Find Tension and mass of painting equipment on scuffle board.

[cbarker1]

A uniform 41.0 kg scaffold of length 5.8 m is supported by two light cables, as shown below. A 78.0 kg painter stands 1.0 m from the left end of the scaffold, and his painting equipment is 1.6 m from the right end. If the tension in the left cable is twice that in the right cable, find the tensions in the cables (in N) and the mass of the equipment in kgs.

View attachment 7617

Works
Givens:

The mass of the scaffold=41.0 kg
Length of the scaffold=5.8 m
the mass of the painter= 78 kg
The length betwen left end and the painter= 1 m
The length between painting equipment and painter= 1.6 m
The Tension of the Left= twice in the right cable.
the mass of the painting equipment=?

$\Sigma F_y=0$
$-m_b*g-m_p*g-m_s*g+T_L+T_R=0$
$\Sigma\tau=0$
$-m_p*g*L_1-m_b*g*L_2+T_R*L_3+(m_s*g*(L1+L2)/2)(?)=0$

If I place the pivot for the torque on left end of the scaffold, then would the weight of the scaffold be not needed?

View attachment 7616

 

[skeeter]

$T$ = tension in the right cable
$P$ = mass of paint equipment

forces up = forces down

$2T + T = 78g + 41g + Pg$

about the left end, torques CW = torque CCW (yes, the scaffold weight exerts a torque)

$78g \cdot 1.0 + 41g \cdot 2.9 + Pg \cdot 4.2 = T \cdot 5.8$

solve the system of equations for $T$ and $P$

 

[cbarker1]

$T$ = tension in the right cable
$P$ = mass of paint equipment

forces up = forces down

$2T + T = 78g + 41g + Pg$

about the left end, torques CW = torque CCW (yes, the scaffold weight exerts a torque)

$78g \cdot 1.0 + 41g \cdot 2.9 + Pg \cdot 4.2 = T \cdot 5.8$

solve the system of equations for $T$ and $P$

how to get the length for the torque of mass of the painting?

 

[skeeter]

how to get the length for the torque of mass of the painting?

5.8 - 1.6 = 4.2