18816 Nm is the amount of torque applied to the bolt by the worker and the weight of the beam.
The given parameters;
length of steel, L = 6.00 m
mass of steel, m = 500 kg
mass of work, m = 70 kg
The center of gravity of the beam Equals The torque caused by the worker and the weight of the beam is computed as follows:
= [tex]\frac{6.00}{2}[/tex]= 3 m
[tex]T_{t} = T_{1} + T_{2}\\T_{t} = F_{1} r_{1} + F_{2}r_{2}\\T_{t} = m_{1}gr_{!} + m_{2}gr_{2}\\[/tex]
[tex]T_{t} =[/tex](500 ×9.8×3) + (70×9.8×6.00)
[tex]T_{t} =[/tex]18816 Nm
This results in a torque of 18816 Nm being applied on the bolt by the worker and the weight of the beam.
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