1. SF=0; no linear (translational) acceleration
Equilibrium
SF=0; no linear (translational) acceleration
St=0; no rotational acceleration
ti = ri x Fi
for 2 dimensional arrangements;
ti = riFi sin(q);
if q=90 deg
ti = riFi r F q
Fsin(q)

2. mb m2 m1 m3
What’s missing?

3. F(+) Lb x(+) ? xcg mb m2 m1 x1 m3 x2 x3
xcg is the location where a support can be placed to balance (two conditions for equilibrium, SF=0 and St=0. To find it, sum the torques of all the masses, then divide by the sum of the masses.
xcg= (x1(m1g)+x2(m2g)+x3(m3g)+ (Lb/2)(mbg))
g(m1+m3+m3+mb)
The g factors out, so you can use weight or mass, just be consistent. In this example, the beam is assumed to be uniform, so the weight acts at it’s midpoint (Lb/2)

4. What else looks like this?Wm Fa Fc Wb
What else looks like this?
Wb = 9800 N
Wm = 800 N
Length of beam = 5 m
Cg of beam at midpoint = 2.5 m
Position of Man = 3 m from left end of beam
Solve for the two support forces Fa and Fc

7. Wm Fa Fc Wb
We have two conditions for equilibrium: Sum of forces = 0 ( no linear acceleration),
and sum of torques = 0 ( no rotational acceleration)
Since we have two unknown forces, we start with the torque equation:
St = 0 = , let’s build a torque table, taking our reference point as the left support and call it zero, all distances will be positive. Downward forces will be negative.

8. Wm Fa Fc Wb
We now have for the sum of torques;
0 = -XbWb – XmWm +XcFc
Solve for Fc;
Fc = (XbWb +XmWm)/Xc

9. Wm Fa Fc Wb
Substitute given values;
Fc = ( )/5.0
Fc = 5380

10. Wm Fa Fc Wb
Now use the sum of forces condition
SF = 0 = Fa – Wb – Wm + Fc
Solve for Fa = Wb + Wm - Fc;
Fa = – 5380
Fa = 5220 N