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A 3.50-m-long, 380kg steel uniform beam extends horizontally from the point where it has been bolted to the framework of a new building under construction. A 74.0kg construction worker stands at the far end of the beam.



What is the magnitude of the total torque due to the worker and the force of gravity on the beam about the point where the beam is bolted into place?

Answer :

MechEngineer

Answer:

9064.44 Nm

Explanation:

Since the steel is uniform, the center of mass is also at the geometric center, which is 3.5 / 2 = 1.75 m from the bolted point.

Let g = 9.81 m/s2. The weight of the beam and the worker is

[tex]W_b = m_bg = 380*9.81 = 3727.8 N[/tex]

[tex]W_w = m_wg = 74*9.81 = 725.94N[/tex]

So the total torque about the bolted point is the sum of product of weights and distances from weight to the bolted point

[tex]W_b*1.75 + W_w*2.4 = 3727.8 *1.75 + 725.94 * 3.5 = 9064.44 Nm[/tex]

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