Moment & Distance Homework: Steel Beam & 80 kg Person

[Attis]

Homework Statement

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.

Homework Equations

M = F * r
F=mg
m1*x1 =m2*x2 or m1g*x1 =m2g*x2

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

 

[Chestermiller]

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?

 

[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?

 

[Andrew Mason]

Homework Statement

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.

Homework Equations

M = F * r
F=mg
m1*x1 =m2*x2 or m1g*x1 =m2g*x2

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

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

 

[voko]

You cannot use the centre of mass of the beam like that to find the torque.

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.

 

[SteamKing]

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

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.

 

[Attis]

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?

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.

 

[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.

 

[Attis]

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

 

[voko]

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.

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?

 

[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.

 

[Attis]

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?

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

 

[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.

 

[Attis]

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.

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.

 

[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?

 

[Andrew Mason]

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.

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

 

[Chestermiller]

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

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

 

[Attis]

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?

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.

 

[voko]

Morning.
Ok, so if I measure all distances from the "free end" (10 m)

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

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.

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

Correct.

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??

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

 

[Attis]

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.

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 ?

 

[voko]

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 ?

What is the "far end"? Is that the opposite of the "free end"?

Does that mean that I know one of the radiuses which I need for the formula R = 1/(m1 + m2)*∑mi*ri ?

You know both of them, as I stated in the previous message.

 

[Attis]

What is the "far end"? Is that the opposite of the "free end"?
You know both of them, as I stated in the previous message.

No, I meant the free end: that the edge is 0-4,5 m from the free end.

Yes: the coordinates are x and 5, but since both x and R are unknown, I don´t really know what to do.
I thought about maybe letting R1=R2 where R1 = 1/(700 +80) * 700*5
and R2= 1/(700 + 80) * 80x
and solving for x, which results in x= 43,75... which is wrong by just one decimal point. Is this a coincidence, or am I finally on the right track?

 

[Attis]

does the man affect the center of mass of the whole man-beam system? Is that why we don´t divide it by two as was the case for the beam´s center of mass?

 

[voko]

No, I meant the free end: that the edge is 0-4,5 m from the free end.

In that case, it cannot be negative. All the distance on the beam, measured from the free end, are positive. For example, the beam's centre of mass is at 5 m, not at -5 m.

Yes: the coordinates are x and 5, but since both x and R are unknown, I don´t really know what to do.

x is what you must find. So it is normal that it is unknown.

R is a different story. It is the entire system's centre of mass. Think: at what values of R is the entire system stable? Is there a maximum or a minimum R for a stable system?

 

[Attis]

If I´ve interpreted my notes correctly, the whole systems´s center of mass needs to lie between x and 5. X lies somewhere along the free lying part of the beam, so anywhere along 0 + 4,5 m, as measured from the free end (if what I wrote previously is correct).

 

[voko]

That is all correct. But you did not answer the questions from my previous post.

 

[Attis]

I didn´t answer cause I wasn´t sure. If x can be max 4,5 m as measured from the free end, and the beams center of mass is 5, then I suppose the whole system´s center of mass should lie between 5 and 4,5 - so (5 + 4,5)/2 = 4,75, but using this in the equation for R results in the wrong answer when solving for x.

 

[voko]

You are not answering the question I asked. You are answering some other question. My question was about the condition for the beam-man system to be stable. Where must its centre of mass be?

 

[Attis]

In the middle of the beam-man system? i.e (5 + (x + 4,5))/2. The beam has a center of mass which is 5, and the mans center of mass is unknown, but lies somewhere between x + 4,5.
I don´t know how to answer your question. Could you formulate it some other way?

 

[voko]

The beam-man system is continuously supported by the roof from 4.5 to 10 m. If the system's centre of mass lies between 4.5 and 10 m, will it tip over? If it is outside that range?

 

[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!