# Lifting one end of a beam

1. Jan 15, 2007

### osuaaron

Hello,

This is my first question, I've been pondering for awhile now.

Example:

I have a beam, it is 21' in length and weighs 397.5 lbs total. I am about to lift one end of this beam. I would like to know how to figure the weight I am lifting on this one end. Assume the weight is constant throughout the beam.

Thank you for any assistance.

2. Jan 15, 2007

### Hootenanny

Staff Emeritus
Welcome to the Forums,

If this is a homework question, could I please remind you that homework questions should be posted in the homework forums.

With respect to the question, tell me what you know of torques.

3. Jan 15, 2007

### osuaaron

Well, I work for a major oil company, and I hurt by back attempting to do this. As a result of the injury, I am working on a presentation for my peers that explains how much weight we are lifting in this way.

I know what torque is, aside from that not very much. I'm sort of a grunt, you could say. I would like to learn this universally so I can apply it to other beams as well, to give the guys a good idea of what they're lifting.

4. Jan 15, 2007

### Hootenanny

Staff Emeritus
I shouldn't be contacting my lawyer before advising you should I?

5. Jan 15, 2007

### osuaaron

LOL.

Well, if you want to, that's your perogative :). I'm just looking for some help to figure this out, it's been a long time since I've done this kind of thing. Basically it looks like I'm trying to convert Torque into weight.

6. Jan 15, 2007

### Staff: Mentor

If you lift one end of a beam that has a uniform mass distribution, you are lifting half the weight.

7. Jan 15, 2007

### disregardthat

What if there where no one to carry it on the other side?

8. Jan 15, 2007

### KingNothing

I'm not 100% sure of this, someone confirm:

If there is no one on the other end, and it is just on the ground, you will be lifting one-half the weight the instant you pick it up. As the angle increases and you lift it, you lift less and less, until eventually the instantaneous force to get it vertical is zero.

I'm assuming these beams are long enough, however, that the angle might have very little impact.

9. Jan 15, 2007

### Staff: Mentor

It depends on where the other supports are. If you are just lifting it from the floor, then the other support is the floor itself (the other end of the beam rests on the floor). In that case, as Russ said, as you begin to lift you would be lifting half of the beam's weight.

(Looks like KingNothing beat me too it.)

10. Jan 15, 2007

### osuaaron

Angles

I was going to say, I believe there is more to it than simply half the weight.

When picking up the pipe, it feels to me like it is possibly half the weight at the point where it is at the ground, then gradually getting lighter as the pipe is lifted to a vertical position.

So I've got an angle, a weight and a length. Can someone help me to figure out an equation to get the weight (based on it's position angle-wise?)

11. Jan 15, 2007

### DaveC426913

Wouldn't that make it simply total weight * cos(angle) /2?

So:

- horizontal: w*(1)/2 = half the weight
- vertical: w*(0)/2 = zero weight
- 45 degrees: w*(.707)/2 = .354 weight

Last edited: Jan 15, 2007
12. Jan 16, 2007

### disregardthat

What if you had to pick it up from the ground at one end, and hold it straigt out with no supporters.

what force would you have to push up and down on it, if you pushed up 10 cm from the end of the beam, and pushed down at the end, and the beam was 1 meter long?

--------------------------_-----'

_ = upwards force
' = downwards force.

Wouldn't the total force be much greater than just lifting it?

13. Jan 16, 2007

### DaveC426913

Perhaps we could stick to the OP's question?

14. Jan 16, 2007

### disregardthat

Well, it's pretty much answered isnt it?

15. Jan 18, 2007

### LHarriger

Not quite, since he asked for an equation.
Also, you are only initially lifting half the weight in an equilibrium condition. If you want to actually raise the beam then this will require an extra force. The equation can be derived by equating the following torques:
$I\alpha=\tau_{person}+\tau_{gravity}$
Solving this gives:
$F=\frac{1}{3}Ma+\frac{1}{2}Mgcos(\theta)$
where g is gravity, a is the tangential acceleration of the very tip of the beam, F is the force applied by the person lifting the beam, and theta is the angle the beam makes with the ground. It is assumed that the foces is exerted perpendicular to the beam.
If you want to use lbs instead of kg then mg is replaced by the pound weight and m is replace by the slug equivelent (just divide the beam weight by 32). If you use SI then acceleration (a) is in meters per second squared. If you use british it is in ft per second squared.

Last edited: Jan 18, 2007
16. Jan 18, 2007

### DaveC426913

Surely this is overkill. The OP likely isn't interested in the force required to move the beam at a certain rate. I think we can assume that weight is the operative concern, and that the additional force required for movement is effectively negligible.

17. Jan 18, 2007

### LHarriger

Depends, If you are lifting at half the acceleration of gravity then this would account for 25% of the force exerted when you start. Though clearly not the leading contributor, I wouldn't quite call it neglible. In fact, the reason I took time to calculate the acceleration term was because I was interested in its contribution to a stress injury. Considering he was planning on giving a presentation I thought it might be useful.