MHB Find Tension and mass of painting equipment on scuffle board.

AI Thread Summary
The discussion focuses on calculating the tensions in the cables supporting a scaffold and the mass of the painter's equipment. Given the uniform scaffold mass of 41.0 kg, a painter weighing 78.0 kg, and their respective positions, the left cable tension is stated to be twice that of the right cable. The equations of equilibrium for vertical forces and torques are established to solve for the tensions and the mass of the equipment. The critical lengths for torque calculations are clarified, particularly the distance from the left end to the painting equipment. The solution involves solving a system of equations derived from these principles.
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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
 
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$T$ = 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 $P$
 
skeeter said:
$T$ = 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?
 
Cbarker1 said:
how to get the length for the torque of mass of the painting?

5.8 - 1.6 = 4.2
 
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