support at all. if you look at your png again, you can see that the beam is lifted left of the the blue point.

The beam is lifted left of the blue point and lowered on the right of the blue point. so the beam rotates around the blue point.

unless the beam is attached to the blue structure it wont supply any support at all. if you look at your png again, you can see that the beam is lifted left of the the blue point.

The beam is lifted left of the blue point and lowered on the right of the blue point. so the beam rotates around the blue point.
Alright, that is what I think, so my last approach (and picutre) are the correct.

Gold Member
you want to know what the mass ##M## must be to balance the beam on the blue pivot point. That means the sum of torques must be zero or, since the the torques act on either side of the pivot point, they must be equal in magnitude.

So you have one mass at ##3m## to the right of the pivot point creating a torque ##T = 100 N \times 3 m = 300 Nm##. Also, you have a mass ##M## at ##r = 2m## to the left of the pivot point that creating a torque in the opposite direction. ##T = r F## with ##F = Mg##.

that should give you enough information to solve for ##M##.

correction: Mass ##M## is at ##r = 1m##.

In your method 2 you've got two torques ##M## and ##R## but there should only be one. Torque on the red point should balance out the torque by the person

Homework Helper
Gold Member
once the beam starts to turn over, the normal force (reaction force) will vanish. I meant, I will only have a reaction force in the "pivot ( blue point
Correct. More generally, you cannot assume that the normal force from a flat support is evenly distributed (or, equivalently, at the midpoint of the contact length). E.g. if you increase the balancing mass beyond the minimum necessary then, effectively, the normal force will be at whatever point balances all the forces and torques.

CWatters