Torque/equilibrium problem?

Question

an 8 meter scaffold supports a mason, bricks, and itself. The scaffold weighs 427N, mason A weighs 895 N, mason B weighs 1291 N and the bricks weigh 1620 N. From the left end of the scaffol, Mason B is 2 M the bricks are 5m and mason A is 7.25M. The scaffold is uniform. In order to establish equilibrium what must the upward force on each end of the scaffold be?

Answer 1

Let FL and FR be the forces acting upward on the ends of the scaffold.

For a static system the torque balance around an arbitrary pivotal point of the system equals zero:
Σ (F×d) = Σ (|F|·|d|·sin(φ))= 0
× is the cross product,
F the force vector with the absolute value |F|
d the distance vector (the leverage) between acting point of force and the pivotal point with the absolute value |d|
φ the angel between F and d

The best choice is the left or the right end of the scaffold because the force acting on this end does not act as torque at this end and the other forces acting perpendicular to the distance vector. Therefore: F×d = |F|·|d|
For the torque balance take the torques acting clockwise positive and the torques acting counterclockwise negative.
The balance for the left end of the scaffold is given by:
- FR ·dR + FMA ·dMA + FMB·dMB + FB·dB + FS·dS = 0
FR ·8m = 895N ·7.25m + 1291N·2m + 1620N·5m + 427N·4m
(you can take the weight of the scaffold as force acting on its center of mass)
=>
FR = 2359.84375N ≈ 2360 N

Next step is the balance of the vertical acting forces, which must cancel to zero:
Σ Fv = 0
<=>
FL + FR - FMA - FMB - FB - FS = 0
<=>
FL = FMA + FMB + FB + FS - FR
= 895N + 1291N + 1620N + 427N - 2360N = 1873N