How do the forces in fluids allow heavy airplanes to fly?
2 answers:
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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
