Reply With QuoteFrom my viewpoint, for an equal loading the magnitude of the two torques should be equal; however, since the screw must rotate in the opposite direction, as viewed from the same end of the screw, the torque is also in the opposite direction.
Hi
I'm new to this forum. I work in industry and am currently studying part time for my BEng aswell. We're doing a short design project based on powerscrew mechanics, and I was wondering if someone could shed some light on mechanics of horizontal powerscrews?
Specifically, I'm having difficulty with the Torque required to move a load along a horizontal power screw. We've already done a vertical powerscrew, and the vertical calculations are all sorted (thanks to Shigley's Mechanical Design). I've used the information in that book to derive equations for Torque required for the horizontal powerscrew, but one comes out negative. I have trialled different FBDs and derivations but one always ends up negative.
I've attached an image with a sketch of the scenario, in case I haven't explained it clearly enough.
I've also attached my derivations of the horizontal power screw torque equations from FBDs. This document also contains the values I'm using, showing that Torque(right) comes out negative.
I've checked everything I know to and am completely stuck! Any ideas?!
Thanks in advance!
From my viewpoint, for an equal loading the magnitude of the two torques should be equal; however, since the screw must rotate in the opposite direction, as viewed from the same end of the screw, the torque is also in the opposite direction.
That would make sense.. I have however taken into account the direction of rotation in the FBDs for Torque(left) and Torque(right). This might suggest something else is a-miss with my FBDs.
I think I see the problem. in your first diagram you show an arrow for the applied force F pointing to the right and with a positive direction. Then in the second diagram you display the F pointing in the opposite direction which would imply, correctly that is a force in the opposite direction; but, in your equation for that case you then assign a negative value to that F force which essentially reverses its applied direction such that it is now equal to an positive force arrow pointing the same direction as the arrow in the first diagram; and, the same issue applies to your reversed fn arrow. If you simply remove the negative signs for F and fn in the second case then you will get the positive value you are expecting.
