hi everyone, suppose i have a system of gears where one shaft contains the initial pinion. a drive torque is applied to the pinion shaft so that the pinion then applies a force to the gear its connected to. how would i calculate the forces on the gear shaft? it is a static system (contant velocity and load torque).
Just take the torques about the shafts and relate the displacements by [tex]x=r \theta[/tex] Both gears travel the same arc length.
thanks for the quick response. what im wondering is how do i calculate the torques? the system layout is attached i have the drive torque (the bottom shaft). the load torque (the top shaft) is constant but not given. my end goal is to calculate a suitable intermediate shaft (the middle shaft) diameter but i think i can do that. first i need the torques in the intermediate shaft though. will the load torque simply be equal to the drive torque in order to maintain static equilibrium? i dont really know where to start here. i presume there will be two torques acting on the intermediate shaft as there is a gear and a pinion on it.. i know the gear and pinion sizes. thanks for the help.
so i just work out the force applied by the pinion on the drive shaft and then put the force on the gear of the intermediate shaft times the distance? sounds obvious now. do i baisically treat the system as not moving (static) under the constant velocity condition?
no, none of the shaft radii are given. i am to work out minimum shaft diameter of the intermediate shaft given the information above. the gears sizes ARE given. i assume that in order to start i should work out the torques on the intermediate shaft.
Well, I have to do a prelab writeup right now. Errr, I can help you later tonight or tomorrow, but I have to get work done. Some things to do in the meantime: Torque = Force times distance Use that knowledge to find the force as you move from one gear to the next. When you know the force, it is equal and opposite at the gear in mesh. Use that same equation to find the next torque or radii, depending on what you need to solve for. Ill be back later, sorry.
okay thats fine. thanks a lot for your help. if i come across any more problems i'll post them and wait for your reply. cheers