Repositioning to take more weight

[lemon]

1.Two men lift the ends of a 10m beam onto their shoulders, so that it is horizontal. The stronger of the two suggests he reposition himself to take 50% more of the weight than his colleague. Where should he put his shoulder
2.clockwise moments=anti-clockwise moments
3.50w+25w=25w
Where w=weight
so the man should be in the center of the beam because he has 50% of the beam to his left and 25% to his right. Leaving 25% to the weaker man

Am I on target here please?

 

[lemon]

I'm finding this difficult to know how to do this. Maybe like this:
The ratio we need to be 50% more is 3:1
(F_strongxdistance%)=(F_weakxdistance%)
3:1
This is 75:25

So now I need to take moments around either the center of the beam or around each of the men to get this ratio. But none of my workings are giving this ratio. Not sure

So, for example: Taking moments from the center of the beam:

75% of 5m = 3.75
F_strongx3.5m=F_weakx5m
F_strong=F_weak5/3.5
=1.4

F_weak=F_strong3.5/5
=0.7

This is a 2:1 ratio. A 100% difference - not what i want
But is this the right technique?

 

[lemon]

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[nasu]

50% more does not mean a ratio 3:1.
If one force is 1, the other is 1+0.5=1.5 so the ratio is 1.5:1 or 3:2.

Just write the equations for the balance of the forces and the balance of the torques.

 

[lemon]

Why would I consider torque, as this is about turning effects. In this problem there is no net force acting in any direction, right? And there is no net turning effect about any point, as the ladder is still in equilibrium, and is still being carried by the two men. Confused.
When you say balance of the forces and torques, do you mean put them to zero, 0?

 

[nasu]

Why would I consider torque, as this is about turning effects. In this problem there is no net force acting in any direction, right? And there is no net turning effect about any point, as the ladder is still in equilibrium, and is still being carried by the two men. Confused.
When you say balance of the forces and torques, do you mean put them to zero, 0?

Yes, that's why there is no rotation: the sum of the torques is equal to zero.
and the sum of the forces is also zero

 

[lemon]

How can I take the sum of something when I don't know where he is going to stand?

 

[lemon]

How can I take the sum of something when I don't know where he is going to stand?
How are these equations going to look?

 

[lemon]

How can I take the sum of something when I don't know where he is going to stand?

 

[nasu]

For the forces:
F1+F2-W=0
where W is the weight and F1, F2 are the two forces.

For the torques, assume that we use the end of the beam as the pivot point and that one of the guys is at this end and the other one is at a distance x from the end.

W*(L/2)-F2*x=0

where L is the length of the beam.
You can also use the middle of the beam as pivot and we have
F1*(L/2)-F2*(x-L/2)=0

In any case, the force that acts right at the pivot point have zero torque.