# Moment and distance

1. Feb 13, 2014

### Attis

1. The problem statement, all variables and given/known data

A 10 m long steel beam which weighs 700 kg lies on a roof. The beam sticks out 4,5 m over the edge of the roof. How far can a person weighing 80 kg go out on the beam (on the part sticking out over the edge of the roof) without the beam tipping over?
The answer is supposed to be 4,375 m
I translated this question from Swedish to English, so once again - please excuse any grammar mistakes.

2. Relevant equations
M = F * r
F=mg
m1*x1 =m2*x2 or m1g*x1 =m2g*x2

3. The attempt at a solution
I started by trying to figure out where the center of gravity (I think that´s what it´s called in English) is for the beam. Let this point = X1/2, assuming this point is in the middle of the beam.
Also let m1 = the weight of the beam = 700 kg
m2= the weight of the man = 80 kg
x2= the length of the beam exceeding the edge of the roof

I then also assumed that the man wasn´t moving, and set up an equation as follows:
m1*(X1/2) = m2*X2
Plugging in the numbers above and solving for X1
results in X1 = 1,03 m
Is this reasonable?

I then figured the man would be moving anywhere between 4,5 - X2, and that the force acting on the man is 80 kg * 9,82 =785,6 N, such that the momentum is given by 785,6 (4,5 - X2) = 3535,2 - 785,6X2

The force acting on the beam is Fbeam = 700 * 9,82 = 6874 N, and under the assumption that 1,03 m is correct, we get a momentum of 6874 N * 1,03 m=7080,22

Setting 3535,2 - 785,6X2 = 7080,22 and solving for X2 results in X2 = -4,673 m, which is obviously wrong!
Help!
Thanks,
Astrid

2. Feb 13, 2014

### Staff: Mentor

Where is the center of mass of the beam located along the beam?

How far from the fulcrum is the center of mass of the beam?

3. Feb 13, 2014

### voko

I do not understand the logic of your approach. If you assume that the centre of gravity is in the middle of the beam, then why do you calculate it? What is X1? What is the system whose centre of gravity you tried to find?

4. Feb 13, 2014

### Andrew Mason

You cannot use the centre of mass of the beam like that to find the torque. You have to use the moment of inertia. This is because the mass that is farther from the fulcrum produces more torque than the mass closer to the fulcrum.

AM

5. Feb 13, 2014

### voko

I do not think torques are even needed here. Finding the centre of mass of the beam-man system is enough to solve this problem.

6. Feb 13, 2014

### SteamKing

Staff Emeritus
You would use mass moment of inertia only if the question asked something about the dynamics of the beam rotating, which reading the OP clearly shows it does not.

7. Feb 13, 2014

### Attis

Since English isn´t my mother tongue, I wasn´t sure if "center of gravity" was the right term. I think I was looking for center of mass, and that X1 is supposed to be the center of mass for the beam.

8. Feb 13, 2014

### Attis

I would assume that the center of mass of the beam (which includes the part which extends over the roof) is 10 m/2 = 5 m? But I do not know how that helps me.
We haven´t gone through moment of inertia in class yet, so I don´t think the question has anything to do with it.

9. Feb 13, 2014

### Attis

Or no, sorry my mistake. I was a bit hasty there.

10. Feb 13, 2014

### voko

These two terms mean the same thing in most practical cases, including this one.

So I repeat: I do not understand the logic of your approach. If you assume that the centre of mass is in the middle of the beam, then why do you calculate it? What is X1? What is the system whose centre of mass you tried to find?

11. Feb 13, 2014

### Attis

Would you use the formula R = 1/(m1 + m2)*∑mi*ri?
So I assume we would get 1/(700 + 80)*∑(80*4,5 + 700*5) = 2,57 m, which is also wrong.

12. Feb 13, 2014

### Attis

You´re right. It doesn´t make sense. I´ll have to try a new approach to finding X1: the center of mass of the beam

13. Feb 13, 2014

### voko

That formula gives you the centre of mass of the beam-man system. But why do you have 4.5 m for the man? And what is it measured from? All the distances in that formula must be measured from the same point.

14. Feb 13, 2014

### Attis

I don´t really know how to account for the fact that the man could be anywhere along the 4,5 m section of the beam into that formula. I figured maybe one could let x be the point from which all the distances in the formula are to be measured from, and that the man could then move x + 4,5 m in the forward direction? which would then maybe mean that the distance from the point to the end of the beam (in the direction away from the roof) could be x - 5,5?
Argh, this is so complicated.

15. Feb 13, 2014

### voko

So, first things first. Let us measure all the distances from the "free end" (the one sticking out). The distance of the man from that point is x (unknown). What is the distance of the edge of the roof from that point? What is distance of the beam's centre of mass from that point? What is the distance of the beam-man system's centre of mass from that point? Finally, what condition should the beam-man system's centre of mass satisfy so that the beam does not tip over?

16. Feb 13, 2014

### Andrew Mason

Letting L be the displacement of a mass element of the beam, dm from the fulcrum, the torque contributed by that mass element is dmgL = μdLgL where μ = mass/unit length of the beam = 70kg/m. So integrate that to work out the torque contributed by the part of the beam whose end is a distance S from the fulcrum. Do that for each side and then you can work out what torque the man contributes to the short side to make the torques equal.

AM

17. Feb 13, 2014

### Staff: Mentor

This integration really isn't necessary, since we know that the center of mass of the beam is at its very center, and this is located 0.5 m from the fulcrum, so the moment of the beam around the fulcrum is 700g (0.5). The integration will, of course, give this same result.

Chet

18. Feb 14, 2014

### Attis

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.

19. Feb 14, 2014

### voko

If you measure from the "free end", then the coordinate of the free end is not 10 m, but 0 m.

How so? The edge of the roof is a fixed. It does not depend on where the man is standing. How far is edge from the free end? That is its coordinate.

Correct.

Yes you can. You have two masses, and their coordinates are x and 5 m.

20. Feb 14, 2014

### Attis

If the free end is 0, then the edge is 0-4,5 m = -4,5 m from the far end, assuming that left is the negative direction ? Does that mean that I know one of the radiuses which I need for the formula R = 1/(m1 + m2)*∑mi*ri ?