Solving a Confusing Torque Problem: Lever Equilibrium in Top View

[Mech King]

Hello,

I think i may be over analyzing this basic problem, but if i have a lever similar to that attached, and i apply a load at one end and a counter moment at the over end (at the pivot center) to balance the load then the system is in equilibrium.

Now if the load applied on the end of lever was not parrallel with gravity, but perpendicular to it, so the image attached became a tope view as opposed to a side view, then how do you quantify the torque, because although gravity is always present, it now won't be tending to push the load around anti-clockwise, so how can i equate the load to a force and resolve this problem? Clearly, if the attached image shows a top/plan view of the system then the lever will be easier to turn.

Can someone please help me understand my madness?

Cheers

 

[maverick280857]

Now if the load applied on the end of lever was not parrallel with gravity, but perpendicular to it, so the image attached became a tope view as opposed to a side view, then how do you quantify the torque, because although gravity is always present, it now won't be tending to push the load around anti-clockwise, so how can i equate the load to a force and resolve this problem? Clearly, if the attached image shows a top/plan view of the system then the lever will be easier to turn.

Okay, so what exactly is the problem? The force which is applied to the end of the lever is perpendicular to gravity. It exerts a torque about the pivot. And unless you want to consider a yanking of the lever due to gravity (a kind of cantilevered effect), gravity has no major role to play here, except enter the vertical equilibrium condition for the lever + pivot. That is, assuming that the vertical torque due to gravity about the pivot, is not allowed to bend/yank the road. If the rod can yank, then the problem is much more complicated because the rod won't be horizontal all the way from the pivot.

 

[PTM19]

The way I see it if the picture presents a top down view then there is no force to counter the "counter torque" and the lever will simply start spinning in it's direction, constantly accelerating. The mass at the end of the lever will only matter as part of the overall mass of the system determining it's inertia.

 

[Mech King]

Thanks for the reponses,

i basically am making a little torque driver calibration device (approximate) and thought it may be easier from an ergonomics point of view to have the calibration device sitting face up on a workbench surface, and then upon further investigtion i thought..."but gravity is not tending to turn the mass, so how can i use th,e set up to calibrate torque if there isn't any torque on the lever, just inertia"

The lever is rigid and sat in bearings, so there is no deflection, and friction can be ignored.

Would it only be possible to do this torque calibration set up if the load is parrallel and not perpendicular to gravity>?

Cheers again

 

[PTM19]

"Would it only be possible to do this torque calibration set up if the load is parrallel and not perpendicular to gravity"

Yes, it has to be parallel to gravity or in other words the arrow on your picture has to point down so that the weight of the mass attempts to rotate the lever in the direction opposite to that of your torque.

If you want the lever to be horizontal you can add something to the design that will transfer the force exerted by the load changing it's direction, a rope and a pulley or something like that.