Question

A 4.40-m-long, 500 kg steel uniform beam extends horizontally from the point where it has been bolted to the framework of a new building under construction. A 70 kg construction worker stands at the far end of the beam.

What is the magnitude of the torque about the bolt due to the worker and the weight of the beam?

Answer 1

Answer:

Torque=13798.4 N.m

Explanation:

Given data

Mass of beam m₁=500 kg

Mass of the person m₂=70 kg

length of steel r₁=4.40m

center of gravity of the beam is at r₂=r₁/2 =4.40/2 = 2.20m

To find

Torque

Solution

Torque due to beam own weight

$T_{1}=m_{1}gr_{1}\\ T_{1}=500*2.2*9.8\\T_{1}=10780N.m$

Torque due to person

$T_{2}=m_{2}r_{2}g\\T_{2}=70*(4.40)*(9.8)\\T_{2}=3018.4 N.m$

Now for total torque

$T_{total}=T_{1}+T_{2}\\T_{total}=10780+3018.4\\T_{total}=13798.4N.m$

Answer 2

The magnitude of the torque about the bolt due to the worker and the weight of the beam is 13,798.4 Nm.

The given parameters;

• length of the steel, L = 4.4 m
• mass of the steel, m = 500 kg
• mass of the work, m = 70 kg

The magnitude of the torque about the bolt due to the worker and the weight of the beam is calculated as follows;

the center of gravity of the beam $= \frac{4.4}{2} = 2.2 \ m$

$\tau _t = \tau _1 + \tau _2\\\\\tau _t = F_1\times r_1 \ + F_2 \times r_2\\\\ \tau _t = m_1 gr_1 + m_2 gr_2\\\\\tau _t = (500 \times 9.8 \times 2.2) \ + \ (70 \times 9.8 \times 4.4)\\\\\tau _t = 13,798.4 \ Nm$

Thus, the magnitude of the torque about the bolt due to the worker and the weight of the beam is 13,798.4 Nm.

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