A 2.60-m-long, 340 kg steel beam extends horizontally from the point where it has been bolted to the framework of a new building under construction. A 75.0 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?
2007-11-02 11:09:53 · 3 answers · asked by rochesmk 2 in Science & Mathematics ➔ Physics
Beam weight = 340kg * 9.8m/s² = 3332N. The beam weight applies 3332N along its entire length, averaging out as though it is from its center. The torque is then 3332N*1.3m = 4331.6Nm from the beam alone.
The 75kg weighs 75kg*9.8m/s² = 735N. Applied at the beam far end, 735N*2.6m = 1911Nm from the worker.
At the bolts, the net torque is 4331.6Nm + 1911Nm = 6242.6Nm.
2007-11-02 11:44:01 · answer #1 · answered by Adam Zampino 3 · 0⤊ 0⤋
If the worker stands over the place the beam of mass m1 is bolted down the torque T about bolted down point is
T= (1/2) L m1g
T=(1/2) x 2.60 x 340 x 9.81= 4340 Nm
If the worker of mass m2 stands on the other end then
T= (1/2) L m1g + Lm2g
T=6250 Nm
2007-11-02 11:21:19 · answer #2 · answered by Edward 7 · 0⤊ 0⤋
I get 6242.6 Newton*meters. It's been a while but I think that is right
2007-11-02 11:22:13 · answer #3 · answered by Owen 1 · 0⤊ 0⤋