The problem involves a 10 m long steel beam weighing 700 kg, which extends 4.5 m over the edge of a roof. The question is how far an 80 kg person can walk out on the beam without it tipping over. The original poster mentions a supposed answer of 4.375 m and provides equations related to moments and forces.
The discussion is ongoing, with participants questioning the original poster's reasoning and calculations. Some guidance has been offered regarding the use of the center of mass and the need for consistent measurement points. Multiple interpretations of the problem are being explored, indicating a productive dialogue.
There is mention of the original poster's uncertainty with terminology due to language differences. Additionally, some participants note that the concept of moment of inertia has not been covered in class, which may affect the understanding of the problem.
Attis said:Yes, it will. But if it lies between 4,5 and 10, the center of mass is calculated to be 7,25
Attis said:Yes, but the limit ought to be greater than the center of mass.
The center of mass of the beam-man system is going to have to be right above the fulcrum. Otherwise, the beam is going to rotate.Attis said:Morning.
Ok, so if I measure all distances from the "free end" (10 m) and let x be the point where the man is standing, then the distance of the edge of the roof from that point is (10 - x) m?
If I once again assume that the beams center of mass is situated at half of the beam, than the center of mass of the beam is 10 - 5 = 5 m
Is this correct so far? If so, can I somehow than use the formula R = 1/(m1 + m2)*∑mi*ri to find out what the distance is of the beam-man system´s center of mass from that point?? I assume that the center of mass is somewhere between 5 m and 10 m along the beam.
Andrew Mason said:Sorry if I confused anyone by my post about needing to use the moment of inertia.
We can use the centre of mass x the distance to the fulcrum to determine the net torque but it obscures the physics somewhat, which I think is the OP's problem. It may be clearer to divide the beam at the fulcrum and calculate the torques from each side. If the difference between those two torques is made up by the man, the beam is perfectly balanced (net torque = 0) and moving any farther out will tip the beam. That really was my point. Ignore that comment about the moment of inertia, which as Steamking correctly pointed out is not applicable here.
AM
Chestermiller said:I agree with AM that the method he is describing is easier to understand and implement, even though Voko's method is correct also. I believe that the OP is studying moments in his course, so he should be using moments to solve this problem.
chet
That's what I was trying to get you to do in posts #2 and #17.Attis said:I managed to solve the problem yesterday. My main problem was that I didn´t really get what X1 was supposed to be, so I asked my teacher. X1 was supposed to be the distance between the fulcrum and the beam´s center of mass, so if the beam has a total distance of 10 m, then the fulcrum starts at 10-4,5 m=5,5 m, and we also know that the beam´s center of mass is 5, which means that X1= 0,5 m.
I then simply used the formula m1x1=m2x2 which in this case was 700 kg* 0,5 m= 80 kg * x2 and solved for x2.
Chestermiller said:That's what I was trying to get you to do in posts #2 and #17.
Chet