Question

I've got a quick question; pardon my ambiguous and perhaps incorrect phrasing.

Ok....you apply x force to the end of a lever perpendicular to the lever itself so that 1 rotational force is produced. This would produce so much rotational force on the lever. If you were to apply the same force parallel to the lever, it would result in 0 rotational force.

Now, does the amount of rotational force exerted vary linearly with the angle it was applied at? For instance, would applying x force at 45 degrees apply 0.5 rotational force?

Also, would there be a difference between applying the force at 45 degrees and 135 degrees?

This is assuming that the force's angle never changed relative to the lever. Any help would be greatly appreciated.

i dunno lol

It changes as a trigonometric function of the angle.  You can break the force-at-an-angle into a perpendicular (x) and parallel (y) component.  The amount of rotational force will be linearly proportional to perpendicular component.



For example.


Say the lever is at 12 o'clock, and you push the lever head down and to the right = 120 degrees (0 degrees is straight up, 90 is right, 180 is down) with a force of F = 1.

Well F is actually the hypotenuse (h) of a right angle triangle with angle (t) of 30, with the other two sides of the triangle being either parallel or perpendicular to the direction your lever arm is pointing (0 degrees).  You want to find the length of the side of the triangle that is perpendicular to this lever arm, i.e. the adjacent side (a)

That's just simple trigonometry, cos(t) = a/h; thus a = cos(t)/h = cos(30)/1 = cos(30)

http://progressiveboink.com/archive/robliefeld3.html

>>3
Ok, many thanks. Was trying to remember back to the first resultant problems we did in trig, but couldn't quite carry it over.....thanks again.