How to calculate this force if torque is known?

[onemilimeter]

Given the torque T, what is the amount of force F2 at point B?

F1 = T / x1

Is it F2 = T / x3 ?

[Mechatrommer]

T = F x r, where r is perpendicular distance of the F (vector) from a reference point, in your case is any point lies in the vertical line of A.

T = F2 x x3, hence...
F2 = T / x3

[Excavatoree]

Your formulas were backwards.  (but have since been corrected, along with this post.)

T=F1*x1

T=F2*x3

So, if you want the same torque, F1*x1=F2*x3

F2=F1*x1/x3

[Mechatrommer]

ok now, your diagram is confusing and explanation incomplete. where is the external force coming from? (the torque?) what is the F1 & F2? if its external forces, then the torque will add up. if F2 is supporting force, then you got its direction backward. if F1 and F2 are supporting, then both tork from F1 n F2 = T etc etc.

[onemilimeter]

Your formulas are backwards.

T=F1*x1

T=F2*x3

So, if you want the same torque, F1*x1=F2*x3

F2=F1*x1/x3

Thanks. They're corrected already.

[Lawsen]

One writer asks where the external force is coming from is important here.  Torque varies in where along the lever arm the external force is at.  That is why breaker bar on a wrench is long and the longer is better and more force by torque to break the stubborn bolt loose.  Delicate items like plastic electronics cases, where the screw requires less torque and less force, then grabbing the wrench nearer to the center or right at the center is best to prevent breaking the screw, bolt, or case.  Torque is in Newton meters or foot pound. The resulting units have to be like this naturally to be the correct flow and answer.

[onemilimeter]

ok now, your diagram is confusing and explanation incomplete. where is the external force coming from? (the torque?) what is the F1 & F2? if its external forces, then the torque will add up. if F2 is supporting force, then you got its direction backward. if F1 and F2 are supporting, then both tork from F1 n F2 = T etc etc.

My apology for the confusion. Point 'O' is attached to the shaft of a motor (not shown in the figure) exerting a torque T.

I wish to calculate F1 and F2 due to the torque T.

F1 can be calculated as F1 = T / x1

What's about F2?

[Mechatrommer]

still confused, where the F1 is coming from? is it to counter act the tourqe? or just to add up the T so ther F2 become greater, coz the T is clockwise, and F1 will generate more clockwise! and then F2 is also clockwise! with reference to O.

edit: if F1 is to counter act the T to put the shaft in equilibrium, then F2 will be zero = F1 in reverse direction! from what i understand so far.

[onemilimeter]

still confused, where the F1 is coming from? is it to counter act the tourqe? or just to add up the T so ther F2 become greater, coz the T is clockwise, and F1 will generate more clockwise! and then F2 is also clockwise! with reference to O.

Sorry again... F1 and F2 are not "applied" forces. In fact, a load cell will be placed underneath the level arm at point B to measure the force F2 (resulted from the torque).

[Mechatrommer]

ok i wont ask further, lets just assume T and F2 are external forces. so ok F2= T / x3, but your diagram still not in equilibrium. pls google "Free Body Diagram" and "Newton's Motion Law".
ps: I've lost my Static book

[onemilimeter]

ok i wont ask further, lets just assume T and F2 are external forces. so ok F2= T / x3, but your diagram still not in equilibrium. pls google "Free Body Diagram" and "Newton's Motion Law".
ps: I've lost my Static book

I wish to know the relationship between the T and F2 because I wish to calibrate the load cell which will be placed underneath the level arm at point B.

Thanks. I will have a look at "Free Body Diagram" topic.

[Mechatrommer]

yeah. coz i'm sure that's not only there is something at point B, but far to left of metal bar, or somewhere there should be something as well. you cannot hang it just like that. the general rules are:
1) sum of force = 0
2) sum of torque = 0
for a static or constantly moving/rotating body. Cheers

[Time]

This is an EE forum not a bar benders forum!

haha jk


Thats what I called Mechanical Engineers when I was in school.  Bar benders.

Civil Engineers were Gravel Spreaders.   Industrial Engineers were Imaginary Engineers. 

Electrical Engineers were Electron Chasers.

[Mechatrommer]

thats why i tried not to elaborate. and at least by teaching, i'll keep the knowledge intact in my brain. fwiw, sometime ee have to deal with this sometime, somewhere, someplace