Question

A 3.80-m-long, 500 kg steel beam extends horizontally from the point where it has been bolted to the framework of a new building under construction. A 67.0 kg construction worker stands at the far end of the beam.

Answer 1

Answer:

Explanation:

The question is ;

3.80 m-long, 500. kg steel beam extends horizontally from the point

where it has been bolted to the framework of a new building under

construction. A 67.0 kg construction worker stands at the far end of

the beam. What is the magnitude of the torque about the point where the beam is bolted into place?

Solution:

Here two torques are in action

a) torque T1 due to weight of worker at the edge of the beam - at center of the beam

b) torque T2 due to weight of the uniform beam - at the point where the beam is bolted

So we first calculate the torque produced due to weight of the worker;

We can see that;

Distance of worker from the center of the beam = 1.9m

Mass of the worker = 67 kg

Value of g= 9.8 m/sec2

T1=force × distance from point of rotation

Here force is weight of the worker which is = mass × g=67×9.8=656.6 N

So the torque is

T1= 656.6×1.9=12225.892 Nm

or

T1 = 12225.892 Nm

Now torque by the beam itself ;

Length of the beam from its center point to the bolt point = 1.9 m

Weight force of beam acting at center point= mass× g= 500×9.8=4900 N

Torque T2 at bolt point by the beam weight = weight of beam × length of beam from its center to the bolt point = 4900×1.9=9310 Nm

T2=9310Nm

So total torque = T1+T2= 12225.892+9310=21565.892 Nm