Moment & Distance Homework: Steel Beam & 80 kg Person

[Attis]

Yes, it will. But if it lies between 4,5 and 10, the center of mass is calculated to be 7,25, which also results in the wrong answer (x is then equal to 26,93 when solving for x in the equation 7,25 = (1/780)*3500 + (1/780)*80x

 

[voko]

Yes, it will. But if it lies between 4,5 and 10, the center of mass is calculated to be 7,25

What makes you think that if the centre of mass is between 4.5 and 10 m, you just take the average? What good is this average?

Recall the formulation of the problem. How far can the man go without the beam tipping over?

"How far" means there is a range of positions, but the problem asks about the limit of this range.

 

[Attis]

Yes, but the limit ought to be greater than the center of mass. I just simply presumed that the center of mass should lie right in the middle of these two points, which is why I took the average.

 

[voko]

Yes, but the limit ought to be greater than the center of mass.

I cannot make sense out of this statement.

We are talking about the range of positions where the centre of mass of the entire system can be.

If the centre of mass is inside this range, the beam-man system stays on. If its centre of mass is outside this range, it tips over.

What is this range? And what are its limits?

 

[Attis]

The range for the center of mass is between 4,5 and 10? such that the center of mass must lie between these two points. I´m not sure how to find the limits. I presume the limits lie within the above range.

 

[voko]

I think we have a terminology problem here. The limits of a range where the range is between 4.5 and 10 are 4.5 and 10.

 

[Attis]

Does it make a difference if the range and limits are the same in this case?

 

[voko]

A range and its limits cannot be the same. A range is a set of numbers. A limit is a number.

 

[Chestermiller]

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.

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.

 

[Andrew Mason]

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]

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

 

[Attis]

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

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.

 

[voko]

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

Possibly. I may have been mislead by the mention of the centre of mass/gravity in the original message.

The beam-man system is stable when its combined centre of mass is within the 4.5 - 10 m range. So we must have $$ {700 \ \text{kg} \cdot 5 \ \text{m} + 80 \ \text{kg} \cdot x \over 700 \ \text{kg} + 80 \ \text{kg}} \ge 4.5 \ \text{m} .$$

 

[Chestermiller]

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.

That's what I was trying to get you to do in posts #2 and #17.

Chet

 

[Attis]

That's what I was trying to get you to do in posts #2 and #17.

Chet

Yes, I see that now, but I was majorly confused before. Thanks for all the help!